Houses in an real estate are all identical however a person can purchase a new house with some all or none of a set of options as indicated by the set below
how many options are there for purchasing a house in this community

{ pool , alarm , lake view , balcony }

The ratio is given by 98:0.34.

The ratio of (1/5 of 245ml) to (0.2 of 0.84ml).

The ratio can be defined as the comparison of the fraction of one quantity towards others. e.g.- water in milk.

we have  (1/5 of 245ml) to (0.2 of 0.84ml).

⇒ [tex]\frac{1}{5} 245 :0.2*0.84[/tex]

⇒[tex]98:0.34[/tex]

Thus, the required ratio is 98:0.34.

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Answer:

you can buy 98 beds and buy dresses by rest of the money

Answer:

61

Step-by-step explanation:

the question is asking to find the number of element of set b that are not in set a

to find this one must know the common elements

the question has given the common elements to be 5

now subtract the number of elements in set b with the common elements

i.e. 66-5

=61

Answer:

9

Step-by-step explanation:

This problem is represented below.

=3(4+2)-9

=3(6)-9

=18-9

=9

Answer:

thats not even understandable

The solution set is {3, -3} and is represented by the number line shown below.

It is the representation of numbers in real order.

The difference between the consecutive numbers in a number line is always positive.

We have,

To solve the equation -2|x| = -6,

We can begin by dividing both sides by -2:

|x| = 3

This means that the absolute value of x is equal to 3.

The solutions to this equation are x = 3 and x = -3.

The number line that represents these solutions is:

<--(+)===================(+)--->

   -3                                       3

The open circles indicate that the solutions are not included in the solution set since the absolute value of x cannot be negative.

Therefore,

The solution set is {3, -3} and is represented by the number line shown above.

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Pauline has 371 stickers. Karl, Serene, and Pauline have 1281 stickers altogether.

A mathematical statement consisting of an equal symbol between two algebraic expressions with the same value is known as an equation.

Let Karl, Serene and Pauline have x, y, and z stickers.

Given condition;

Karl, Serene, and Pauline have 1281 stickers altogether;

x+y+z = 1281

Pauline had 3 times as many stickers as Serene.

z = 3 y

y=z/3

Karl had 175 stickers fewer than Pauline.

x=z -175

Substitute all the values in the terms of z;

x+y+z = 1281

z -175 + z/3 + z = 1281

8z/3 = 1281 +175

z=546

Pauline have;

x = z -175

x =546 -175

x = 371

Hence Pauline has 371 stickers.

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The solution of the logarithmic expression [tex]\log_2(17 -\sqrt{19}) + \log_2(17 +\sqrt{19})[/tex] is [tex]\log_2[270][/tex]

The expression is given as:

[tex]\log_2(17 -\sqrt{19}) + \log_2(17 +\sqrt{19})[/tex]

Apply the product law of logarithm

[tex]\log_2[(17 -\sqrt{19}) *(17 +\sqrt{19})]\\[/tex]

Apply the difference of two squares

[tex]\log_2[(17^2 -(\sqrt{19})^2][/tex]

Evaluate the squares

[tex]\log_2[(289 -19][/tex]

Evaluate the difference

[tex]\log_2[270][/tex]

Hence, the solution of the logarithmic expression [tex]\log_2(17 -\sqrt{19}) + \log_2(17 +\sqrt{19})[/tex] is [tex]\log_2[270][/tex]

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f(0) = -3(0) + 5 = 5

f(-4) = -3(-4) + 5 = 17

f(6) = ½(6) - 4 = -1

f(9) = 2

f(2) = ½(2) - 4 = -3

Using the restraints, find which equation the f(x) function validates and plug it in.

Answer:

Linear Equations In One Variable =

Next step :

in the equation y/6 = 156, it would be equivalent to y = 156 × 6.

so that :

y = 156 × 6

y = 936

this equation solved. (Answer : 936)

Answer:

2.48 seconds

Step-by-step explanation:

See attached

The probability that Jamie buys a telecaster or a Less Paul is 45% or forty-five percent.

1. Find the individual probabilities:

Probabilities to buy a telecaster:

Probabilities to buy a Les Paul:

2. Add the probabilities:

3. Multiply by 100:

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A line is a one-dimensional shape that is straight, has no thickness, and extends in both directions indefinitely. The equation of the line will be y =4x+5. Thus, the correct option is C.

A line is a one-dimensional shape that is straight, has no thickness, and extends in both directions indefinitely. The equation of a line is given by,

y =mx + c

where,

x is the coordinate of the x-axis,

y is the coordinate of the y-axis,

m is the slope of the line, and

c is the y-intercept.

Given the slope of the line is 4 and it goes through (-1,1), therefore, the equation of the line will be,

y = mx + c

y = 4x + c

Substitute the value of points,

1 = 4(-1) + c

1 = -4 + c

5 = c

Hence, the equation of the line will be y =4x+5. Thus, the correct option is C.

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Step-by-step explanation:

please mark me as brainlest

Answer:

B

Step-by-step explanation:

Range is all the possible values of 'y'. As seen in the graph, the smallest value of 'y' is 20, where the point is (6, 20), and the largest value of 'y' is 60, where the point is (1, 60). Therefore, the range is B. 20≤y≤60

The justification for whether Skyler, Robert, or Kaitlyn was right or wrong stems from how they went about their processes:

1) SKYLER'S PROCESS:

Skyler wants to resolve the following issue;

Reduce 729 to the lowest base possible as a first step; [3^(6)]^x = 9

729^(x) = 9

The next thing she did was convert 6x to 9.

Since both sides must have the same base when dealing with exponent equality, she may have transformed 9 to have the same base of 3 as the left hand side before equating the exponents.

So Skyler is wrong.

2) ROBERT'S PROCESS:

Skyler failed to capture both bases equally, as we witnessed, but Robert has now made the right decision by;

9^(3x) = 9¹

After which he equated the exponents to get;

3x = 1

x = 1/3

Thus; Robert is correct

3) KAITLYN'S PROCESS:

Kaitlyn is also accurate since she adhered to the same method, which calls for comparing bases before comparing exponents.

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Answer:

[tex]\frac{4}{3}[/tex] or [tex]1\frac{1}{3}[/tex]

Step-by-step explanation:

[tex]1 + \frac{1}{3} \\= \frac{3}{3} + \frac{1}{1}\\= \frac{4}{3}[/tex]

(if the question is asking for the simplified form , it is [tex]1\frac{1}{3}[/tex]

Answer:

[tex]\text{C.} \ \ \ {\left(\textit{AB}\right)}^{2} \ = \ \left(\textit{AC}\right)\left(\textit{AD}\right)[/tex]

Step-by-step explanation:

This problem uses the concept of the tangent-secant theorem which describes the relationship of the segments a secant line and a tangent line with the associated circle. This theorem is found as Proposition 36 in Book 3 of Euclid's Elements.

As shown in the figure attached below, segment AB (in blue) forms a tangent with the circle BCD and segment AD (in orange) is the secant where it intersects the circle at point C.

Furthermore, let two segments (in green) be drawn one from point C and point D.

To show that [tex]\triangle ABC[/tex] is similar to [tex]\triangle ADB[/tex], notice that both triangles share a common angle [tex]\angle BAC[/tex]. Additionally, by the alternate segment theorem, [tex]\angle ABC[/tex] is equal to [tex]\angle ADB[/tex]. Therefore, [tex]\angle ACB[/tex] is also equal to [tex]\angle ABD[/tex].

Hence, [tex]\triangle ABC[/tex] is indeed similar to [tex]\triangle ADB[/tex]. This implies the ratio of the sides of both triangles is the same. Particularly,

                                                  [tex]\displaystyle{\frac{AB}{AD} \ \ = \ \ \frac{AC}{AB}}[/tex].

Then, performing cross multiplication yields

                                             [tex]{\left(AB\right)}^{2} \ \ = \ \ \left(AC\right)\left(AD\right)[/tex].

Therefore, the product of the lengths of the secant segment and its external segment is equal to the square of the length of the tangent segment.

Answer: A. [tex]\frac{12}{13}[/tex]

Step-by-step explanation:

The best way to remember this is SOH-CAH-TOA. For sin, it would be the opposite angle over the hypotenuse. The opposite angle of ∠C is 24 and the hypotenuse is 26. But that isn't in the answer choices, so you'll simplify the fraction.

[tex]\frac{24}{26} =\frac{12}{13}[/tex], therefore, the answer is A.

I hope this helps! Pls mark brainliest!! :)

Answer:

huh

Step-by-step explanation:

Step-by-step explanation:

please mark me as brainlest

The value of the integration is

[tex]\int {\frac{-7x^{3} }{x^{3}+1} } \, dx =7x - \frac{7}{3} \ln(x+1) + \frac{7}{6} ln (x^{2} -x+1)-\frac{7\sqrt{3} }{3} tan^{-1} (\frac{2x-1}{\sqrt{3} })+C[/tex], where C is an integrating constant.

We know that finding the area of the curve's undersurface is the process of integration. To do this, cover the area with as many tiny rectangles as possible, then add up their areas. The sum gets closer to a limit that corresponds to the area under a function's curve. Finding an antiderivative of a function is the process of integration. If a function can be integrated and its integral over the domain is finite with the given bounds, then the integration is definite.

Here, we have to determine the value of the given integral

[tex]\int {\frac{-7x^{3} }{x^{3}+1} } \, dx[/tex].

So,

[tex]\int {\frac{-7x^{3} }{x^{3}+1} } \, dx = 7 \int {\frac{1-x^{3}-1 }{x^{3}+1} } \, dx[/tex][tex]= 7 \int [{1-\frac{1 }{x^{3}+1} } ]\, dx =7\int \, dx - 7 \int {\frac{1 }{x^{3}+1} } \, dx[/tex] ...(1)

Now,

[tex]\frac{1}{x^{3} +1} = \frac{1}{(x+1)(x^{2} -x+1)}[/tex]

We can write

[tex]\frac{1}{(x+1)(x^{2} -x+1)}=\frac{A}{(x+1)} + \frac{Bx+C}{(x^{2} -x+1)}[/tex]

i.e. [tex]1 = A(x^{2} -x+1)+(Bx+C)(x+1)[/tex]

i.e. [tex]1 = Ax^{2} -Ax + A+Bx^{2} +Bx+Cx+C[/tex]

i.e. [tex]1=(A+B)x^{2} +(-A+B+C)x +(A+C)[/tex]

Comparing both sides, we get

[tex]A+B=0 \implies B = -A[/tex] ...(2)

[tex]-A+B+C = 0 \implies -A+(-A)+C=0 \implies -2A+C=0[/tex] ...(3)

[tex]A+C=1[/tex] ...(4)

Subtracting (3) from (4),

[tex]A+C+2A-C=1-0 \implies 3A=1 \implies A=\frac{1}{3}[/tex]

From (4), [tex]C= 1-\frac{1}{3} =\frac{3-1}{3} =\frac{2}{3}[/tex]

From (2), [tex]B =-\frac{1}{3}[/tex]

We get

[tex]\frac{1}{x^{3}+1 } =\frac{1}{3} \frac{1}{(x+1)} - \frac{1}{3} \frac{x-2}{x^{2} -x+1}[/tex]

[tex]=\frac{1}{3} \frac{1}{(x+1)} - \frac{1}{6} \frac{2x-4}{x^{2} -x+1}[/tex]

[tex]=\frac{1}{3} \frac{1}{(x+1)} - \frac{1}{6} \frac{2x-1-3}{x^{2} -x+1}[/tex]

[tex]=\frac{1}{3} \frac{1}{(x+1)} - \frac{1}{6} \frac{2x-1}{x^{2} -x+1}+\frac{3}{6} \frac{1}{x^{2} -x+1}[/tex]

[tex]=\frac{1}{3} \frac{1}{(x+1)} - \frac{1}{6} \frac{2x-1}{x^{2} -x+1}+\frac{1}{2} \frac{1}{x^{2} -x+1}[/tex]

Integrate both sides,

[tex]\int \frac{1}{x^{3}+1 }\, dx =\frac{1}{3} \int \frac{1}{(x+1)} \, dx - \frac{1}{6} \int \frac{2x-1}{x^{2} -x+1}\, dx+\frac{1}{2} \int \frac{1}{x^{2} -x+1}\, dx[/tex]

i.e. [tex]\int \frac{1}{x^{3}+1 }\, dx =\frac{1}{3} \ln(x+1) - \frac{1}{6} ln (x^{2} -x+1)+\frac{\sqrt{3} }{3} tan^{-1} (\frac{2x-1}{\sqrt{3} })[/tex]

From (1),

[tex]\int {\frac{-7x^{3} }{x^{3}+1} } \, dx =7x - \frac{7}{3} \ln(x+1) + \frac{7}{6} ln (x^{2} -x+1)-\frac{7\sqrt{3} }{3} tan^{-1} (\frac{2x-1}{\sqrt{3} })+C[/tex]

Therefore, the value of the integration is

[tex]\int {\frac{-7x^{3} }{x^{3}+1} } \, dx =7x - \frac{7}{3} \ln(x+1) + \frac{7}{6} ln (x^{2} -x+1)-\frac{7\sqrt{3} }{3} tan^{-1} (\frac{2x-1}{\sqrt{3} })+C[/tex], where C is an integrating constant.

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In a study of birth orders and intelligence, IQ tests were given to 18 and 19 years old men to estimate the size of the difference, we have that  [tex]|t|=0.618\leq t_c =2.101[/tex] and for this reason, we cannot reject the null hypothesis.

Null and Alternative Hypotheses

[tex]H_o: \mu_{1} = \mu_{2}\\\\Ha: \mu_1 \neq \mu_2 .[/tex]

Since the population standard deviations are unknown, we will apply a t-test on the means of the two populations, using two independent samples.

The degrees of freedom are df = 18, and the significance threshold is set at alpha = 0.05, based on the data presented. If we assume that the population variances are the same, we can calculate the degrees of freedom in this way:

The rejection region for this two-tailed test is R = \{t : |f| > 2.101}

Therefore the Test Statistics is

[tex]t= \frac{X 1 -X 2 }{\frac{\sqrt{ (n_1 -1)s_1 ^ 2 +(n_2 -1)s_2 ^2}} {n_1 +n_2 -2} (( 1/n_1 + 1 n_2 )}}[/tex]

[tex]t= \frac{X 1 -X 2 }{\frac{\sqrt{ (n_1 -1)s_1 ^ 2 +(n_2 -1)s_2 ^2}} {n_1 +n_2 -2} (( 1/n_1 + 1 n_2 )}}[/tex]

t=0.618

In conclusion. we have that  [tex]|t|=0.618\leq t_c =2.101[/tex]

For this reason, we cannot reject the null hypothesis.

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The Medication A mixed this week should be 10 milligrams.

We have,

Medications to be mixed in the proper proportions.

Last week's proportions as the doctor indicated ;

Medication A = 90 milligrams

Medication B = 72 milligrams

This week's proportions as doctor indicated;

Medication B = 8 milligrams

Let, Medication A = x milligrams

As we are instructed to mix the Medication in the proper proportions,

So,

According to the question,

90/72 = x/8

10/8 = x/8

⇒ x = 10 milligrams

Here,  the Medication A mixed is 10 milligrams.

Therefore, the Medication A mixed this week should be 10 milligrams.

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Answer:

See below

Step-by-step explanation:

Longest is the down tube.....it is opposite the largest angle (95°) angle in the triangle .    This is a SCALENE triangle ( all sides of different length)

The longest tube that we have here is the down tube. The triangle that is formed id the scalene triangle

A scalene triangle is a type of triangle where all three sides have different lengths. In other words, no two sides of a scalene triangle are equal in length. Additionally, the angles of a scalene triangle are also different from each other.

The term "scalene" comes from the Greek word "skalenos," which means "uneven" or "unequal." This name reflects the defining characteristic of the scalene triangle, which is the absence of any equal sides or angles.

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Answer:

Step-by-step explanation:

Answer:

Step-by-step explanation:

we find the volume as if it were a parallelepiped and we remove the volume of the missing part.

8 * 6 * 10 = 480 cm³

480 - 2 * 3 * 10 = 420cm³

Step-by-step explanation:

1) zeros of the given function:

6x²-3=0; ⇔ 6(x²-0.5)=0; ⇔ x²=0.5; ⇔

[tex]\left[\begin{array}{ccc}x=-\sqrt{0.5} \\x=\sqrt{0.5} \end{array}[/tex]

2) relationship:

if to see the equation x²-0.5=0 (ax²+bx+c=0 is standart form!), then the sum of the zeros is '0' (it is 'b' of the standart form), the product of equation roots is '-0.5' (it is 'c' of the standart form).

[tex]{\large{\textsf{\textbf{\underline{\underline{Given :}}}}}}[/tex]

★ The polynomial

f(x) = 6x² - 3

[tex]{\large{\textsf{\textbf{\underline{\underline{To \: find :}}}}}}[/tex]

★ Zeroes of the polynomial f(x) = 6x² - 3

[tex]{\large{\textsf{\textbf{\underline{\underline{Solution :}}}}}}[/tex]

We have,

[tex]f(x) = \tt 6 {x}^{2} - 3[/tex]

Which can also be written as

[tex] \implies f(x) = \tt {(\sqrt{6} x)}^{2} - { (\sqrt{3}) }^{2} [/tex]

Using a² - b² = (a - b) (a + b)

[tex] \implies f(x) = \tt ( \sqrt{6} x - \sqrt{3} )( \sqrt{6} x + \sqrt{3} )[/tex]

To find the zeroes, solve f(x) = 0

[tex] \longrightarrow \tt ( \sqrt{6} x - \sqrt{3} )( \sqrt{6} x + \sqrt{3} ) = 0[/tex]

either [tex] \tt \sqrt{6} x - \sqrt{3} = 0 \: or \: \sqrt{6} x + \sqrt{3} = 0[/tex]

[tex] \implies \tt \sqrt{6} x = \sqrt{3 \: } \: or \: \: \sqrt{6} x = - \sqrt{3}[/tex]

[tex] \implies \tt x = \dfrac{ \sqrt{3} }{ \sqrt{6} } \: or \: x = - \dfrac{ \sqrt{3} }{ \sqrt{6} }[/tex]

[tex] \implies \tt x = \dfrac{ \sqrt{3} }{ \sqrt{2 \times 3} } \: or \: x = - \dfrac{ \sqrt{3} }{ \sqrt{2 \times 3} }[/tex]

[tex]\implies \tt x = \dfrac{ \cancel{ \sqrt{3} }}{ \sqrt{2} \: \cancel{\sqrt{3}} } \: or \: x = - \dfrac{ \cancel{ \sqrt{3} }}{ \sqrt{2} \: \cancel{\sqrt{3}} }[/tex]

[tex]\implies \tt x = \dfrac{1}{ \sqrt{2} } \: \: or \: \: - \dfrac{1}{ \sqrt{2} }[/tex]

Hence, the zeroes of f(x) = 6x² - 3 are:

[tex] \tt \alpha =\sf \boxed {{ \red{ \dfrac{1}{ \sqrt{2} } } }}\: \: and \: \: \beta =\sf \boxed {{ \red{ - \dfrac{1}{ \sqrt{2} } } }}[/tex]

• Verification

Sum of zeroes = [tex] ( \alpha + \beta )[/tex]

[tex] = \tt \dfrac{1}{ \sqrt{2} } + \bigg(- \dfrac{1}{ \sqrt{2} } \bigg)[/tex]

[tex] = \tt \dfrac{1}{ \sqrt{2} } + - \dfrac{1}{ \sqrt{2} } [/tex]

[tex]= \tt 0[/tex]

and, [tex]\tt - \dfrac{Coefficient \: of \: x}{Coefficient \: of \: {x}^{2} }[/tex]

[tex] \tt = - \dfrac{0}{6} [/tex]

[tex] \tt = 0[/tex]

[tex] \therefore \tt \: Sum \: of \: zeroes = {\boxed{ \red{\dfrac{ \tt Coefficient \: of \: x}{ \tt Coefficient \: of \: {x}^{2}}}}}[/tex]

Also,

Product of zeroes = [tex] \alpha \beta [/tex]

[tex] = \dfrac{1}{ \sqrt{2} } \times - \dfrac{1}{ \sqrt{2} } [/tex]

[tex] = - \dfrac{1}{ 2 } [/tex]

and, [tex]\tt - \dfrac{Constant \: term}{Coefficient \: of \: {x}^{2} }[/tex]

[tex] \tt = \dfrac{ - 3}{6} [/tex]

[tex] \tt = \dfrac{ - 1}{2} [/tex]

[tex] \therefore \tt \: Product \: of \: zeroes = {\boxed{ \red{\dfrac{ \tt Constant \: term}{ \tt Coefficient \: of \: {x}^{2}}}}}[/tex]

[tex]\rule{280pt}{2pt}[/tex]

Answer:

x + 3y = 30

Step-by-step explanation:

We are given that a line contains the points (12,6) and (-3,11)​.

We want to write the equation of this line in standard form.

Standard form is written as ax+by=c, where a, b, and c are free integer coefficients, however a and b cannot be 0, and a cannot be negative.

Regardless, before we write an equation in slope-intercept form, we must first write the equation in a different form, such as slope-intercept form.

Slope-intercept form is given as y=mx+b, where m is the slope and b is the value of y at the y-intercept.

So first, let's find the slope of the line.
The slope (m) can be found using the formula [tex]\frac{y_2-y_1}{x_2-x_1}[/tex], where [tex](x_1, y_1)[/tex] and [tex](x_2, y_2)[/tex] are points.

Even though we already have 2 points, let's label their values to avoid any confusion and mistakes when calculating.

[tex]x_1=12\\y_1=6\\x_2=-3\\y_2=11[/tex]

Now substitute these values into the formula.

m=[tex]\frac{y_2-y_1}{x_2-x_1}[/tex]

m=[tex]\frac{11-6}{-3-12}[/tex]

Subtract

m=[tex]\frac{5}{-15}[/tex]

Simplify

m=[tex]-\frac{1}{3}[/tex]

The slope is -1/3

Here is the equation of the line so far in slope-intercept form:

[tex]y=-\frac{1}{3} x + b[/tex]

We need to solve for b.

As the equation passes through (12,6) and (-3,11)​, we can use either one to help solve for b.

Taking (12, 6) for example:

[tex]6=-\frac{1}{3}(12) + b[/tex]

Multiply

[tex]6=-\frac{12}{3} + b[/tex]

Divide

6 = -4 + b

Add 4 to both sides.

10 = b

Substitute 10 as b in the equation.

[tex]y = -\frac{1}{3} x + 10[/tex]

Here is the equation in slope-intercept form, but remember, we want it in standard form.

In standard form, the values of both x and y are on the same side, so let's add -1/3x to both sides.

[tex]\frac{1}{3} x + y = 10[/tex]

Remember that a (the coefficient in front of x) has to be an integer, 1/3 is not an integer.

So, let's multiply both sides by 3 to clear the fraction.

[tex]3(\frac{1}{3} x + y) = 3(10)[/tex]
Multiply.

x + 3y = 30

Topic: finding the equation of the line (standard form)

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Answer:

[tex]\frac{3}{32}[/tex]

Step-by-step explanation:

-> Simplify

[tex]\frac{6}{64} =\frac{6/2}{64/2} =\frac{3}{32}[/tex]

6/64 simplified.