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3 years ago

Line I passes through (4, 5) and is perpendicular to the line shown on the coordinate grid

Mathematics

1 answer:

Dennis_Churaev [7]3 years ago

7 0

The equation of liner in standard form is 4 1/2

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What is the answer to 15 x 4.2

Mnenie [13.5K]

Answer:

Step-by-step explanation:

15 x 4.2

15 x 42 (get rid of the decimal by moving it one to the right)

15 x 42 = 630

630 = 63.0 (put the decimal back in by placing it one spot to the left)

Answer: 63

hope this helps :)

3 0

3 years ago

The graph shows the scores of an exam. About what percent of students scored above 86%?

Gemiola [76]

Attach shows the scores of students in an exam

$\begin{gathered} \text{(students scored above 86\%) = }\frac{\text{ number of students score above 86\%}}{\text{Total scored }}\text{ x }\frac{100}{1} \\ \text{(students scored above 86\%) = }\frac{\text{8+5.9+2.2+2}}{1.4+2.2+4.5+6.2+8.5+16.5+18+16.3+8+8+5.9+2.2+2}X\frac{100}{1} \\ \\ \text{(students scored above 86\%) = }\frac{18.1}{99.7}\text{ X }\frac{100}{1} \\ \text{(students scored above 86\%) = }18.15\text{ \%} \\ \text{(students scored above 86\%) = 18\%} \end{gathered}$

Hence the correct answer = 18% Option A

8 0

1 year ago

-4x = 20 Please can you help anything can help

enot [183]

Answer:

the answer is x=-5. a negative times a negative is a positive.

3 0

3 years ago

Read 2 more answers

The volume of a volleyball is about 288 cubic inches. what is the radius of the volleyball, to the nearest tenth of an inch? use

Len [333]

Answer:

The radius of the volleyball is 8.3 inches

Step-by-step explanation:

Given

$r = \sqrt{\frac{3v}{4\pi}}$

$v = 288$

Required

Determine the value of r

To do this, we simply substitute 288 for v and 3.14 for π in the given equation.

This gives

$r = \sqrt{\frac{3v}{4\pi}}$

$r = \sqrt{\frac{3 * 288}{4 * 3.14}}$

$r = \sqrt{\frac{864}{12.56}}$

$r = \sqrt{68.7898089172}$

$r = 8.29396219651$

$r = 8.3\ inch$ (Approximated)

Hence;

The radius of the volleyball is 8.3 inches

3 0

3 years ago

Read 2 more answers

which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

8 0

3 years ago