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11111nata11111 [884]

3 years ago

9

The scatter plot shows the number of hours worked and money spent on entertainment by each of 24 students. Also shown is the lin

e of best fit for the data. Fill in the blanks below.

1 answer:

dangina [55]3 years ago

4 0

Answer:

a) increase

b) strong positive

c) $22.50

Step-by-step explanation:

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What is 2-1 reversed? A.1 B.-1 C.2 D. -2

vovikov84 [41]

2-1 reversed is turning it around : 1-2 which equals -1 because there is one left over. B is your answer and hoped I helped!

5 0

3 years ago

Read 2 more answers

What is the slope of the line shown below?

Anton [14]

Answer:

Slope = 3

Step-by-step explanation:

2 - 8 = -6

2 - 4 = -2

Slope = 3

8 0

3 years ago

When the points below are connected, they will form triangle ABC. Show and state whether this triangle has a right angle or not.

cluponka [151]

Answer:

See explanartions below

Step-by-step explanation:

Given the coordinates A(-6,3) B(0,5) C(2,-1)

We first need to get the slope of AB, BC and AC

For AB;

A(-6,3) B(0,5)

Mab = 5-3/0-(-6)

Mab = 2/6

Mab = 1/3

For BC;

B(0,5) C(2,-1)

Mbc = -1-5/2-0

Mbc = -6/2

Mc = -3

For AC:

A(-6,3) C(2,-1)

Mac = -1-3/2-(-6)

MAc = -4/8

Mac = -1/2

Taking the product of Mab and Mbc

Mab * Mbc = 1/3 + -3

Mab * Mbc = -1

<em>Since the product of Mab and Mbc is -1, hence AB abd BC are perpendicular showing that one of the angle of triangle ABC is 90 degrees</em>

8 0

3 years ago

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

HELPP<br> find the area of the shaded figure

Aliun [14]

Answer:

24 because 8×3. .................

5 0

3 years ago