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3 years ago

6

HELP ASAP!!!!!!!!!!!

2 answers:

LenaWriter [7]3 years ago

7 0

Answer: 7/32

Step-by-step explanation:

Just add 9 + 12 + 7 + 4 = 32

And then take the 7 and put it as the numerator and the 32 as the denominator.

jeka57 [31]3 years ago

3 0

Answer:

D.

Step-by-step explanation:

Total students=9+12+7+4

=32 students

Students who chose fish=7

So

The fraction is =7/32

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9. (4,10) and (-2,-5)<br>Slope:

marshall27 [118]

Answer:

-7/2

Step-by-step explanation:

y2-y1 / x2-x1

-5-10 / -2-4

-15 / -6

-7 / -2

= -7/2

4 0

2 years ago

Which inequality is equivalent to absolute value x-4 >9

hoa [83]

Answer:

x> 13

Step-by-step explanation:

x-4 >9

Add 4 to each side

x-4+4 >9+4

x> 13

7 0

3 years ago

20 POINTS !!! ONLY HELP IF UR 100% SURE !!!

Arte-miy333 [17]

Your answer is C for sure I am!

8 0

2 years ago

Read 2 more answers

elect answers from the drop-down menus to correctly complete the statements. A scale factor of was applied to the first triangle

snow_tiger [21]

The scale factor is what u multiply the first image by to get the second image. So by dividing the second image by the first image will give u the scale factor.

6 / 1.5 = 4....so there is a scale factor of 4 because 1.5 * 4 = 6

7 0

2 years ago

Read 2 more answers

Find the angle between the given vectors. Round your answer, in degrees, to two decimal places. u=⟨2,−6⟩u=⟨2,−6⟩, v=⟨4,−7⟩

NISA [10]

Answer:

$\theta = 108.29$

Step-by-step explanation:

Given

$u =$

$v =$

Required:

Calculate the angle between u and v

The angle $\theta$ is calculated as thus:

$cos\theta = \frac{u.v}{|u|.|v|}$

For a vector

$A =$

$A = a * b$

$cos\theta = \frac{u.v}{|u|.|v|}$ becomes

$cos\theta = \frac{.}{|u|.|v|}$

$cos\theta = \frac{2*6+4*-7}{|u|.|v|}$

$cos\theta = \frac{12-28}{|u|.|v|}$

$cos\theta = \frac{-16}{|u|.|v|}$

For a vector

$A =$

$|A| = \sqrt{a^2 + b^2}$

So;

$|u| = \sqrt{2^2 + 6^2}$

$|u| = \sqrt{4 + 36}$

$|u| = \sqrt{40}$

$|v| = \sqrt{4^2+(-7)^2}$

$|v| = \sqrt{16+49}$

$|v| = \sqrt{65}$

So:

$cos\theta = \frac{-16}{|u|.|v|}$

$cos\theta = \frac{-16}{\sqrt{40}*\sqrt{65}}$

$cos\theta = \frac{-16}{\sqrt{2600}}$

$cos\theta = \frac{-16}{\sqrt{100*26}}$

$cos\theta = \frac{-16}{10\sqrt{26}}$

$cos\theta = \frac{-8}{5\sqrt{26}}$

Take arccos of both sides

$\theta = cos^{-1}(\frac{-8}{5\sqrt{26}})$

$\theta = cos^{-1}(\frac{-8}{5 * 5.0990})$

$\theta = cos^{-1}(\frac{-8}{25.495})$

$\theta = cos^{-1}(-0.31378701706)$

$\theta = 108.288386087$

<em></em> $\theta = 108.29$ <em> (approximated)</em>

4 0

2 years ago