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3 years ago

15

Choose the equations that are equvialent. Select all that apply.

1 answer:

Fed [463]3 years ago

6 0

Answer:a,b,c

Step-by-step explanation:

The are all equivalent

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spayn [35]

So remember that percent means parts out of 100 so
52%=52/100=0.52

26%=26/100=0.26
'of' means multily

0.52 times x=65
divide both sides by 00.52
x=125

26% of a number is 39
0.26 times y=39
divide both sides by 0.26
y=150

x=125
y=150
y is greater

you know that becuase we know that 52% is exatly twice of 26% so therefor, if x=y then 52% of x is half of 26% of y so to compare them, we just divide the 65 number by 2 to find 26% of that number to get 65/2=32.5
32.5<39 so x is smaller number

7 0

2 years ago

Find the probability of getting a head when you flip a coin, and then getting a vowel when you draw from a bag of letters with t

lesantik [10]

Answer:

UM I am not sure how you want this but a coin it is 50 50 or 1/2 and with the probability of getting a vowel is 5/26.

Step-by-step explanation:

Pls tell me if I am wrong I have no learned this yet if I am wrong I will edit it to be right.

6 0

3 years ago

Read 2 more answers

I need help assap

Bad White [126]

It’s 2=x because you have start with 2 then 5

5 0

2 years ago

Please help me I will give you bring me if it’s right

Otrada [13]

Answer:x

Step-by-step explanation:

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