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2 years ago

What is the measure of the missing side of the right triangle?

Mathematics

2 answers:

finlep [7]2 years ago

4 0

B or 12 would be the correct choice!
Hope this helps and mark as brainliest please!

zloy xaker [14]2 years ago

4 0

Something's wrong with the question ... like where's the other half

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A recipe take 90 minutes to make.25% of the time is spent preparing the ingredients.How many minutes are spent preparing

joja [24]

It will take 22.5 min cause 25 divided by 100 wuld be 4 wich then divide 90 with

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2 years ago

125

Marina CMI [18]

Answer:

Step-by-step explanation:

$2250 \div 25 = 90$

$90 \div 45 = 2$

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2 years ago

I attached an image. someone help me asap. i'm not the smartest person so i cont understand it :(

vampirchik [111]

Answer:

for the second column it is 12.5

for the third column it is 20%

for the fourth column it is 120%

for the fifth column it is 33%

for the sixth column it is 11.35

for the seventh column it is 900

for the 8th column it is 60%

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Step-by-step explanation:

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2 years ago

Read 2 more answers

This is a math question from highschool (LOOK AT PICTURE)

Murljashka [212]

Answer: The answer is 12q-1/2

Step-by-step explanation:

4 0

1 year ago

Which is the inverse of the function a(d)=5d-3? And use the definition of inverse functions to prove a(d) and a-1(d) are inverse

Drupady [299]

Answer:

$a'(d) = \frac{d}{5} + \frac{3}{5}$

$a(a'(d)) = a'(a(d)) = d$

Step-by-step explanation:

Given

$a(d) = 5d - 3$

Solving (a): Write as inverse function

$a(d) = 5d - 3$

Represent a(d) as y

$y = 5d - 3$

Swap positions of d and y

$d = 5y - 3$

Make y the subject

$5y = d + 3$

$y = \frac{d}{5} + \frac{3}{5}$

Replace y with a'(d)

$a'(d) = \frac{d}{5} + \frac{3}{5}$

Prove that a(d) and a'(d) are inverse functions

$a'(d) = \frac{d}{5} + \frac{3}{5}$ and $a(d) = 5d - 3$

To do this, we prove that:

$a(a'(d)) = a'(a(d)) = d$

Solving for $a(a'(d))$

$a(a'(d)) = a(\frac{d}{5} + \frac{3}{5})$

Substitute $\frac{d}{5} + \frac{3}{5}$ for d in $a(d) = 5d - 3$

$a(a'(d)) = 5(\frac{d}{5} + \frac{3}{5}) - 3$

$a(a'(d)) = \frac{5d}{5} + \frac{15}{5} - 3$

$a(a'(d)) = d + 3 - 3$

$a(a'(d)) = d$

Solving for: $a'(a(d))$

$a'(a(d)) = a'(5d - 3)$

Substitute 5d - 3 for d in $a'(d) = \frac{d}{5} + \frac{3}{5}$

$a'(a(d)) = \frac{5d - 3}{5} + \frac{3}{5}$

Add fractions

$a'(a(d)) = \frac{5d - 3+3}{5}$

$a'(a(d)) = \frac{5d}{5}$

$a'(a(d)) = d$

Hence:

$a(a'(d)) = a'(a(d)) = d$

7 0

2 years ago