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Jet001 [13]
3 years ago
7

Solve for v -7 v+3 = -63

Mathematics
2 answers:
denis23 [38]3 years ago
8 0

Answer:

v = 66/7

Step-by-step explanation:

Step 1: Write equation

-7v + 3 = -63

Step 2: Solve for <em>v</em>

<u>Subtract 3 on both sides:</u> -7v = -66

<u>Divide both sides by -7:</u> v = 66/7

Step 3: Check

<em>Plug in x to verify if it's a solution.</em>

-7(66/7) + 3 = -63

-66 + 3 = -63

-63 = -63

dusya [7]3 years ago
3 0

Answer:

Step-by-step explanation:

-7v+3=-63

-7v=-63-3

-7v=-66

-7v-7=-66÷-7

V=9.4

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f(x) = 20x² - 18x - 25

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3 years ago
50= p• 0.05•2<br> Solve
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Answer:

P=500

Step-by-step explanation:

My reasoning is multiply 0.05 and 2 and you get 0.1. Then you divide 0.1 out of 50 and get 500. Too check the answer you take 500 and multiply it by 0.05 and then  multiply it by 2 and you should get an answer of 50 like it says in the equation.

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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
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