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3 years ago

Evaluate the expression when a=−15 and b=−5. b2/a + 5

Mathematics

1 answer:

schepotkina [342]3 years ago

3 0

Answer:

-2.5

Step-by-step explanation:

Evaluate the expression:

b²/(a + 5)

When a= −15 and b=−5

Substitute values:

b²/(a + 5) =
(-5)² / (-15 + 5) =
25/(-10) =
-2.5

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In case the expression is b²/a + 5, the value is:

b²/a + 5 =
(-5)²/(-15) + 5 =
25/(-15) + 5 =
- 5/3 + 5 =
10/3

-------------------------------------------------------------------

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hodyreva [135]

Answer:

x ≈ 11.7

Step-by-step explanation:

Using the cosine ratio in the right triangle

cos26° = $\frac{adjacent}{hypotenuse}$ = $\frac{x}{13}$ ( multiply both sides by 13 )

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2 years ago

The pregnancy length in days for a population of new mothers can be approximated by a normal distribution with a mean of 269 day

fenix001 [56]

Answer:

a) 279 b)254

Step-by-step explanation:

Mean = 269 days

standard deviation = 8 days

let n be the pregnancy lenght.

using normal distribution formula

z = (n – mean)/ standard deviation

n = z * standard deviation + mean

Now, z values for the top 12% has to be found using a z table.

The top 12% means 88% of area under the curve. In this case z = 1,176

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n = 278.4 = 279

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3 years 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:

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

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Answer:

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Step-by-step explanation:

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Hope that helped!

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