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never [62]
3 years ago
15

Theirs a picture on top.. :) 15. simplify 16. simplify

Mathematics
1 answer:
notsponge [240]3 years ago
6 0

Answer:

15(b)

16(a)

Step-by-step explanation:

15) 2x+10/2

Taking 2 common

2(x+5)/2

x+5

16) -2(-x-3)

(-2*-x)+(-2*-3)

2x+(6)

2x+6

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Find two positive real numbers whose product is a maximum. the sum is 80
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Let's say the numbers are "a" and "b"

thus \bf y=ab\qquad 
\begin{cases}
a+b=80\\
b=80-a
\end{cases}\implies y=a(80-a)\implies y=80a-a^2\\\\
-----------------------------\\\\
\cfrac{dy}{da}=80-2a

set the derivative to 0, and check the critical points, there's only one anyway

and do a first-derivative test, to see if it's a maximum
4 0
3 years ago
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A quality control engineer is testing the battery life of a new smartphone. The company is advertising that the
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The question requires stating the Null and Alternative hypothesis for the scenario described :

Answer:

H0 : μ = 24

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The claim or hypothesis to be tested in a statistical problem or analysis is the alternative hypothesis. This is usually the opposite of the Null hypothesis which is the usually the initial truth until there is significant evidence provided using statistical test to prove otherwise.

The null hypothesis, H0 in the scenario above is :

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3 years ago
Construct a triangle ABC in which BC-8 cm, B = 45° and AB-AC-3.5cm​
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Mr. Harmin is sitting in his office on the third floor of a building. He is at an elevation of 30 feet in relation to street lev
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To find the product of 42.12 and 10³, move the decimal point in 42.12 __ places to the right because 10³ has __ zeros.
Yakvenalex [24]
<h3>To find the product of 42.12 and 10^3,  move the decimal point in 42.12 3 places to the right because 10^3 has 3 zeros</h3>

<em><u>Solution:</u></em>

Given that,

\text{ product of } 42.12 \text{ and } 10^3

Which means,

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Here, the exponent of 10 is positive ( which is 3)

When the exponent is positive, we have to move the decimal point to right

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Here, exponent is 3 , therefore move the decimal point right 3 places in 42.12

Therefore,

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