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3 years ago

Write down a pair of integers whose

1 answer:

3 0

<span>The pair of integers that I chose are:

(a) sum is –3
5 + (-8) = -3

(b) difference is –5
2 - 7 = -5

(c) difference is 2
14 -12 = 2

(d) sum is 0
2 - 2 = 0</span>

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Perform the indicated operation g(x)=-x-3 h(x) = x 2+X Find (goh) (x)

Klio2033 [76]

Answer:

$(g\circ h)(x)=-x^2-x-3$

Step-by-step explanation:

So we have the two functions:

$g(x)=-x-3\text{ and } h(x)=x^2+x$

And we want to find:

$(g\circ h)(x)$

This is the same thing as:

$=g(h(x))$

So, substitute h(x) into g(x):

$g(x)=-x-3\\g(h(x))=-(x^2+x)-3$

Distribute the negative:

$g(h(x))=-x^2-x-3$

And we're done!

So:

$(g\circ h)(x)=-x^2-x-3$

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3 years ago

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PLS HELP ME WITH GEOMETRY PLS :)

MArishka [77]

Answer:

-4, -24

Step-by-step explanation:

The location of Z is either +10 or -10 units away from Y, since YZ = 10. Then Z could be located at either -4 or -24.

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3 years ago

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What to plot on the graph

Tems11 [23]

just put a dot on the number line where 0, 3, and -3 are

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3 years ago

Given: AE ≅ CE ; DE ≅ BE Prove: ABCD is a parallelogram. Parallelogram A B C D is shown. Diagonals are drawn from point A to poi

nikklg [1K]

Answer:

The proof is below

Step-by-step explanation:

Given a parallelogram ABCD. Diagonals AC and BD intersect at E. We have to prove that AE is congruent to CE and BE is congruent to DE i.e diagonals of a parallelogram bisect each other.

In ΔACD and ΔBEC

AD=BC (∵Opposite sides of a parallelogram are equal)

∠DAC=∠BCE (∵Alternate angles)

∠ADC=∠CBE (∵Alternate angles)

By ASA rule, ΔACD≅ΔBEC

By CPCT(Corresponding Parts of Congruent triangles)

AE=EC and DE=EB

Hence, AE is conruent to CE and BE is congruent to DE

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3 years ago

The city of Anvil is currently home to 21000 people, and the population has been growing at a continuous rate of 4% per year. Th

Yanka [14]

Answer:

They'll reach the same population in approximately 113.24 years.

Step-by-step explanation:

Since both population grows at an exponential rate, then their population over the years can be found as:

$\text{population}(t) = \text{population}(0)*(1 + \frac{r}{100})^t$

For the city of Anvil:

$\text{population anvil}(t) = 21000*(1.04)^t$

For the city of Brinker:

$\text{population brinker}(t) = 7000*(1.05)^t$

We need to find the value of "t" that satisfies:

$\text{population brinker}(t) = \text{population anvil}(t)\\21000*(1.04)^t = 7000*(1.05)^t\\ln[21000*(1.04)^t] = ln[7000*(1.05)^t]\\ln(21000) + t*ln(1.04) = ln(7000) + t*ln(1.05)\\9.952 + t*0.039 = 8.8536 + t*0.0487\\t*0.0487 - t*0.039 = 9.952 - 8.8536\\t*0.0097 = 1.0984\\t = \frac{1.0984}{0.0097}\\t = 113.24$

They'll reach the same population in approximately 113.24 years.

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3 years ago