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3 years ago

Thanks for the help pt.1

Mathematics

2 answers:

erma4kov [3.2K]3 years ago

4 0

(jk)(x)= j (k(x)) = j(9-2x)

= -7(9-2x) = -63 + 14x

but i see there is no such option. may be second option is typo.

so you can select 2nd option

Alla [95]3 years ago

4 0

Answer:

The correct answer is $(jk)(x) = -63+14x$ .

Step-by-step explanation:

We are given the following two functions, $j ( x ) = - 7 x$ and $k ( x ) = 9 - 2 x$ and we are to find the value of $( j k ) ( x )$ .

Applying the function j on function k to get:

$(jk)(x) = -7(9-2x)$

$(jk)(x) = -63+14x$

You might be interested in

Consider the following least squares specification between test scores and the student - teacher ratio: TestScore = 557.8 + 36.4

GenaCL600 [577]

Answer:

$Test score_i= 557.8 +36.42 Income$

If we increase the income by 1% that means that the new income would be 1.01 the before one and if we replace this we got:

$Test score_f = 557.8 + (36.42* 1.01 Income)= 557.8 +36.7842 Income$

And the net increase can be founded like this:

$Test score_f -Tet score_i = 557.8 +36.7842 Income- [557.8 +36.42 Income] = 36.7842 Income -36.42 Income = 0.3642$

So then the net increase would be:

C. 0.36 points

Step-by-step explanation:

For this case we have the following linear relationship obtained from least squares between test scores and the student-teacher ratio:

$Test score_i= 557.8 +36.42 Income$

If we increase the income by 1% that means that the new income would be 1.01 the before one and if we replace this we got:

$Test score_f = 557.8 + (36.42* 1.01 Income)= 557.8 +36.7842 Income$

And the net increase can be founded like this:

$Test score_f -Tet score_i = 557.8 +36.7842 Income- [557.8 +36.42 Income] = 36.7842 Income -36.42 Income = 0.3642$

So then the net increase would be:

C. 0.36 points

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Step-by-step explanation:

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Step-by-step explanation:

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$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$