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3 years ago

The width of a rectangle is 3 more than twice its length. the area is 35 square meters what is the length of the rectangle

1 answer:

defon3 years ago

6 0

35=3L+3
3L=32
L=10 $\frac{2}{3}$
THE LENGTH OF THE RECTANGLE IS 10 AND TWO THIRD METERS.
10 $\frac{2}{3}$ +2(10 $\frac{2}{3}$ )+3=35

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horrorfan [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

$g(h(x)) = \sqrt[4]{x^2 + x +5} + 3$

$h(g(x)) = \sqrt{x} + 7\sqrt[4]{x} + 17$

$h(h(x)) = [x^2 + x + 5 ]^2 + x^2 + x + 10$

Step-by-step explanation:

From the question we are told that

$h(x) = x^2 + x + 5$

and

$g(x) = \sqrt[4]{x} + 3$

Considering first question

Now we are told g(h(x))

i.e

$g(h(x)) = [x^2 + x + 5 ]^{\frac{1}{4} } + 3$

=> $g(h(x)) = \sqrt[4]{x^2 + x +5} + 3$

Considering second question

Now we are told h(g(x))

i.e

$h(g(x)) = [x^{\frac{1}{4} } + 3]^2 + x^{\frac{1}{4} } + 3 + 5$

=> $h(g(x)) = x^{\frac{1}{2} } + 6x^{\frac{1}{4} } + 9+ x^{\frac{1}{4}} + 8$

=> $h(g(x)) = x^{\frac{1}{2}} + 7x^{\frac{1}{4}} + 17$

=> $h(g(x)) = \sqrt{x} + 7\sqrt[4]{x} + 17$

Considering third question

$h(h(x))= [x^2 + x + 5]^2 + [x^2 + x + 5 ] + 5$

=> $h(h(x)) = [x^2 + x + 5 ]^2 + x^2 + x + 10$

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3 years ago

Find the value of y. images attached.

telo118 [61]

Answer:

<h3>The answer is option D</h3>

Step-by-step explanation:

Since it's a right angled triangle we can use trigonometric ratios to find the value of y

To find the value of y we use tan

$\tan(28) = \frac{y}{35} \\ y = 35 \tan(28) \\ y = 18.6098301...$

We have the final answer as

Hope this is helps you

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