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3 years ago

Find the value of x. Round to the nearest tenth.

Mathematics

1 answer:

dusya [7]3 years ago

5 0

Answer:

x = 10.2

Step-by-step explanation:

Here we have a right triangle with hypotenuse 11, angle 22° and adjacent side x. Our task is to find x to the nearest tenth.

The cosine function nicely connects the hypotenuse and the adjacent side:

adjacent side

cos 22° = ------------------------

hypotenuse

We solve for the adjacent side (x) by multiplying both sides of this equation by "hypotenuse:"

x = adjacent side = 11*cos 22 = 11(0.9272 radian).

Performing the indicated multiplication, we arrive at:

x = 10.1990, or, rounded to the nearest tenth, x = 10.2.

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Iteru [2.4K]

Answer:

Step-by-step explanation:

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3 years ago

Rashaj has $37.25 in his wallet. He would

Butoxors [25]

I agree. From the calculation, the total <em>amount spent </em><em>is </em><em>less than</em><em> (<)the amount in his pocket</em>, this means that <em>Rashaj can buy 5 boxes of candy. </em>

Inequalities are expressions that are not always equal to each other. The inequality signs are <, > ≤ and ≥

Given the following

Amount Rashaj has in his wallet = $37.25

Amount spent

A movie ticket = $7.25

5 boxes of candy at $3.75 per box = 5($3.75) = $18.75

Taking the sum of the total amount spent

Amount spent = $7.25 + $18.75

Amount spent = $26

Since the total <em>amount spent </em><em>is </em><em>less than</em><em> (<)the amount in his pocket</em>, this means that <em>Rashaj can buy 5 boxes of candy. </em>

<em>Learn more here: brainly.com/question/17448505</em>

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2 years ago

Imagine telling a guy you liked him. He tells you he has feelings for you but he likes your best friend. But, your best friend l

kirza4 [7]

Answer:get with your older brother

Step-by-step explanation:

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3 years ago

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Alborosie

I am pretty sure you can only make one.
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3 years ago

Finding values or<br> k(x) = x - 2<br> h(x) = x2 +1<br> (h + k)(2) =

Serjik [45]

Answer:

Step-by-step explanation:

$(h+k)(x)=h(x)+k(x)\\\\\bold{METHOD\ 1:}\\\\k(x)=x-2,\ h(x)=x^2+1\\\\(k+h)(2)=h(2)+k(2)\\\\\text{Calculate}\ h(2)\ \text{and}\ k(2).\\\text{Put}\ x=2\ \text{to the equations of the functions}\\\\h(2)=2^2+1=4+1=5\\k(2)=2-2=0\\\\\text\\\\\text{Substitute:}\\\\(h+k)(2)=5+0=5$

$\bold{METHOD\ 2:}\\\\k(x)=x-2,\ h(x)=x^2+1\\\\(h+k)(x)=h(x)+k(x)=(x^2+1)+(x-2)=x^2+x-1\\\\(h+k)(2)-\text{put}\ x=2\\\\(h+k)(2)=2^2+2-1=4+2-1=6-1=5$

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3 years ago