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3 years ago

9

Solve for x. Round your answer to 2 decimal places. A right triangle is shown with the hypotenuse labeled x. One angle has a mea

sure of 32 degrees, and the opposite leg has a length of 7.

1 answer:

arsen [322]3 years ago

4 0

Answer:

x would be 58°

Step-by-step explanation:

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ANSWER: "There is not enough information to determine the answer". Option D is correct.

Step-by-step explanation:

For a case that was ruled by a court to be reversed by a higher court, they must be conditions in the ruling of the lower court that has been reconsidered by the higher court. This is regardless of the jurisdiction of the court or judge that ruled the case.

A case from all the courts in the options can be reversed by a higher court if the courts judgement lacks merit by the higher court, at any given times, irrespective of the court.

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4 years ago

What would be the linear equation for (0,2) and (4,-4)

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Step-by-step explanation:

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3 years ago

Help me out guys. Got stuck over this algebra question

vovikov84 [41]

Answer:

Compare LHS and RHS to prove the statement.

Step-by-step explanation:

Given: a + b + c = 0

We have to show that $\frac{1}{1 + x^b + x^{-c}} + \frac{1}{1 + x^c + x^{-a}} + \frac{1}{1 + x^a + x^{-b}} = 1$

We take LCM, simplify the terms and compare LHS and RHS. We will see that LHS = RHS and the statement will be proved.

Taking LCM, we get:

$\frac{(1 + x^c + x^{-a})(1 + x^a + x^{-b}) + (1 + x^b + x^{-c})(x^a + x^{-b} + 1) + (x^b + x^{-c} + 1)(x^c + x^{-a} + 1)}{(1 + x^a + x^{-b})(1 + x^b + x^{-c})(1 + x^c + x^{-a})}$ = 1

⇒ $(1 + x^c + x^{-a})(1 + x^a + x^{-b}) + (1 + x^b + x^{-c})(x^a + x^{-b} + 1) + (x^b + x^{-c} + 1)(x^c + x^{-a} + 1) = (1 + x^a + x^{-b})(1 + x^b + x^{-c})(1 + x^c + x^{-a})$

We simplify each term and then compare LHS and RHS.

Simplifying the first term:

$(1 + x^c + x^{-a})(1 + x^a + x^{-b})$

= $x^{c + a} + x^{c - b} + x^c + 1 + x^{-a - b} + x^{-a} + x^{a} + x^{-b} + 1$

= $x^{c + a} + x^{c - b} + x^{c} + x^{-a - b} + x^{-a} + x^{a} + x^{-b} + 2 \hspace{5mm} \hdots (A)$

Now, we simplify the second term we have:

$(1 + x^b + x^{-c})(x^a + x^{-b} + 1)$

= $x^{a + b} + 1 + x^{b} + x^{a - c} + x^{-b - c} + x^{-c} + x^{a} + x^{-b} + 1$

= $x^{a + b} + x^{b} + x^{a - c} + x^{- c - b} + x^{-c} + x^a + x^{-b} + 2 \hspace{5mm} \hdots (B)$

Again, simplifying $(x^b + x^{-c} + 1)(x^c + x^{-a} + 1)$ ,

= $x^{b + c} + x^{b - a} + x^b + 1 + x^{-c -a} + x^{-c} + x^c + x^{-a} + 1$

= $x^{b + c} + x^{b - a} + x^b + x^{-c -a} + x^{-c} + x^c + x^{-a} + 2$ (C)

Therefore, LHS = A + B + C

= $x^{c + a} + x^{c - b} + 2x^c + x^{-a - b} + 2x^{-a} + 2x^{a} + 2x^{-b} + x^{a + b} + 2x^b + x^{a - c} + x^{-c -b} + x^{b + c} + x^{b - a} + x^{-c -a} + 6$

Similarly. RHS

= $(x^{b + c} + x^{b - a} + x^b + 1 + x^{-c -a} + x^{-c} + x^c + x^{-a} + 1)(x^a + x^{-b} + 1)$

Note that if a + b + c = 0, $\implies x^{a + b + c} = x^0 = 1$

So, RHS = $x^{c + a} + x^{c - b} + 2x^c + x^{-a - b} + 2x^{-a} + 2x^{a} + 2x^{-b} + x^{a + b} + 2x^b + x^{a - c} + x^{-c -b} + x^{b + c} + x^{b - a} + x^{-c -a} + 6$

We see that LHS = RHS.

Therefore, the statement is proved.

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4 years ago

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Step-by-step explanation:

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8.84 would be your answer

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