Ask question

Ask question

All categories

3 years ago

6

HELP! Simon invests in a medical company. He buys the stock at $189.35 and sells it for $167.78. What is the percent of decrease

? Round your answer to the nearest whole percent.

1 answer:

myrzilka [38]3 years ago

3 0

167.78 / 189.35 = .88608 * 100 = 88.6083971481 round it you get 89

You might be interested in

Hoy many pairs of parallel lines does a Square have

otez555 [7]

A square has 2 pairs of parallel line because all four sides are equal

7 0

3 years ago

Read 2 more answers

Last week Mark ran 31 miles more than Jack. Mark ran 46 miles .How many miles did Jack run

Levart [38]

You take what Mark ran, 46 miles, and you see that Mark ran 31 more miles than Jack, so the equation would be 46-31, which would be 15 miles. So, Jack ran 15 miles.

8 0

2 years ago

Read 2 more answers

Lucy is playing a game at the school carnival. There is a box of marbles, and each box has a white, a green, a red, and an orang

Dmitriy789 [7]

<span>4 marbles times 6 die faces = 24</span>

6 0

2 years ago

**8. At a buffet restaurant, the price for dinner for an adult is $6.95 and the price for dinner for a child is $3.95. A

qaws [65]

Answer is A

Step-by-step explanation:

3 0

2 years ago

[tex]cos {}^{4} α+sin {}^{4} α= \frac{1}{4} (3+cos4α)<br>Prove:<br>

asambeis [7]

Given:

$\cos^4 \alpha+\sin^4\alpha=\dfrac{1}{4}(3+\cos 4 \alpha)$

To prove:

The given statement.

Proof:

We have,

$\cos^4 \alpha+\sin^4\alpha=\dfrac{1}{4}(3+\cos 4 \alpha)$

$LHS=\cos^4 \alpha+\sin^4\alpha$

$LHS=(\cos^2 \alpha)^2+(\sin^2 \alpha)^2$

$LHS=(\cos^2 \alpha+\sin^2\alpha)^2-2\sin ^2\alpha\cos^2 \alpha$ $[\because a^2+b^2=(a+b)^2-2ab]$

$LHS=(1)^2-2(1-\cos^2 \alpha)\cos^2 \alpha$ $[\because \cos^2 \alpha+\sin^2\alpha=1]$

$LHS=1-2\cos^2 \alpha+2\cos^4 \alpha$

Now,

$RHS=\dfrac{1}{4}(3+\cos 4 \alpha)$

$RHS=\dfrac{1}{4}[3+(2\cos^2 2\alpha-1)]$ $[\because \cos 2\theta=2\cos^2\theta -1]$

$RHS=\dfrac{1}{4}[2+2\cos^2 2\alpha]$

$RHS=\dfrac{1}{4}[2+2(2\cos^2 \alpha-1)^2]$ $[\because \cos 2\theta=2\cos^2\theta -1]$

$RHS=\dfrac{1}{4}[2+2(4\cos^4 \alpha-4\cos \alpha+1)]$ $[\because (a-b)^2=a^2-2ab+b^2]$

$RHS=\dfrac{1}{4}[2+8\cos^4 \alpha-8\cos \alpha+2]$

$RHS=\dfrac{1}{4}[4+8\cos^4 \alpha-8\cos \alpha]$

$RHS=1+2\cos^4 \alpha-2\cos \alpha$

$RHS=1-2\cos^2 \alpha+2\cos^4 \alpha$

$LHS=RHS$

Hence proved.

8 0

3 years ago