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2 years ago

12

An athlete goes on a diet. He loses 3 pounds each month for 8 months. He lost a total of 24 pounds. If he wants to lose 30 pound

s total, then he needs to lose | more pounds to reach his goal.

1 answer:

sukhopar [10]2 years ago

5 0

If he wants to loose 30 points total, subtract what he lost already, 24, from 30. He needs to lose 6 more pounds.

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Which of the following equations is the result of completing the square on x^2 - 6x - 9 = 0?

lina2011 [118]

X² - 6x - 9 = 0
x² - 3x - 3x - 9 = 0
x (x-3) -3(x + 3) = 0
(x - 3)² = 0

In short, Your Answer would be Option A

Hope this helps!

6 0

2 years ago

Read 2 more answers

Evaluate 20/y - 8/x + 3xy/x+y when x=2 and y=5

Lesechka [4]

20(5)-8(2)+3(2-5)+5
100-16-9= 79

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3 years ago

If mZDEF = 115, then what are mZFEG and mZHEG? The diagram is not to scale.

stepan [7]

Answer:

the answer is c

Step-by-step explanation:

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6 0

3 years ago

-6+7x =-2x+57 What is this?

crimeas [40]

Answer:

x = 7

Step-by-step explanation:

-6+7x = -2x+57

+6+2x +2x+6

9×/9 = 63/9

× = 7

8 0

3 years ago

Answer the question in the picture

nekit [7.7K]

Recall the angle sum identities:

$\sin(x+y)=\sin x\cos y+\cos x\sin y$

$\cos(x+y)=\cos x\cos y-\sin x\sin y$

Now,

$\tan(x+y)=\dfrac{\sin(x+y)}{\cos(x+y)}=\dfrac{\sin x\cos y+\cos x\sin y}{\cos x\cos y-\sin x\sin y}$

Divide through numerator and denominator by $\cos x\cos y$ to get

$\tan(x+y)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}$

Next, we use the fact that $x,y$ lie in the first quadrant to determine that

$\sin x=\dfrac12\implies\cos x=\sqrt{1-\sin^2x}=\dfrac{\sqrt3}2$

$\cos y=\dfrac{\sqrt2}2\implies\sin x=\sqrt{1-\cos^2x}=\dfrac1{\sqrt2}$

So we then have

$\tan x=\dfrac{\sin x}{\cos x}=\dfrac{\frac12}{\frac{\sqrt3}2}=\dfrac1{\sqrt3}$

$\tan y=\dfrac{\sin y}{\cos y}=\dfrac{\frac1{\sqrt2}}{\frac{\sqrt2}2}=1$

Finally,

$\tan(x+y)=\dfrac{\frac1{\sqrt3}+1}{1-\frac1{\sqrt3}}=\dfrac{1+\sqrt3}{\sqrt3-1}=2+\sqrt3\approx3.73$

4 0

2 years ago