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3 years ago

Please help me with this problem

Mathematics

1 answer:

juin [17]3 years ago

8 0

My guess would be less than $25

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Solve using substitution:<br> y = 22x + 10<br> y = 20x + 16

stealth61 [152]

Answer:

x=3, y=76

Step-by-step explanation:

You set them equal because you plug in "y" as 22x+10 into "y=20x+16"

22x+10=20x+16

2x=6

x=3

20(3)+16=76

4 0

3 years ago

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(21.) A writer earns $3400 a month. Last month she spent $204 on food.

Sergeu [11.5K]

6 percent was spent

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2 years ago

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Sara Blake's grandmother gave her $3000.00 to save for college. She put it in a savings account that earns 6% interest per year.

Elina [12.6K]

• P is the principal amount, $3000.00.

• r is the interest rate, 6% per year, or in decimal form, 6/100=0.06.

• t is the time involved, 8 years time periods.

• So, t is 8 year time periods.

To find the simple interest, we multiply 3000 × 0.06 × 8 to get that:

The interest is: $1440.00

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3 years ago

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xeze [42]

Answer:

Step-by-step explanation:

I wrote it super fast. lol

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3 years ago

Sinx = 1/2, cosy = sqrt2/2, and angle x and angle y are both in the first quadrant.

Leviafan [203]

Answer:

Option D. 3.73

Step-by-step explanation:

we know that

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

and

$sin^{2}(\alpha)+cos^{2}(\alpha)=1$

step 1

Find cos(X)

we have

$sin(x)=\frac{1}{2}$

we know that

$sin^{2}(x)+cos^{2}(x)=1$

substitute

$(\frac{1}{2})^{2}+cos^{2}(x)=1$

$cos^{2}(x)=1-\frac{1}{4}$

$cos^{2}(x)=\frac{3}{4}$

$cos(x)=\frac{\sqrt{3}}{2}$

step 2

Find tan(x)

$tan(x)=sin(x)/cos(x)$

substitute

$tan(x)=1/\sqrt{3}$

step 3

Find sin(y)

we have

$cos(y)=\frac{\sqrt{2}}{2}$

we know that

$sin^{2}(y)+cos^{2}(y)=1$

substitute

$sin^{2}(y)+(\frac{\sqrt{2}}{2})^{2}=1$

$sin^{2}(y)=1-\frac{2}{4}$

$sin^{2}(y)=\frac{2}{4}$

$sin(y)=\frac{\sqrt{2}}{2}$

step 4

Find tan(y)

$tan(y)=sin(y)/cos(y)$

substitute

$tan(y)=1$

step 5

Find tan(x+y)

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

substitute

$tan(x+y)=[1/\sqrt{3}+1}]/[{1-1/\sqrt{3}}]=3.73$

7 0

3 years ago