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3 years ago

John made $20 for 2 hours of work. Sally made $50 for 4 hours of work. How do John and Sally's hourly pay rate compare?

2 answers:

6 0

20/2= 10

John makes 10 dollars an hour.

50/4= 12.5

Sally makes 12.50 dollars an hour.

Sally makes more than John.

I hope this helps!
~kaikers

vlabodo [156]3 years ago

3 0

John gets $10 per hour and Sally makes $12.50 per hour so Sally makes $2.50 more then John per hour.

We can tell John makes $10 per hour because we divid how much he made by the amount of hours it took to make to get 10 and we did the same thing for Sally ($50/4=12.5) and then we subtract Sally's pay per hour by John's pay per hour to see how much more Sally makes per hour.

I hope that helped and made sense. :)

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Help please !! easy question i believe :))

kolezko [41]

Solve for x given two sides

Solution

x is an angle hence we are gonna use the formula Sin

Sin 0 = opp/hyp

0 represents theta.

Sin 0 = opp/hyp
Sin 0 = 27/33

Put 27/33 in decimal form

= 0.8181

Transpose to make 0 the subject

Sin 0 = 0.8181

0 = Sin-1 0.8181

When adding Sin from one side to another it becomes Sin inverse.

0 = Sin-1 08181

Now we find the answer

0 = Sin-1 0.8181

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2 years ago

Two Dimensional Nets and Surface Area - Quiz - Level

Virty [35]

Answer:

$18\ ft^2$

Step-by-step explanation:

The complete question in the attached figure

we know that

The surface area of a cube is equal to the area of its 6 squares faces

The area of one square face is equal to

$A=b^2$

where

b is the length side of the square

we have

$b=1\ ft$

substitute

$A=(1)^2=1\ ft^2$ ---> area of one square face

Find the surface area of three cubes

Multiply the area of one square face by 6 (one cube has 6 faces) and then multiply the result by 3 (3 cubes)

$SA=3[6(1)]=18\ ft^2$

6 0

3 years ago

Perform the following operations s and prove closure. Show your work.

nadezda [96]

Answer:

1. $\frac{x}{x+3}+\frac{x+2}{x+5}$ = $\frac{2x^2+10x+6}{(x+3)(x+5)}\\$

2. $\frac{x+4}{x^2+5x+6}*\frac{x+3}{x^2-16}$ = $\frac{1}{(x+2)(x-4)}$

3. $\frac{2}{x^2-9}-\frac{3x}{x^2-5x+6}$ = $\frac{-3x^2-7x-4}{(x+3)(x-3)(x-2)}$

4. $\frac{x+4}{x^2-5x+6}\div\frac{x^2-16}{x+3}$ = $\frac{1}{(x-2)(x-4)}$

Step-by-step explanation:

1. $\frac{x}{x+3}+\frac{x+2}{x+5}$

Taking LCM of (x+3) and (x+5) which is: (x+3)(x+5)

$=\frac{x(x+5)+(x+2)(x+3)}{(x+3)(x+5)}\\=\frac{x^2+5x+(x)(x+3)+2(x+3)}{(x+3)(x+5)} \\=\frac{x^2+5x+x^2+3x+2x+6}{(x+3)(x+5)} \\=\frac{x^2+x^2+5x+3x+2x+6}{(x+3)(x+5)} \\=\frac{2x^2+10x+6}{(x+3)(x+5)}\\$

Prove closure: The value of x≠-3 and x≠-5 because if there values are -3 and -5 then the denominator will be zero.

2. $\frac{x+4}{x^2+5x+6}*\frac{x+3}{x^2-16}$

Factors of x^2-16 = (x)^2 -(4)^2 = (x-4)(x+4)

Factors of x^2+5x+6 = x^2+3x+2x+6 = x(x+3)+2(x+3) =(x+2)(x+3)

Putting factors

$=\frac{x+4}{(x+3)(x+2)}*\frac{x+3}{(x-4)(x+4)}\\\\=\frac{1}{(x+2)(x-4)}$

Prove closure: The value of x≠-2 and x≠4 because if there values are -2 and 4 then the denominator will be zero.

3. $\frac{2}{x^2-9}-\frac{3x}{x^2-5x+6}$

Factors of x^2-9 = (x)^2-(3)^2 = (x-3)(x+3)

Factors of x^2-5x+6 = x^2-2x-3x+6 = x(x-2)+3(x-2) =(x-2)(x+3)

Putting factors

$\frac{2}{(x+3)(x-3)}-\frac{3x}{(x+3)(x-2)}$

Taking LCM of (x-3)(x+3) and (x-2)(x+3) we get (x-3)(x+3)(x-2)

$\frac{2(x-2)-3x(x+3)(x-3)}{(x+3)(x-3)(x-2)}$

$=\frac{2(x-2)-3x(x+3)}{(x+3)(x-3)(x-2)}\\=\frac{2x-4-3x^2-9x}{(x+3)(x-3)(x-2)}\\=\frac{-3x^2-9x+2x-4}{(x+3)(x-3)(x-2)}\\=\frac{-3x^2-7x-4}{(x+3)(x-3)(x-2)}$