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3 years ago

6

Angelo makes x dollars for y hours of work. Sarah makes the same amount of money working 1 hour less. Which of the following exp

ressions represents the positive difference between the two people's hourly wage? Enter a, b, c, d, or e. a. -1 - b. - y-1 c. y 1 + d. Y- 1 y-1 w х у e. y х Enter

1 answer:

Orlov [11]3 years ago

8 0

Answer:

A

Step-by-step explanation:

x/y - x/y-1

But remember the fraction with the bigger denometer is the smaller number

x/y is the smaller number

therefore the answer is:

x/y-1 - x/y or A

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Find all pairs of real numbers (a,b) such that (x-a)^2+(2x-b)^2=(x-3)^2+(2x)^2

Mrrafil [7]

Answer:

$(3,0)$

$(\frac{-9}{5},\frac{12}{5})$

Step-by-step explanation:

Let's expand both sides.

I'm going to use the following identity to expand the binomial squared expressions: $(u+v)^2=u^2+2uv+v^2$ or $(u-v)^2=u^2-2uv+v^2$ .

Left-hand side:

$(x-a)^2+(2x-b)^2$

$(x^2-2ax+a^2)+((2x)^2-2b(2x)+b^2)$

$x^2-2ax+a^2+4x^2-4bx+b^2$

Reorder so $x^2$ 's are together and that $x[tex]'s are together.[tex](x^2+4x^2)+(-2ax-4bx)+(a^2+b^2)$

$5x^2+(-2a-4b)x+(a^2+b^2)$

Right-hand side:

$(x-3)^2+(2x)^2$

$x^2-2(3)x+9+4x^2$

$x^2-6x+9+4x^2$

Reorder so $x^2$ 's are together and that $x[tex]'s are together.[tex](x^2+4x^2)+(-6x)+9$

$5x^2-6x+9$

Now let's compare both sides.

If we want both sides to appear exactly the same we need to choose values $a$ and $b$ such the following are true equations:

$-2a-4b=-6$

$a^2+b^2=9$

So if we solve the system we can find the values $a$ and $b$ such that the left=right.

Let's solve the first equation for $a$ in terms of $b$ .

Add $2a$ on both sides:

$-4b=-6+2a$

Divide both sides by -4:

$b=\frac{-6+2a}{-4}$

Reduce (divide top and bottom by -2):

$b=\frac{3-a}{2}$

Now let's plug this into second equation:

$a^2+b^2=9$

$a^2+(\frac{3-a}{2})^2=9$

$a^2+\frac{9-6a+a^2}{4}=9$ (I used the identity $(u-v)^2=u^2-2uv+v^2$ )

Multiply both sides by 4 to clear the fractions from the problem:

$4a^2+(9-6a+a^2)=36$

Combine like terms on left hand side:

$4a^2+a^2-6a+9=36$

$5a^2-6a+9=36$

Subtract 36 on both sides:

$5a^2-6a-27=0$

Now let's try to factor.

We are going to try to find two numbers that multiply to be 5(-27) and add to be -6.

5(-27)=(5*3)(-9)=15(-9)=-15(9) while -15+9=-6.

So let's replace $-6a$ with $-15a+9a$ and factor by grouping.

$5a^2-15a+9a-27=0$

$5a(a-3)+9(a-3)=0$

$(a-3)(5a+9)=0$

This implies $a-3=0$ or $5a+9=0$ .

Solving the first is easy. Just ad 3 on both sides to get: $a=3$ .

The second requires two steps. Subtract 9 and then divide by 5 on both sides.

$5a=-9$

$a=\frac{-9}{5}$ .

So let's go back to finding $b$ now that we know the $a$ values.

If $a=3$ and $b=\frac{3-a}{2}$ ,

then $b=\frac{3-3}{2}=0$ .

So one ordered pair $(a,b)$ that satisfies the equation is:

$(3,0)$ .

If $a=\frac{-9}{5}$ and $b=\frac{3-a}{2}$ ,

then $b=\frac{3-\frac{-9}{5}}{2}$ .

Let's multiply top and bottom by 5 to clear the mini-fraction.

$b=\frac{15-(-9)}{10}$

$b=\frac{24}{10}$

$b=\frac{12}{5}$

So one ordered pair $(a,b)$ that satisfies the equation is:

$(\frac{-9}{5},\frac{12}{5})$ .

5 0

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5 over 10 + 3 over 100 equals?

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The answer is 53 over 100

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Solve the following equation for x: z=x^2+wx

Alja [10]

Minus z both sides
0=x^2+wx+z
we can use quadratic formula or manually complete the square
let's use quadratic

for
0=ax^2+bx+c
$x= \frac{-b+/- \sqrt{b^2-4ac} }{2a}$
given
0=1x^2+wx-z
a=1
b=w
c=-z

$x= \frac{-w+/- \sqrt{w^2-4(1)(-z)} }{2(1)}$
$x= \frac{-w+/- \sqrt{w^2+4z} }{2}$

$x= \frac{-w+ \sqrt{w^2+4z} }{2}$ or $x= \frac{-w- \sqrt{w^2+4z} }{2}$

4 0

3 years ago

Which expression is equivalent to r^m divided by r^n

wel

The expression is equivalent to r^(m-n)

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4 years ago

Four more than twice a number is the same as 8 less than three times the same

romanna [79]

Answer:

Step-by-step explanation:

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3 years ago