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podryga [215]
4 years ago
11

How do you solve 125 - (5-9)^3÷2

Mathematics
1 answer:
hoa [83]4 years ago
8 0
Solve what is in parenthesis first (5-9) = -4

Solve exponents next -4^3 = -64

Divide next
- 64 \div 2
= -32

Lastly, subtract and resolve the negatives which become a positive number 125 - (-32) = 125 + 32 = 157
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Find x when f(x)=1 . (These means find the input that gives an output of 1.)<br><br> f(x)=1/x-2
Andreyy89

Answer:

x = 1/3

Step-by-step explanation:

1 = 1/x -2

add 2 to both sides

3 = 1/x

x = 1/3

5 0
3 years ago
Factor −5x2 + 10x.<br><br> −5x(x + 2)<br> 5(−x2 + 10x)<br> 5x(−x + 2)<br> x(5x + 10)
Hitman42 [59]

{ \red{ \bold{option \: (c)}}}

Step-by-step explanation:

{ \purple{ \tt{ { - 5x}^{2}  + 10x}}}

Take { \purple{ \tt{5x}}}as common. Then,

{ \purple{ \tt{5x( - x + 2)}}}

4 0
1 year ago
A fitness club charges $25 per month as well as a one-time $80 fee of all new members which equation represents the cost in doll
madam [21]

Answer: So if you want to find the total cost (c) for a certain number of months (m) and it costs $35 per month, you just need to multiply the number of months by the cost per month. So

35m=c

If you want to see the units cancel

($35/month)(months)=$

The months will cancel out, and you will be left with total dollars.

Not sure about the second part of your question, but if you wanted to figure out how much it would cost to be a member for 6 months, you would just plug in 6 for m.

35(6)=c

c=210

So $210

Step-by-step explanation:

4 0
4 years ago
There are 10 girls and 8 boys in Mr Augellis class. If 5 more girls and 5 more boys join the class, will the ratio of girls to b
zimovet [89]
Yes they will stay the same
8 0
3 years ago
Help!! 50 points and brainliest!
Viktor [21]

Answer:

Second choice:

x=2t

y=4t^2+4t-3

Fifth choice:

x=t+1

y=t^2+4t

Step-by-step explanation:

Let's look at choice 1.

x=t+1

y=t^2+2t

I'm going to subtract 1 on both sides for the first equation giving me x-1=t. I will replace the t in the second equation with this substitution from equation 1.

y=(x-1)^2+2(x-1)

Expand using the distributive property and the identity (u+v)^2=u^2+2uv+v^2:

y=(x^2-2x+1)+(2x-2)

y=x^2+(-2x+2x)+(1-2)

y=x^2+0+-1

y=x^2

So this not the desired result.

Let's look at choice 2.

x=2t

y=4t^2+4t-3

Solve the first equation for t by dividing both sides by 2:

t=\frac{x}{2}.

Let's plug this into equation 2:

y=4(\frac{x}{2})^2+4(\frac{x}{2})-3

y=4(\frac{x^2}{4})+2x-3

y=x^2+2x-3

This is the desired result.

Choice 3:

x=t-3

y=t^2+2t

Solve the first equation for t by adding 3 on both sides:

x+3=t.

Plug into second equation:

y=(x+3)^2+2(x+3)

Expanding using the distributive property and the earlier identity mentioned to expand the binomial square:

y=(x^2+6x+9)+(2x+6)

y=(x^2)+(6x+2x)+(9+6)

y=x^2+8x+15

Not the desired result.

Choice 4:

x=t^2

y=2t-3

I'm going to solve the bottom equation for t since I don't want to deal with square roots.

Add 3 on both sides:

y+3=2t

Divide both sides by 2:

\frac{y+3}{2}=t

Plug into equation 1:

x=(\frac{y+3}{2})^2

This is not the desired result because the y variable will be squared now instead of the x variable.

Choice 5:

x=t+1

y=t^2+4t

Solve the first equation for t by subtracting 1 on both sides:

x-1=t.

Plug into equation 2:

y=(x-1)^2+4(x-1)

Distribute and use the binomial square identity used earlier:

y=(x^2-2x+1)+(4x-4)

y=(x^2)+(-2x+4x)+(1-4)

y=x^2+2x+-3

y=x^2+2x-3.

This is the desired result.

3 0
4 years ago
Read 2 more answers
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