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Mashutka [201]
3 years ago
15

The town of Worman Grove collected 9,645 blue pens and 18, 836 black pens for a school supplies drive. Their goal is to have 30,

000 pens. How many more pens do they need o reach their goal? *
Your answer
Mathematics
1 answer:
Sonbull [250]3 years ago
6 0

Answer:

1519 pens

Step-by-step explanation:

add 9,645 to 18,836 and then subtract that number from 30,000

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Suppose that there are two types of tickets to a show: advance and same-day. The combined cost of one advance ticket and one sam
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We can set up 2 equations given the information.
Let the price of advance ticket be represented by A and same day ticket by S

A + S = 50
20A + 40S = 1700

Solve for A in the first equation by subtracting S on both sides.
U will get A = 50 - S

Now substitute 50 - S for A in the second equation.

20 (50 - S) + 40S = 1700
1000 - 20S + 40S = 1700
20S = 700
S = 35

Same day ticket costs $35 and advance ticket costs $15
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Type of angle:<br>Angle measures:
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Type: acute angle
measurement: 85°
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Find f^-1(x) and it’s domain.
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Answer: A

Step-by-step explanation:

Letting f(y)=x,

x=\sqrt{y}-5\\\\x+5=\sqrt{y}\\\\y=(x+5)^{2}

Also, the domain of an inverse is the same as the range of the original function, so the range is x \geq 0

7 0
2 years ago
2.1 = -0.6m<br><br>What is the solution to m?​
Anon25 [30]

Answer:

m=-3.5

Good Luck!!!

8 0
3 years ago
Read 2 more answers
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
3 years ago
Read 2 more answers
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