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mash [69]
3 years ago
10

What's the answer ? I been having trouble finding the answer

Mathematics
1 answer:
Fiesta28 [93]3 years ago
5 0
What's the answer to what?
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The school that Ted goes to is selling tickets to a choral performance. On the first day of ticket sales the school sold 7 adult
cestrela7 [59]

Answer:

Equations

Step-by-step explanation:

x=adult y=child

Day 1= 7x+8y= $168

day 2= 14x+13y= $294

7 0
3 years ago
Question is in the photo.
pashok25 [27]

Answer:

is there anymore info in it

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Help me please! I have no idea what I’m doing! <br><br> Solve : 3p+ 25(8) = -21p -40
tatyana61 [14]

Answer:

3p + 200 = -21p - 40

     -200             -200

3p = -21p - 240

+21p  -21p

24p = -240

p = 10

Hope this helps

6 0
3 years ago
Read 2 more answers
I need help please answer these two questions for me! Will give points.​
jarptica [38.1K]

Answer:

7.) 7

10.)  0

Step-by-step explanation:

When it means "evaluate the function", it's in essence asking us to see what the function spits out when we feed it a certain input. Our inputs are our x values, which spit out a y value.

Evaluating the function when x = 1:

Let's look at where the function has an x value of 1. We see it near the bottom of the table and see the y value associated with the input is 7. So when the function is fed 1 as an input, it spits out 7.

Evaluating the function when f(x) = - 2:

This one is a weird because of the new notation. Just think of it as some value of f, which we don't know (so we represent it as an x-variable) must equal -2. So let's look at our table to find out where our output is -2. We find that when f(x) = -2 the input is 0. So the input which gives -2 is 0.

7 0
3 years ago
Suppose that 2 ≤ f '(x) ≤ 3 for all values of x. what are the minimum and maximum possible values of f(5) − f(1)?
zzz [600]

Assuming f is differentiable over the interval (1, 5), by the mean value theorem we know that there is some 1 such that

f'(c)=\dfrac{f(5)-f(1)}{5-1}

We're given that 2\le f'(x)\le3 for all x, so

2\le\dfrac{f(5)-f(1)}4\le3\implies8\le f(5)-f(1)\le12

that is, the minimum possible value is 8, and the maximum possible value is 12.

8 0
3 years ago
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