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nikklg [1K]
3 years ago
13

Last question and i can go!!! this is due today and im giving breainlies t !

Mathematics
2 answers:
olga55 [171]3 years ago
8 0

Answer:

A

Step-by-step explanation:

nataly862011 [7]3 years ago
6 0

Answer:

Hey There!

Here,

we've a circle with a radius =. 5(10/2)m

And a square with side =10m

The area of shaded region

➠ Area of Square- Area of Circle

➠10*10-(3.14*5*5)

➠100-78.5

➠21.5m² is the answer

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At the concession stand, medium sodas cost $1.25 and hot dogs cost $2.50. Natalie's group brought in pizzas, but is buying the d
Shtirlitz [24]
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Read 2 more answers
Find so that the point (-1,2) is on the graph f(x)=ax^2+4. What does A equal?
natali 33 [55]

We have to find the parameter a so that (-1,2) is part of the function f(x) = ax²+4.

To check if a point is part of a function, we can replace the values of x and y = f(x) with the coordinates of the point and then, if the equation stays true, then the point is part of the function.

So for (x,y) = (-1,2) to be part of the function y = f(x), this equation has to stands true:

\begin{gathered} y=f(x) \\ y=ax^2+4 \\ 2=a(-1)^2+4 \\ 2=a+4 \\ a=2-4 \\ a=-2 \end{gathered}

Then, the function would have to be f(x) = -2x² + 4.

We can check with a graph as:

Answer: a = -2

8 0
1 year ago
What is the formula for volume of a square?
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3 0
3 years ago
X+3<8 and 3(x+4)-11<10
kondaur [170]

Answer:

<h3>X<5 and X<3</h3>

Step-by-step explanation:

To solve this problem, first, you have to isolate it on one side of the equation. Remember to solve this problem, isolate x on one side of the equation.

x+3<8 and 3(x+4)-11<10

x+3<8

x+3-3<8-3 (First, subtract 3 from both sides.)

8-3 (Solve.)

8-3=5

x<5

3(x+4)-11<10

3(x+4)-11+11<10+11 (Add 11 from both sides.)

10+11 (Solve.)

10+11=21

3(x+4)<21

3(x+4)/3<21/3 (Next, divide by 3 from both sides.)

21/3 (Solve.)

21/3=7

x+4<7

x+4-4<7-4 (Then, subtract 4 from both sides.)

7-4=3

x<3

The correct answer is x<5 and x<3.

4 0
3 years ago
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