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weqwewe [10]
2 years ago
14

I have to do this project see look and I don't understand what I'm supposed to do pls help​

Mathematics
2 answers:
Blababa [14]2 years ago
6 0
Maybe you need to plot some points on the graph. Do you have any instructions for the assignment?
Masteriza [31]2 years ago
3 0

Answer:

you need to plot points on the graph. did you get the water park one? if you did you need to design a Waterpark and plot the points

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Can I please get some help on this...I have tried to many times...
Bezzdna [24]

Answer: A

Step-by-step explanation:

First, the problem is g(f(x)). You would plug in f(x) wherever you see an x in g(x). To find the domain, you take the bottom function, and set it equal to 0.

\sqrt{x-2} =0

When you solve that, you get x=2. You know your domain is x≥2, but there is as asymptote at x=11. That means the graph never reaches x=11, but gets very close. You find that by setting the entire equation equal to 0 and solve from there.

5 0
3 years ago
List them from least to greatest
joja [24]

Answer:

the answer is √24,5.1,√33

3 0
2 years ago
Factor 200t+83 - 80t2.<br> A. 8t(t - 5)(t + 5)<br> B. 8t(t - 5)2<br> C. -8t(t - 5)(t + 5)
Mashutka [201]

Step-by-step explanation:

8t³-80t²+200t

= 8(t³-10t+25)

= 8t(t²-10t+25)

= 8t(t-5)(t-5)

= 8t(t-5)²

the answer is B.

3 0
2 years ago
Simplify the radical expression.<br><br> 1. √45<br><br> 2. √180x^2<br><br> 3.√490y^5 w^6
lara [203]
<span>#1) 3√5
#2) 6x</span><span>√5
#3) 7w^3y^2</span>√10y
8 0
3 years ago
Read 2 more answers
For what value of k will the lines x+2y=0, 3x-4y-10=0 and 5x+ky-7=0 are concurrent?​
konstantin123 [22]

Answer:

After solve the equations we get value of k=3

Step-by-step explanation:

We need to find value of k for which the lines x+2y=0, 3x-4y-10=0 and 5x+ky-7=0 are concurrent.

If the lines are concurrent, they pass through same point.

Let:

x+2y=0--eq(1)\\ 3x-4y-10=0--eq(2)\\ 5x+ky-7=0--eq(3)

First solving equation 1 and 2 to find values of x and y

From eq(1) we find value of x and put it in eq(2)

From \ eq(1) x+2y=0\\x=-2y\\Put x=-2y \ in \ eq(2)\\3x-4y-10=0\\3(-2y)-4y-10=0 \\-6y-4y=10\\-10y=10\\y=\frac{10}{-10}\\y=-1

After solving we get value of y=-1

Now putting in eq(1) to get value of x

x+2y=0\\x+2(-1)=0\\x-2=0\\x=2

So, Value of x= 2

Now put value of x=2 and y=-1 into eq(3) to find value of k

5x+ky-7=0\\5(2)+k(-1)-7=0\\10-k-7=0\\-k+3=0\\-k=-3\\k=3

So, After solve the equations we get value of k=3

4 0
2 years ago
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