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S_A_V [24]
3 years ago
13

Based on the graph what are the solutions to ax^2 + bx+c=0 Select all that apply

Mathematics
2 answers:
AlexFokin [52]3 years ago
8 0

Answer:

x=-2 and x = 5

Step-by-step explanation:

The solutions to the equation are where the graph crosses the x axis

We can see that the graph crosses at x=-2 and x = 5

max2010maxim [7]3 years ago
4 0

Answer:

x = -2

x = 5

Step-by-step explanation:

The answer is found when the graph crosses the x-axis

Hope this helps :D

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You can mow one acre in 126 minutes. i can mow one acre in 117 minutes. how long will it take you to mow a 0.17-acre lawn? 13 in
Morgarella [4.7K]

a) (0.17 ac)×(126 min/ac) = 21.42 min

b) I can mow 1/126 ac/min; you can mow 1/117 ac/min. Together, we can mow at a rate that is the sum of these:

... (1/126 + 1/117) ac/min = (117 + 126)/(126×117) = 243/14,742 ac/min

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5 0
3 years ago
Use the information to answer the following question.
Zanzabum

Answer:

C

Step-by-step explanation:

Since the difference between 0 and -10 is 10, this means that the difference between the diving board and the pool also has to be 10. Therefore, the diving board is 10 feet above the water since 0+10 is 10.

6 0
3 years ago
Read 2 more answers
<img src="https://tex.z-dn.net/?f=%20%5Crm%20%5Cint_%7B0%7D%5E2%20%5Cleft%28%20%5Csqrt%5B3%5D%7B%20%7Bx%7D%5E%7B2%7D%20%20%2B%20
Verizon [17]

If f(x) has an inverse on [a, b], then integrating by parts (take u = f(x) and dv = dx), we can show

\displaystyle \int_a^b f(x) \, dx = b f(b) - a f(a) - \int_{f(a)}^{f(b)} f^{-1}(x) \, dx

Let f(x) = \sqrt{1 + x^3}. Compute the inverse:

f\left(f^{-1}(x)\right) = \sqrt{1 + f^{-1}(x)^3} = x \implies f^{-1}(x) = \sqrt[3]{x^2-1}

and we immediately notice that f^{-1}(x+1)=\sqrt[3]{x^2+2x}.

So, we can write the given integral as

\displaystyle \int_0^2 f^{-1}(x+1) + f(x) \, dx

Splitting up terms and replacing x \to x-1 in the first integral, we get

\displaystyle \int_1^3 f^{-1}(x) \, dx + \int_0^2 f(x) \, dx = 2 f(2) - 0 f(0) = 2\times3+0\times1=\boxed{6}

7 0
2 years ago
Let x^2+15x=49
Slav-nsk [51]
<span>x^2 + 15x + 56.25 = 105.25

    "Completing the square" is one of many different techniques for solving a quadratic equation. What you do is add a constant to both sides of the equation such that the lefthand side can be factored into the form a(x+d)^2. For instance, squaring (X+D) = X^2 + 2DX + D^2. Notice the 2DX term. That is the same term as the 15x term in the problem. So 2D = 15, D = 7.5. And D^2 = 7.5^2 = 56.25.
So we have
 x^2 + 15x + 56.25 = 49 + 56.25
 Which is
 x^2 + 15x + 56.25 = 105.25
  Which is the answer desired.

   Now the rest of this is going beyond the answer. Namely, it's answering the question "Why does complementing the square help?" Well, we know that the left hand side of the equation can now be written as
 (x+7.5)^2 = 105.25

   Now take the square root of each side
 (x+7.5) = sqrt(105.25)

   And let's use both the positive and negative square roots.
So x+7.5 = 10.25914226
 and
 x+7.5 = -10.25914226

   And let's find X.
 x+7.5 = 10.25914226
 x = 2.759142264

   x+7.5 = -10.25914226
 x = -17.75914226

   So the roots for x^2 + 15x - 49 is 2.759142264, and -17.75914226</span>
3 0
3 years ago
Read 2 more answers
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