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xxTIMURxx [149]
2 years ago
5

I need help please no need for step by step.

Mathematics
2 answers:
goldenfox [79]2 years ago
7 0

Answer:

sorry but there is no question/pic

Step-by-step explanation:

Klio2033 [76]2 years ago
4 0
Sorry you dont have a picture so i can not help
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Need help with this one
Vedmedyk [2.9K]
The answer is 20% since the percentages cancels each other out.
8 0
3 years ago
The following triangles are similar. Find side AB.
tankabanditka [31]

Answer:

4

Step-by-step explanation:

Set up ratios of corresponding sides.  x+2 is to 6 as 2 is to 3:

\frac{x+2}{6}=\frac{2}{3}

Now cross multiply to get

3(x + 2) = 12 and

3x + 6 = 12 so

3x = 6 and

x = 2.  That means that side AB, x + 2, is 2 + 2 which is 4

4 0
3 years ago
Y=x^2-6x-16 in vertex form
satela [25.4K]

Answer:

y=(x-3)^{2} -25

Step-by-step explanation:

The standard form of a quadratic equation is y=ax^{2} +bx+c

The vertex form of a quadratic equation is y=a(x-h)^{2} +k

The vertex of a quadratic is (h,k) which is the maximum or minimum of a quadratic equation. To find the vertex of a quadratic, you can either graph the function and find the vertex, or you can find it algebraically.

To find the h-value of the vertex, you use the following equation:

h=\frac{-b}{2a}

In this case, our quadratic equation is y=x^{2} -6x-16. Our a-value is 1, our b-value is -6, and our c-value is -16. We will only be using the a and b values. To find the h-value, we will plug in these values into the equation shown below.

h=\frac{-b}{2a} ⇒ h=\frac{-(-6)}{2(1)}=\frac{6}{2} =3

Now, that we found our h-value, we need to find our k-value. To find the k-value, you plug in the h-value we found into the given quadratic equation which in this case is y=x^{2} -6x-16

y=x^{2} -6x-16 ⇒ y=(3)^{2} -6(3)-16 ⇒ y=9-18-16 ⇒ y=-25

This y-value that we just found is our k-value.

Next, we are going to set up our equation in vertex form. As a reminder, vertex form is: y=a(x-h)^{2} +k

a: 1

h: 3

k: -25

y=(x-3)^{2} -25

Hope this helps!

3 0
3 years ago
Given f"(x) = 2, f'(1)=4, and f(2)=-2,<br> find f(x).
Anna [14]

Answer:

Step-by-step explanation:

f"(x)=2

integrating

f'(x)=2x+c

f'(1)=2+c=4

c=4-2=2

f'(x)=2x+2

integrating

f(x)=2x^2/2+2x+a

f(x)=x^2+2x+a

f(2)=-2

(2)^2+2(2)+a=-2

4+4+a=-2

a=-2-8=-10

f(x)=x^2+2x-10

6 0
2 years ago
Is the solution of this equation extraneous?
UkoKoshka [18]

Answer:

x = - 3 is extraneous

Step-by-step explanation:

Given

\sqrt{x+7} - 1 = x  ( add 1 to both sides )

\sqrt{x+7} = x + 1 ( square both sides )

x + 7 = (x + 1)² ← distribute right side

x + 7 = x² + 2x + 1 ( subtract x + 7 from both sides )

0 = x² + x - 6 ← in standard form

0 = (x + 3)(x - 2) ← in factored form

Equate each factor to zero and solve for x

x + 3 = 0 ⇒ x = - 3

x - 2 = 0 ⇒ x = 2

As a check

Substitute these values into the equation and if both sides are equal then they are the solutions.

x = - 3 : \sqrt{-3+7} - 1 = \sqrt{4} - 1 = 2 - 1 = 1 ≠ - 3

x = 2 : \sqrt{2+7} - 1  = \sqrt{9} - 1 = 3 - 1 = 2

x = 2 is a solution and x = - 3 is extraneous

6 0
3 years ago
Read 2 more answers
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