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otez555 [7]
3 years ago
9

Can someone plsss help :)

Mathematics
1 answer:
OLga [1]3 years ago
5 0
Set equal to eachother
(2x+1) = 79
Subtract 1 from both sides
2x=78
Divide 2 from both sides
2x/2=78/2
X= 39
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Help me please PLEASEEEEE
Nuetrik [128]
The answer is 25 since 4•4 is 16 and 16 plus 12 is 28. 28 - (2•3=6) + 3= 25
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3 years ago
What are some conclusions you can make ably the distribution of the data?
olganol [36]

Answer: you can say that all of the numbers have been evenly spread through the chart

Step-by-step explanation:

6 0
3 years ago
Given f(x) = 15x + 35, what is the domain of f?
xeze [42]
All real values of x
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Read 2 more answers
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
Identify an appropriate method to solve the equation x²- 14 x+ 48= -10
riadik2000 [5.3K]

(
x
+
6
)
(
x
+
8
)
Is the awnser for this problem
3 0
3 years ago
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