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fomenos
3 years ago
7

How do you solve:6r+7=13+7r

Mathematics
2 answers:
Sergio [31]3 years ago
7 0
You need to get the r by itself on one side, with the constants (numbers without the r) on the other side.

6r+7=13+7r\\6r-7r+7=13+7r-7r\\-r+7=13\\-r+7-7=13-7\\-r=6\\-r/-1=6/-1\\r=-6
densk [106]3 years ago
5 0
6r+7=13+7r \\ -6r\ \ \ \ \ -6r \\ 7=13+r \\ -13\ \ -13 \\ -6=r
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What is the vertex of the quadratic function f(x)=(x-8)(x-2)?
maw [93]

Answer:

(5, -9)

Step-by-step explanation:

f(x)=(x-8)(x-2)

We want the vertex.  So let us get this in Vertex Form.

f(x)  =  (x - 8)(x - 2) =  x^2 + 16 - 10x

f(x) = x^2 - 10x + 16

f(x) = x^2 -10x  + 25  - 25 + 16

f(x) = (x^2 -10x  + 25)  - 25 + 16

f(x) = (x - 5)^2 - 9

Vertex at  (5, -9)

3 0
3 years ago
If the sine of an acute angle is 3/5, what is the cosine of<br> that angle?
Korvikt [17]

Answer:

The cosine of the angle is 4/5.

Step-by-step explanation:

An acute angle is between 0º and 90º, in the first quadrant of the trigonometric circle, in which both the sine and the cosine are positive values.

For each angle \alpha, we have that:

\sin^{2}{\alpha} + \cos^{2}{\alpha} = 1

We have that:

\sin{\alpha} = \frac{3}{5}

So

\sin^{2}{\alpha} + \cos^{2}{\alpha} = 1

(\frac{3}{5})^{2} + \cos^{2}{\alpha} = 1

\cos^{2}{\alpha} = \frac{16}{25}

\cos{\alpha} = \frac{4}{5}

The cosine of the angle is 4/5.

8 0
3 years ago
Help<br> Me ? I’m marking brainlist
dsp73

Answer: I belive the answer is C

Step-by-step explanation:

7 0
3 years ago
Someone please help me on this question ?:)
Whitepunk [10]

C  ,      a relationship between two variables can be expressed by an equation in which one variable is equal to a constant.

8 0
3 years ago
Anderson has $50 in his savings account. He deposits $5 every week. His father also deposits $20 into the account every time And
Sav [38]

Part A: 5 is the coefficient, w is the variable, 50 is the constant

Part B: 10 x 5 = 50 + 2 x 20 = 40+ 50 = $140

Part C: the constant,50,would change and become 85.  Hope this helps :)
8 0
3 years ago
Read 2 more answers
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