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Minchanka [31]
3 years ago
12

Can someone solve 8 = b - 1 + 8b STEP BY STEP PLEASE

Mathematics
1 answer:
andrew11 [14]3 years ago
7 0

Answer:

8 = b - 1 + 8b

group like terms

8 + 1 = b + 8b

9 = 9b cus b standing alone me 1b

divide both side by 9

<u>9</u><u> </u> = <u>9</u><u>b</u>

9 9

b = 1

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Which values of x make the inequality 1 + 5 x &lt; −9 true?
Charra [1.4K]

Answer:

x<-2

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Help plz which one is it?
Alina [70]
The answer is A

Reasoning;

<span>a1 = first term;
an</span><span>= </span>a<span>n-1</span><span> + </span><span>d
</span>In a sequence you add the first term (a1) with the second term in a repeating linear equation. In this one you need to add 4 to each term so, the equation is;

a1 = 2
an = an -1 + 4

creating:
2 6 10 14 18 22 26... 
3 0
3 years ago
F(x)= x+4; D: {-5, -3, 0, 4}
Vika [28.1K]

Step-by-step explanation:

-5+4=-1

-3+4=1

0+4=4

4+4=8

6 0
3 years ago
​
DedPeter [7]

Answer:

The solution is (3/8, -7/8).

Step-by-step explanation:

y = −5x + 1

y = 3x − 2

Since  the 2 expressions in x are both equal to y :

−5x + 1  = 3x - 2

-5x - 3x = -2 - 1

-8x = -3

x = 3/8.

So y = 3x - 2

= 3(3/8) - 2

= 9/8 - 2 = -7/8.

6 0
3 years ago
Ap calc multiple choice question; attached below<br> please explain how, please!
anygoal [31]

Answer:

C. \frac{f(b)-f(1)}{b-1}=20

General Formulas and Concepts:

<u>Calculus</u>

  • Mean Value Theorem (MVT) - If f is continuous on interval [a, b], then there is a c∈[a, b] such that  f'(c)=\frac{f(b)-f(a)}{b-a}
  • MVT is also Average Value

Step-by-step explanation:

<u>Step 1: Define</u>

f(x)=e^{2x}

f'(c) = 20

Interval [1, b]

<u>Step 2: Check/Identify</u>

Function [1, b] is continuous.

Derivative [1, b] is continuous.

∴ There exists a c∈[1, b] such that f'(c)=\frac{f(b)-f(a)}{b-a}

<u>Step 3: Mean Value Theorem</u>

  1. Substitute:                    20=\frac{ f(b)-f(1)}{b-1}
  2. Rewrite:                        \frac{ f(b)-f(1)}{b-1}=20

And we have our final answer!

5 0
2 years ago
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