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2 years ago

Solve for x please help !!

Mathematics

2 answers:

kati45 [8]2 years ago

8 0

Answer:

Step-by-step explanation:

serg [7]2 years ago

5 0

Answer:

it would be 31

Step-by-step explanation:

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Please tell me what 1+1 is ??????

Luden [163]

It is 2...hope it helps!

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2 years ago

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Find the common difference of the arithmetic sequence 13,10,7

Hunter-Best [27]

Answer:

-3 I think

Step-by-step explanation:

well 10-7=3 and 13-10=3

but since it is greater number first it would make it a -3

4 0

2 years ago

Average rate of change help!!

frez [133]

Given that the function is $f(x)=x^{3}-6 x^{2}+4 x+7$

We need to determine the average rate of change over the interval $0 \leq x \leq 5$

Value of f(x) when x = 0:

Substituting x = 0 in the function $f(x)=x^{3}-6 x^{2}+4 x+7$ , we have;

$f(0)=(0)^{3}-6 (0)^{2}+4 (0)+7$

$f(0)=0-0+0+7$

$f(0)=7$

Thus, the value of f(0) is 7.

Value of f(x) when x = 5:

Substituting x = 5 in the function $f(x)=x^{3}-6 x^{2}+4 x+7$ , we have;

$f(5)=(5)^{3}-6 (5)^{2}+4 (5)+7$

$f(5)=125-150+20+7$

$f(5)=2$

Thus, the value of f(5) is 2.

Average rate of change:

The average rate of change can be determined using the formula,

$Rate \ of \ change=\frac{f(b)-f(a)}{b-a}$

where $a=0$ and $b=5$

Thus, we have;

$Rate \ of \ change=\frac{f(5)-f(0)}{5-0}$

$Rate \ of \ change=\frac{2-7}{5-0}$

$Rate \ of \ change=\frac{-5}{5}$

$Rate \ of \ change=-1$

Thus, the average rate of change over the interval $0 \leq x \leq 5$ is -1.

3 0

2 years ago

Y = 2x - 8 3x - 2y = 13

SSSSS [86.1K]

Answer:

{x,y} = {3,-2}

Step-by-step explanation:

// Solve equation [1] for the variable y

[1] y = 2x - 8

// Plug this in for variable y in equation [2]

[2] 3x - 2•(2x-8) = 13

[2] -x = -3

// Solve equation [2] for the variable x

[2] x = 3

// By now we know this much :

x = 3

y = 2x-8

// Use the x value to solve for y

y = 2(3)-8 = -2

Solution :

{x,y} = {3,-2}

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2 years ago

My son is having such a hard time with 6th grade math. and I'm no help at all. someone help please

lisabon 2012 [21]

I'd be more than happy to help. Im a sophmore and Love math

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