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3 years ago

Help with Algebra Questions

1 answer:

7 0

Here is a quadratic equation x^2 + 5x = 14
x^2 + 5x - 14 = 0 - Subtract 14 from both sides
(x + 7)(x - 2) = 0 - Factor
x + 7 = 0 x - 2 = 0 - Set each factor equal to 0
x = -7 x = 2 - solve each piece

If the constant is easy to find factors of then use the factoring method
If the constant is difficult and you have a graphing calculator, then graph
If no graphing calculator then use the quadratic formula

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Lin solved the equation 8 (x - 3) +7= 2x (4 - 17) incorrectly.

svet-max [94.6K]

Answer:

Her final answer should have been x = 1/2.

Step-by-step explanation:

Her 2 errors were that she didn't distribute the 2x to both 4 and -17 and that she made the 13 positive instead of negative when she added 4 and -17 together.

Correct solution:

8(x-3)+7=2x(4-17)

8x-24+7=8x-34x

-17=-34x

The negatives on both sides cancel each other out

x = 1/2

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3 years ago

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Simplify each expression. (7x+x^4+x^3)-(6x-2x^3+8x^4)

aleksandrvk [35]

the answer is x + 3x^3 - 7x^4

8 0

2 years ago

Help pleasee<br> On khan academy under geometry unit test.

Rzqust [24]

X=147
Angle x and 33 are a linear pair meaning they equal 180 and 180-33=147

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2 years ago

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When creating a histogram, the first recorded scale is 0 to 10. What should the second scale indicate?

klio [65]

Answer:

A) 10-20

Step-by-step explanation:

Its A because it started with 0 and it went to 10 so it adds 10 each time cause it gave us the hint with 0 to 10,

I hope this helps

6 0

2 years ago

Q: using the Substitution Method, and the value ory in

GenaCL600 [577]

Answer:

Step-by-step explanation:

To properly apply the substitution method, it will be better for us to rearrange the system of equations to have similar variables on the same side

$44 - 4 = -2y ----equation 1$

$10 = 7x + 4y --- equation 2$

We can simply evaluate equation 1 to have

$40 = -2y\\y = 40/-2 = -20\\$

y = -20

From the first equation alone, we can evaluate the value of y as -20. This is because only one unknown is present in equation one, hence a single equation is sufficient enough to evaluate it. If to unknowns were present, the two equations would have been utilized to evaluate the solution.

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2 years ago