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seropon [69]
2 years ago
14

I’m need the solution ASAP

Mathematics
1 answer:
skad [1K]2 years ago
7 0

Answer:

C and D, x = 5

Step-by-step explanation:

To solve this problem, you add 9.2 to both sides then divide both sides by 5.5. The Addition Property of Equality states that if two expressions are equal to each other, and you add the same value to both sides of the equation, the equation will remain equal. This is the first step to solve the problem. The Division Property of Equality states that when we divide both sides of an equation by the same non-zero number, the two sides remain equal. That is the second step to solve the problem. To solve the problem, add 9.2 to both sides, that gets us 5.5x = 27.5. Now, we divide both sides by 5.5. That gets us x = 5.

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Justin has $500 in a savings account at the beginning of the summer. He wants to have at least $300 in the account by the end of
SSSSS [86.1K]

Answer:

10 weeks

Step-by-step explanation:

1 because 20 times 10 is 200 and if he starts with 500 then he can withdraw 20 dollars a week for 10 weeks to have 300 left by the end of the summer, hope this helps :)

3 0
3 years ago
Question 3 of 10 What is the solution to this equation? 4(x-8) + 10 = -10 O A. x= 3 OB. x = 5 O c. x=2 O D. x= 4​
krek1111 [17]

Answer:

O A. x = 3

Step-by-step explanation:

4(x - 8) + 10 = -10

4x - 32 + 10 = -10

4x = 12

x = 3

5 0
2 years ago
Solve x^3-7x^2+7x+15​
ruslelena [56]

Step-by-step explanation:

\underline{\textsf{Given:}}

Given:

\mathsf{Polynomial\;is\;x^3+7x^2+7x-15}Polynomialisx

3

+7x

2

+7x−15

\underline{\textsf{To find:}}

To find:

\mathsf{Factors\;of\;x^3+7x^2+7x-15}Factorsofx

3

+7x

2

+7x−15

\underline{\textsf{Solution:}}

Solution:

\textsf{Factor theorem:}Factor theorem:

\boxed{\mathsf{(x-a)\;is\;a\;factor\;P(x)\;\iff\;P(a)=0}}

(x−a)isafactorP(x)⟺P(a)=0

\mathsf{Let\;P(x)=x^3+7x^2+7x-15}LetP(x)=x

3

+7x

2

+7x−15

\mathsf{Sum\;of\;the\;coefficients=1+7+7-15=0}Sumofthecoefficients=1+7+7−15=0

\therefore\mathsf{(x-1)\;is\;a\;factor\;of\;P(x)}∴(x−1)isafactorofP(x)

\mathsf{When\;x=-3}Whenx=−3

\mathsf{P(-3)=(-3)^3+7(-3)^2+7(-3)-15}P(−3)=(−3)

3

+7(−3)

2

+7(−3)−15

\mathsf{P(-3)=-27+63-21-15}P(−3)=−27+63−21−15

\mathsf{P(-3)=63-63}P(−3)=63−63

\mathsf{P(-3)=0}P(−3)=0

\therefore\mathsf{(x+3)\;is\;a\;factor}∴(x+3)isafactor

\mathsf{When\;x=-5}Whenx=−5

\mathsf{P(-5)=(-5)^3+7(-5)^2+7(-5)-15}P(−5)=(−5)

3

+7(−5)

2

+7(−5)−15

\mathsf{P(-5)=-125+175-35-15}P(−5)=−125+175−35−15

\mathsf{P(-5)=175-175}P(−5)=175−175

\mathsf{P(-5)=0}P(−5)=0

\therefore\mathsf{(x+5)\;is\;a\;factor}∴(x+5)isafactor

\underline{\textsf{Answer:}}

Answer:

\mathsf{x^3+7x^2+7x-15=(x-1)(x+3)(x+5)}x

3

+7x

2

+7x−15=(x−1)(x+3)(x+5)

\underline{\textsf{Find more:}}

Find more:

6 0
3 years ago
Find the distance between the points with cordanites (-26,0 and (95,0)
Paul [167]
26 + 95? Use absolute values; 121.
3 0
3 years ago
Fill in the blanks then, choose the property of addition you used
amid [387]
13,0,15!!.!:&:&:&;8;&;!,$&: s
7 0
2 years ago
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