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likoan [24]
3 years ago
5

Solve this equation ×(2)-8×+7=0 Solve the equation

Mathematics
2 answers:
yKpoI14uk [10]3 years ago
7 0

Answer:

-7/6

Step-by-step explanation:

2x-8x+7=0

combine like terms

6x+7=0

subtract the 7 from both sides

6x=-7

divide 6 from both sides

-7/6 or -1 and 2/3

aev [14]3 years ago
5 0

Answer:

-7/6

Step-by-step explanation:

2x-8x+7=0

combine like terms

6x+7=0

subtract the 7 from both sides

6x=-7

divide 6 from both sides

-7/6 or -1 and 2/3

plz mark me as brainliest :)

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while shopping for clothes, Tracey spent 38$ less than 3 times what Daniel spent. write and solve an equation to find how much D
alexandr1967 [171]

\bold{\huge{\orange{\underline{Solution}}}}

\bold{\underline{ Given :- }}

  • <u>While </u><u>shopping </u><u>for </u><u>clothes </u><u>Tracey </u><u>spent </u><u>3</u><u>8</u><u>$</u><u> </u><u>less </u><u>than </u><u>3</u><u> </u><u>times </u><u>of </u><u>what </u><u>Daniel </u><u>spent </u>

\bold{\underline{ To\: Find :- }}

  • <u>We </u><u>have </u><u>to </u><u>determine </u><u>the </u><u>total </u><u>cost </u><u>spent </u><u>by </u><u>daniel</u>

\bold{\underline{ Let's \: Begin :- }}

Cost spent by Tracey for her clothes = 38$

Let assume the spending by Daniel is x

\bold{\underline{Therefore ,}}

\sf{ x / 3 = 38 }

\sf{ x  = 38 × 3 }

\sf{ x = 114 dollar }

\sf{\red{ Hence,\: Daniel\: spent\: 114 \: dollars\: on \: His\: clothes }}

7 0
2 years ago
Select all the expressions that are equivalent to (2)^n+³
eimsori [14]

Answer:

The expressions which equivalent to  (2)^{n+3} are:

4(2)^{n+1}  ⇒ B

8(2)^{n} ⇒ C

Step-by-step explanation:

Let us revise some rules of exponent

  • a^{m} × a^{m}  = a^{m+n}
  • (a^{m})^{n} = a^{m*n}

Now let us find the equivalent expressions of  (2)^{n+3}

A.

∵ 4 = 2 × 2

∴ 4 =  2^{2}

∴  (4)^{n+2} =  (2^{2})^{n+2}

- By using the second rule above multiply 2 and (n + 2)

∵ 2(n + 2) = 2n + 4

∴  (4)^{n+2} =  (2)^{2n+4}  

B.

∵ 4 = 2 × 2

∴ 4 =  2²

∴  4(2)^{n+1} = 2² ×  (2)^{n+1}

- By using the first rule rule add the exponents of 2

∵ 2 + n + 1 = n + 3

∴   4(2)^{n+1} =  (2)^{n+3}

C.

∵ 8 = 2 × 2 × 2

∴ 8 =  2³

∴  8(2)^{n} = 2³ ×  (2)^{n}

- By using the first rule rule add the exponents of 2

∵ 3 + n = n + 3

∴  8(2)^{n} =  (2)^{n+3}

D.

∵ 16 = 2 × 2 × 2 × 2

∴ 16 = 2^{4}

∴  16(2)^{n} = 2^{4}  ×  (2)^{n}

- By using the first rule rule add the exponents of 2

∵ 4 + n = n + 4

∴  16(2)^{n} =  (2)^{n+4}

E.

(2)^{2n+3} is in its simplest form

The expressions which equivalent to  (2)^{n+3} are:

4(2)^{n+1}  ⇒ B

8(2)^{n} ⇒ C

3 0
3 years ago
Which statement is true about the polynomial 3x2y2 − 5xy2 − 3x2y2 + 2x2 after it has been fully simplified?
lorasvet [3.4K]

Answer: Varies

Step-by-step explanation:

There would still be some Xs and Ys.

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What is the estimate of 979
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I'm pretty sure that it would be 1,000...
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Circle = 360°

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