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irina1246 [14]
3 years ago
5

Solve for x: 5(x-2)+7=65-4(5x-8) SHOW STEPS PLZZZZZZ HELP

Mathematics
2 answers:
miskamm [114]3 years ago
6 0

5(x-2)+7=65-4(5x-8)

We move all terms to the left:

5(x-2)+7-(65-4(5x-8))=0

We multiply parentheses

5x-(65-4(5x-8))-10+7=0

We calculate terms in parentheses: -(65-4(5x-8)), so:

65-4(5x-8)

determiningTheFunctionDomain

-4(5x-8)+65

We multiply parentheses

-20x+32+65

We add all the numbers together, and all the variables

-20x+97

Back to the equation:

-(-20x+97)

We add all the numbers together, and all the variables

5x-(-20x+97)-3=0

We get rid of parentheses

5x+20x-97-3=0

We add all the numbers together, and all the variables

25x-100=0

We move all terms containing x to the left, all other terms to the right

25x=100

x=100/25

x=4

NemiM [27]3 years ago
4 0

Answer:

x=4

Step-by-step explanation:

1) Expand each side: 5x-10+7=65-20x+32

2) Move x to one side: 25x-10+7=65+32

3) Move everything else to other side: 25x= 65+32+10-7

4) Simplify: 25x=100

5) Solve for x by dividing both sides by 25: x=4

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4 years ago
In the figure, AB║CD and ∠EIA is congruent to ∠GJB Complete the following statements to prove that ∠IKL is congruent to ∠DLH.
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Here it is given that AB || CD

< EIA = <GJB

Now

∠EIA ≅ ∠IKC and ∠GJB is ≅ ∠ JLD (Corresponding angles)

∠EIA  ≅ ∠GJB then ∠IKC ≅ ∠ JLD (Substitution Property of Congruency)

∠IKL + ∠IKC 180° and ∠DLH +  ∠JLD =180° (Linear Pair Theorem)

So

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But ∠IKC  ≅ ∠JLD

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So we have

m∠IKL + m∠JLD = 180°

∠IKL and ∠JLD are supplementary angles.

But ∠DLH and ∠JLD are supplementary angles.

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

We are told that the temperature may vary from 700 degrees Celsius to 1200 degrees Celsius.

And that this temperature is x.

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Let's find the average of the two temperature limits given:

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