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masya89 [10]
2 years ago
13

Write an equation for : one third of a number plus 5 is 8

Mathematics
1 answer:
solong [7]2 years ago
3 0

<em>Hey!</em><em>!</em>

<em>Ques</em><em>tion</em><em>:</em><em> </em><em>one</em><em> </em><em>third</em><em> </em><em>of</em><em> </em><em>a</em><em> </em><em>number</em><em> </em><em>plus</em><em> </em><em>5</em><em> </em><em>is</em><em> </em><em>8</em>

<em>Let</em><em> </em><em>the </em><em>number</em><em> </em><em>be</em><em> </em><em>X</em>

<em>Answer</em><em>:</em><em> </em><em>1</em><em>/</em><em>3</em><em>*</em><em>X+</em><em>5</em><em>=</em><em>8</em>

<em>hope</em><em> </em><em>it</em><em> </em><em>helps</em>

<em>Good</em><em> </em><em>luck</em><em> on</em><em> your</em><em> assignment</em>

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6cm, 15cm, 6cm does it make a triangle
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Answer:

Yes

Step-by-step explanation:

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3 years ago
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Two number cubes each have sides labeled 1 to 6. Ann rolls both number cubes. On the first roll, the sum of the numbers was equa
ad-work [718]

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yes.

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because i know

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Chamblee High School is selling Valentine's Day gifts as a fundraising event. One long stemmed rose costs $3.00 while one long s
Mumz [18]

Answer:  Choice B) 60 roses and 10 carnations

============================================================

Explanation:

  • r = number of roses
  • c = number of carnations

r and c are positive whole numbers.

r+c = total number of flowers = 50, since 50 orders are made.

The first equation to set up is r+c = 50.

This equation can be solved to get r = 50-c.

------------------

3r = cost of all the roses only, in dollars

1.5c = cost of all the carnations only, in dollars

3r+1.5c = total cost of all the flowers = 195 dollars

3r+1.5c = 195

------------------

Let's apply substitution to solve

3r+1.5c = 195

3(50-c)+1.5c = 195

150-3c+1.5c = 195

-1.5c+150 = 195

-1.5c = 195-150

-1.5c = 45

c = 45/(-1.5)

c = -30

That's not good. We shouldn't get a negative value.

It turns out that the condition r+c = 50 should be ignored. Notice how none of the answer choices listed have r+c leading to 50.

So we'll only focus on the equation 3r+1.5c = 195

-----------------

If we plugged in r = 100 and c = 100, then we get

3r+1.5c = 195

3(100)+1.5(100) = 195

300+150 = 195

450 = 195

Which is false. So we can rule out choice A

Let's repeat those steps for choice B

3r+1.5c = 195

3(60)+1.5(10) = 195

180 + 15 = 195

195 = 195

So that works out. I have a feeling your teacher meant to say "70 orders" instead of "50 orders". If so, then the equation r+c = 50 would be r+c = 70 and everything would lead to choice B as the final answer.

Choices C and D are similar to that of choice A, so they can be ruled out.

8 0
2 years ago
How you solve this by using solving proportions
DerKrebs [107]

3/(x + 4) = (x + 2)/5

Multiply both sides by 5 :

5 × 3/(x + 4) = 5 × (x + 2)/5

15/(x + 4) = x + 2

Multiply both sides by x + 4 :

(x + 4) × 15/(x + 4) = (x + 4) × (x + 2)

Provided that x + 4 ≠ 0, we can cancel the factors of x + 4 on the left side:

15 = (x + 4) (x + 2)

Expand the product on the right side:

15 = x² + 6x + 8

Move everything to one side:

0 = x² + 6x - 7

We can factorize easily in this case
0 = (x + 7) (x - 1)

Then

x + 7 = 0   or   x - 1 = 0

⇒   x = -7   or   x = 1

7 0
2 years ago
Solve for X please help
QveST [7]

Answer:

<em>-</em><em>1</em>

Step-by-step explanation:

<em>x</em><em> </em><em>is</em><em> </em><em>an</em><em> </em><em>unknown</em><em> </em><em>and</em><em> </em><em>can</em><em> </em><em>be</em><em> </em><em>known</em><em> </em><em>as</em><em> </em><em><u>1</u></em><em><u> </u></em><em>or</em><em> </em><em><u>-</u></em><em><u>1</u></em><em><u> </u></em><em>based</em><em> </em><em>on</em><em> </em><em>the</em><em> </em><em>circumstance</em><em>.</em><em> </em>

<em>Therefore</em><em>,</em><em> </em><em>in</em><em> </em><em>this</em><em> </em><em>case</em><em>,</em><em> </em><em>x</em><em> </em><em>is</em><em> </em><em>known</em><em> </em><em>as</em><em> </em><em><u>-</u></em><em><u>1</u></em>

3 0
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