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tatuchka [14]
3 years ago
11

Solve for X Question: X/2 +x/3 =5

Mathematics
1 answer:
Sever21 [200]3 years ago
7 0

Answer:

(3x + 2x) \div 6 = 5 \\ 5x = 5 \times 6 \\ x = 30 \div 5 \\ x = 6

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<img src="https://tex.z-dn.net/?f=x%5E%7B2%7D%20-10x%2B5-3%28x-5%29" id="TexFormula1" title="x^{2} -10x+5-3(x-5)" alt="x^{2} -10
stellarik [79]

Answer:

x

2

−

13

x

+

20

x

2

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13

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x

2

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13

x

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20

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8 0
3 years ago
Putting recipe makes 4 1/2 cups of pudding. How many 1/3 cup servings does this equal?
mel-nik [20]

Answer:

C

Step-by-step explanation:

First, write 4 ½ in improper form.  2 × 4 + 1 = 9. So the fraction is 9/2.

Now divide:

9/2 ÷ 1/3

To divide by a fraction, multiply by the reciprocal:

9/2 × 3/1

27/2

13 1/2

Answer C.

4 0
4 years ago
Read 2 more answers
What is larger,10L or 1,000mL  
Vlada [557]
1\ L=1,000\ mL\ \ \ \Roghtarrow\ \ \ 10\ L\ >\ 1,000\ mL\\\\Ans.\ 10\ L\ is\ larger
5 0
3 years ago
The perimeter of a quadrilateral​ (four-sided polygon) is 41 inches. The longest side is twice as long as the shortest side. The
zhenek [66]

Answer:

The length of all four sides of a quadrilateral are 7 inches, 14 inches, 10 inches, 10 inches

Step-by-step explanation:

Let the shortest side be x

As per given information longest side = 2x

and other 2 sides are equal in length but 3 more times that shortest side = 3+x

Perimeter of Quadrilateral = 41 inches given

But perimeter of quadrilateral = sum of all sides

x+2x+x+3+x+3=41\\5x+6 =41\\5x= 41-6\\5x=35\\x = \frac{35}{5}=7

Shortest side = 7\ in

Longest side = 2x= 2\times 7= 14 \ in

Other 2 sides = 3+x=3+7 = 10 \ in

6 0
4 years ago
Harry drove 2 hours on the freeway then decreased his speed by 20mph and drove 6 more hours. total trip was 352 what was his spe
poizon [28]
Recall your d = rt, distance = rate * time

we know the total trip was 352miles, so, let's say he was going for the first 2hrs at "r" speed, he went "d" miles, then he slowed down to "r - 20" and went 6 more hours, since the total trip is 352 miles, the 6hrs take the slack from 352 - d, and that's how many miles he covered on those 6hours

\bf \begin{array}{lccclll}&#10;&distance&rate&time\\&#10;&-----&-----&-----\\&#10;\textit{before slowing}&d&r&2\\&#10;\textit{after slowing}&352-d&r-20&6&#10;\end{array}&#10;\\\\\\&#10;&#10;\begin{cases}&#10;\boxed{d}=2r\\&#10;352-d=(r-20)6\\&#10;----------\\&#10;352-\boxed{2r}=(r-20)6&#10;\end{cases}

solve for "r".
7 0
4 years ago
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