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vredina [299]
3 years ago
13

Find the perimeter of a rectangle with length (x-5) and width (3x+1).

Mathematics
2 answers:
Aleks [24]3 years ago
8 0

Answer:

8x-8

Step-by-step explanation:

P=s+s+s+s or in this case P= 2l+2w

P= 2(x-5)+2(3x+1)

P=(2x-10)+(6x+2)

P=8x-8

Phantasy [73]3 years ago
6 0

Answer:

8x -8 hope this helped

Step-by-step explanation:

3x+1 + 3x + 1= 6x+2

x-5 + x-5 = 2x - 10

6x+2 +2x-10= 8x-8

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15 3/8+756-<br> What is the answer
evablogger [386]

Answer: 771 3/8

Step-by-step explanation:

15 3/8+756

we can apply the assosiative propery and do 15+756+3/8

so 15+756= 771

so finally you do 771+3/8=771 3/8

8 0
3 years ago
Use differentials to find an approximate value for <img src="https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B65%7D%20" id="TexFormul
Sveta_85 [38]
: Let y = f(x) =  x^1/3  
Then dy = 1/3*x^(−2/3) dx 
 Since f(64) = 4.
We take x = 64 and dx = ∆x = 1  
This gives dy = 1/3*(64)^(−2/3)* (1) =  1/48  
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6 0
4 years ago
Read 2 more answers
Solve for a. 7a + 18 = 102
Oliga [24]

Answer:

a=12

Step-by-step explanation:

7a+18=102

7a=102-18

7a=84

a=84/7

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6 0
3 years ago
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Use the distributive property to solve the equation 28 - (3x + 4) = 2(x + 6) + 5
Shalnov [3]
1.) 28-3x-4= 2x+12+x
2.) 24-3x= 3x+12
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7 0
3 years ago
Atul has 2/3 lb of candy. Jose has 3/5 lb and Maria has 1/2 lb less then José. How many more pounds of candy does Atul have than
liraira [26]

Given:

Atul has \dfrac{2}{3} lb of candy.

Jose has \dfrac{3}{5} lb of candy.

Maria has \dfrac{1}{2} lb less than Jose.

To find:

How many more pounds of candy does Atul have than Maria?

Solution:

Since, Maria has \dfrac{1}{2} lb less than Jose, therefore

Maria has = \dfrac{3}{5}-\dfrac{1}{2} lb

Maria has = \dfrac{6-5}{10} lb

Maria has = \dfrac{1}{10} lb

Difference between the candies Atul and Maria have.

Difference = \dfrac{2}{3}-\dfrac{1}{10} lb

Difference = \dfrac{20-3}{30} lb

Difference = \dfrac{17}{30} lb

Therefore,  Atul have \dfrac{17}{30} lb of candy more than Maria.

6 0
3 years ago
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