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USPshnik [31]
3 years ago
10

Solve the following question algebraically, show your work 13 + w/7 <-(fraction) = -18

Mathematics
1 answer:
evablogger [386]3 years ago
8 0

Solve for w: by simplifying both sides of the equation, then isolating the variable.

w=-217

Work: 1.  Subtract 13 from both sides (w/7 = -18 - 13), 2.Simplify -18-13 to -31 (w/​7=-31), 3.Multiply both sides by 7 (w = -31 * 7), 4. Simplify 31*7 to 217 (w=-217)

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A car accelerates at 3.00 m/s2. The car has a mass of 1354kg. What net force will produce the acceleration?
otez555 [7]

Answer:

4062N

Step-by-step explanation:

Using Newton’s Second Law;

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Hope this helps! :D

7 0
2 years ago
0.357 round to 2 decimal places
Andrej [43]
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7 0
3 years ago
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5a - 9 + a = 9
Ne4ueva [31]

Answer:

a = 3

Step-by-step explanation:

1) Simplify 5a - 9 + a to 6a - 9.

6a - 9 = 9

2) Add 9 to both sides.

6a = 9 + 9

3) Simplify 9 + 9 to 18.

6a = 18

4) Divide both sides by 6.

a =  \frac{18}{6}

5) Simplify 18/6 to 3.

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Therefor, the answer is, a = 3.

7 0
3 years ago
The average value of the function y = x² – 1 on [0, 12] is
shusha [124]

Answer:

47.

Step-by-step explanation:

1) the rule:

f_{avr}=\frac{1}{b-a} \int\limits^b_a {f(x)} \, dx ,

where a;b are 0 and 12, f(x)=x²-1.

2) according to the rule above:

f_{avr}=\frac{1}{12-0} \int\limits^{12}_0 {(x^2-1)} \, dx=\frac{1}{12}(\frac{x^3}{3}-x)|^{12}_0=48-1=47.

3 0
2 years ago
A bicycle lock has a four-digit code. The possible digits, 0 through 9, cannot be repeated. What is the probability that the loc
kozerog [31]
First let's find the number of the elements in the sample space, that is the total number of codes that can be produced.

The first digit is any of {0, 1, 2...,9}, that is 10 possibilities
the second digit is any of the remaining 9, after having picked one. 
and so on...

so in total there are 10*9*8*7 = 5040 codes.

a. What is the probability that the lock code will begin with 5?

Lets fix the first number as 5. Then there are 9 possibilities for the second digit, 8 for the third on and 7 for the last digit.

Thus, there are 1*9*8*7=504 codes which start with 5.

so 

P(first digit is five)=\frac{n(first -digit- is- 5)}{n(all-codes)}= \frac{1*9*8*7 }{10*9*8*7 }= \frac{1}{10}=0.1

b. <span>What is the probability that the lock code will not contain the number 0? 

from the set {0, 1, 2...., } we exclude 0, and we are left with {1, 2, ...9}

from which we can form in total 9*8*7*6 codes which do not contain 0.

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Answer:

0.1 ; 0.6</span>
4 0
3 years ago
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