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Bas_tet [7]
3 years ago
5

What is 10 times as much as 130

Mathematics
1 answer:
alexandr1967 [171]3 years ago
7 0
10 times as much as 130 is 1300
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Mary runs 4 laps in 8 mins. Nicole runs 12 laps in 18 mins. Who runs faster
Georgia [21]
Find to LCM of 8 and 18 to compare times

Mary: 4*9laps/8*9min
36 laps/ 72 mins
Nicole: 12*4laps/18*4 mins
48 laps/ 72 mins

Nicole runs more laps in the same amount of time as Mary

Therefore, Nicole runs faster
6 0
3 years ago
Simplify. -5a-3b+8a+b
Misha Larkins [42]
The answer is
3a - 2b
6 0
3 years ago
Prove that (x-2)is factor of p (x)=2x³-3x²-17x+30​
Natasha_Volkova [10]

Answer:

P(2)=0 then x-2 is a factor of P(x)

Step-by-step explanation:

To see if (x-2) is a factor just plug in 2 for x into the polynomial expression.

2(2)^3-3(2)^2-17(2)+30

2(8)-3(4)-34+30

16-12-34+30

4-34+30

-30+30

0

Since P(2)=0 then x-2 is a factor

8 0
3 years ago
Answer some of the questions, please
zloy xaker [14]

Answer:

1)= a 30

2)= b 2¹⁰

3)= d 0.75²

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4 0
3 years ago
Read 2 more answers
Arrange the equations in the correct sequence to rewrite the formula for displacement, , to find a. In the formula, d is displac
OverLord2011 [107]

Answer:

2(d-vt)=-at^2

a=2(d-vt)/t^2

at^2=2(d-vt)

Step-by-step explanation:

Arrange the equations in the correct sequence to rewrite the formula for displacement, d = vt—1/2at^2 to find a. In the formula, d is

displacement, v is final velocity, a is acceleration, and t is time.

Given the formula for calculating the displacement of a body as shown below;

d=vt - 1/2at^2

Where,

d = displacement

v = final velocity

a = acceleration

t = time

To make acceleration(a), the subject of the formula

Subtract vt from both sides of the equation

d=vt - 1/2at^2

d - vt=vt - vt - 1/2at^2

d - vt= -1/2at^2

2(d - vt) = -at^2

Divide both sides by t^2

2(d - vt) / t^2 = -at^2 / t^2

2(d - vt) / t^2 = -a

a= -2(d - vt) / t^2

a=2(vt - d) / t^2

2(vt-d)=at^2

4 0
3 years ago
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