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3 years ago

How to multiply 150*230

Mathematics

2 answers:

Snowcat [4.5K]3 years ago

5 0

$Solution, 150\cdot \:230=34500\\$

Steps:

$150\cdot \:230$

$\mathrm{Multiply\:the\:numbers:}\:150\cdot \:230=34500, =34500$

The correct answer is 34500

Hope this helps!!!

alexdok [17]3 years ago

3 0

1 5 0

x 2 3 0

Step #1:

Multiply the 0 and 0 to get 0.

1 5 0

x 2 3 0

Step #2:

Multiply the 5 and 0 to get 0.

1 5 0

x 2 3 0

0 0

Step #3:

Multiply the 1 and 0 to get 0.

1 5 0

x 2 3 0

0 0 0Step #4:

Multiply the 0 and 3 to get 0.

1 5 0

x 2 3 0

0 0 0

0 0

Step #5:

Multiply the 5 and 3 to get 15.

Since 15 is greater than 9, you'll have to carry the 1.

1 5 0

x 2 3 0

0 0 0

5 0 0

Step #6:

Multiply the 1 and 3 to get 3.

And, add the carried 1 to get 4.

1 5 0

x 2 3 0

0 0 0

4 5 0 0

Step #7:

Multiply the 0 and 2 to get 0.

1 5 0

x 2 3 0

0 0 0

4 5 0 0

0 0 0

Step #8:

Multiply the 5 and 2 to get 10.

Since 10 is greater than 9, you'll have to carry the 1.

1 5 0

x 2 3 0

0 0 0

4 5 0 0

0 0 0 0

Step #9:

Multiply the 1 and 2 to get 2.

And, add the carried 1 to get 3.

1 5 0

x 2 3 0

0 0 0

4 5 0 0

3 0 0 0 0

All done! Just add the blue numbers together to get your answer:

34500

and that's your answer! sorry no blue i forgot

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Answer:

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Step-by-step explanation:

George-G

1/3G+G=24

4/3G=24

3/4*4/3G=24*3/4

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George will be 18 years old there we can find his brothers age by dividing it by three thus 18/3=6

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3 years ago

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The flag of a country contains an isosceles triangle. (Recall that an isosceles triangle contains two angles with the same mea

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31 degrees, 31 degrees, 118 degrees

Step-by-step explanation:

Step 1 :

Let x be the measure of 2 angles of the given isosceles triangle with same measure

Let y be the measure of 3rd angle

So we have x + x + y = 180

Step 2 :

Given that the measure of 3rd angle of triangle is 25° more than three times the measure of either of the other two angles

So we have , y = 3 x + 25

Step 3:

Substituting for y in the first equation we have,

x + x + 3 x + 25 = 180

=> 5 x + 25 = 180

=> 5 x = 180-25 = 155

=> x = 155/5 = 31

Hence the 2 angles of the triangle are 31 degrees.

Step 4:

we have y = 3 x + 25

=> y = 3 * 31 + 25 = 118

Hence the 3rd angle of given triangle is 118 degrees

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3 years ago

A particle moves according to a law of motion s = f(t), t ? 0, where t is measured in seconds and s in feet.

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Answer:

a) $\frac{ds}{dt}= v(t) = 3t^2 -18t +15$

b) $v(t=3) = 3(3)^2 -18(3) +15=-12$

c) $t =1s, t=5s$

d) $[0,1) \cup (5,\infty)$

e) $D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46$

And we take the absolute value on the middle integral because the distance can't be negative.

f) $a(t) = \frac{dv}{dt}= 6t -18$

g) The particle is speeding up $(1,3) \cup (5,\infty)$

And would be slowing down from $[0,1) \cup (3,5)$

Step-by-step explanation:

For this case we have the following function given:

$f(t) = s = t^3 -9t^2 +15 t$

Part a: Find the velocity at time t.

For this case we just need to take the derivate of the position function respect to t like this:

$\frac{ds}{dt}= v(t) = 3t^2 -18t +15$

Part b: What is the velocity after 3 s?

For this case we just need to replace t=3 s into the velocity equation and we got:

$v(t=3) = 3(3)^2 -18(3) +15=-12$

Part c: When is the particle at rest?

The particle would be at rest when the velocity would be 0 so we need to solve the following equation:

$3t^2 -18 t +15 =0$

We can divide both sides of the equation by 3 and we got:

$t^2 -6t +5=0$

And if we factorize we need to find two numbers that added gives -6 and multiplied 5, so we got:

$(t-5)*(t-1) =0$

And for this case we got $t =1s, t=5s$

Part d: When is the particle moving in the positive direction? (Enter your answer in interval notation.)

For this case the particle is moving in the positive direction when the velocity is higher than 0:

$t^2 -6t +5 >0$

$(t-5) *(t-1)>0$

So then the intervals positive are $[0,1) \cup (5,\infty)$

Part e: Find the total distance traveled during the first 6 s.

We can calculate the total distance with the following integral:

$D= \int_{0}^1 3t^2 -18t +15 dt + |\int_{1}^5 3t^2 -18t +15 dt| +\int_{5}^6 3t^2 -18t +15 dt= t^3 -9t^2 +15 t \Big|_0^1 + t^3 -9t^2 +15 t \Big|_1^5 + t^3 -9t^2 +15 t \Big|_5^6$

And if we replace we got:

$D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46$

And we take the absolute value on the middle integral because the distance can't be negative.

Part f: Find the acceleration at time t.

For this case we ust need to take the derivate of the velocity respect to the time like this:

$a(t) = \frac{dv}{dt}= 6t -18$

Part g and h

The particle is speeding up $(1,3) \cup (5,\infty)$

And would be slowing down from $[0,1) \cup (3,5)$

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Fittoniya [83]

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