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3 years ago

I’m sorry, I suck at math, can someone please help me with this?

Mathematics

1 answer:

scZoUnD [109]3 years ago

3 0

Answer:

w¹×w¹×w¹=w¹+¹+¹= w³

formula is a^m×a^n=a^m+n

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Question 8

Alenkasestr [34]

Answer:

The answer is A

Step-by-step explanation:

Using PEMDAS you answer parentheses first so

(8-5)2-(2+4) would then be

3*2-6 then you would multiply

6-6 and find the answer of

6 0

3 years ago

What is the median of this set of data?<br><br> 362, 187, 211, 298, 331, 250, 347, 199

Studentka2010 [4]

Answer:

314.5

Step-by-step explanation:

in the group data will take the middle team

298and 331

=298+331÷2

=629÷2

=314.5

that was the median 314.5

5 0

3 years ago

Read 2 more answers

What is the area of this figure? 4 cm 5cm 5cm Enter your answer in the box. cm?

Kruka [31]

What’s the shape of it?
If it’s a rectangle 20
Bot something with more than 4 sides 100cm

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3 years ago

A particle moves on a straight line and has acceleration a(t)=24t+2. Its position at time t=0 is s(0)=3 and its velocity at time

user100 [1]

Answer:

It's position at time t = 5 is 593.

Step-by-step explanation:

The velocity v(t) is the integral of the acceleration a(t)

The position s(t) is the integral of the velocity v(t)

We have that:

The acceleration is:

$a(t) = 24t + 2$

Velocity:

$v(t) = \int {a(t)} \, dt = \int {24t + 2} \, dt = 12t^{2} + 2t + K$

K is the initial velocity, that is v(0). Since V(0) = 13, K = 13

Then

$v(t) = 12t^{2} + 2t + 13$

Position:

$s(t) = \int {s(t)} \, dt = \int {12t^{2} + 2t + 13} \, dt = 4t^{3} + t^{2} + 13t + K$

Since s(0) = 3

$s(t) = 4t^{3} + t^{2} + 13t + 3$

What is its position at time t=5?

This is s(5).

$s(t) = 4t^{3} + t^{2} + 13t + 3$

$s(5) = 4*5^{3} + 5^{2} + 13*5 + 3$

$s(5) = 593$

It's position at time t = 5 is 593.

3 0

3 years ago

Help pls hurry this test is timed what’s 1+1

larisa [96]

Answer:

Step-by-step explanation:

ayyy66y6yyy6yyyiiiiiiii

6 0

3 years ago