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3 years ago

Which of the following situations would combine to make zero?

Mathematics

1 answer:

ZanzabumX [31]3 years ago

8 0

The answer is D she started out with 250$ and then subtracted 250$
250-250=0

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A block with a mass of 10 kg is being pushed. its acceleration is 5 m/s.s . Calculate the force acting on the block ? plz answer

Hitman42 [59]

Answer:

<h3>The answer is 50 N</h3>

Step-by-step explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

<h3>force = mass × acceleration</h3>

From the question

mass = 10 kg

acceleration = 5 m/s²

We have

force = 10 × 5

We have the final answer as

Hope this helps you

5 0

2 years ago

I hope it’s clear enough but can some please help

Liula [17]

Answer:

Step-by-step explanation:

hope this helps you! have a good day!

6 0

3 years ago

3. The equation of a circle is x^2+y^2-4x+2y-11=0 . What are the center and the radius of the circle? Show your work.

Charra [1.4K]

Rewrite in standard form to find the center (h,k) and redius r.
center: (2,-1)
Radius:4
have a great day!!!

7 0

3 years ago

Can somebody prove this mathmatical induction?

Flauer [41]

Answer:

See explanation

Step-by-step explanation:

1 step:

n=1, then

$\sum \limits_{j=1}^1 2^j=2^1=2\\ \\2(2^1-1)=2(2-1)=2\cdot 1=2$

So, for j=1 this statement is true

2 step:

Assume that for n=k the following statement is true

$\sum \limits_{j=1}^k2^j=2(2^k-1)$

3 step:

Check for n=k+1 whether the statement

$\sum \limits_{j=1}^{k+1}2^j=2(2^{k+1}-1)$

is true.

Start with the left side:

$\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}\ \ (\ast)$

According to the 2nd step,

$\sum \limits_{j=1}^k2^j=2(2^k-1)$

Substitute it into the $\ast$

$\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}=2(2^k-1)+2^{k+1}=2^{k+1}-2+2^{k+1}=2\cdot 2^{k+1}-2=2^{k+2}-2=2(2^{k+1}-1)$

So, you have proved the initial statement

4 0

3 years ago

Find the product for 56.44

miv72 [106K]

Step-by-step explanation:

55.44=2464

mark as brainliest

8 0

2 years ago