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2 years ago

If angle 2 and angle 3 are supplementary, what is the reason

Mathematics

1 answer:

kondaur [170]2 years ago

4 0

They add up to 180 degrees

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Please help u have to factor

enot [183]

Both terms have a $-m^2n$ in common, so we can pull it out. The factored form will be $-m^2n$ times whatever is left after pulling it out.

In the first term, we will have just 8 left since everything else in the term is factored out.

The second term has a $7m^4n^3$ left. Notice that both of these terms no longer have a negative sign since the negative was factored out.

The answer is $(-m^2n)(8+7m^4n^3)$

8 0

2 years ago

Fastest answer will be declared the brainliest

Hatshy [7]

Answer:

All whole numbers are rational numbers.

False

All integers are whole numbers.

True

There are integers that are not rational numbers.

True

There are whole numbers that are not integers.

True.

7 0

2 years ago

-(7p+6)-2(-1-2p) what is that simplified?

charle [14.2K]

Answer:

-3p-40

Step-by-step explanation:

-(7p+6)-2(-1-2p)

-7p-42+2+4p

-3p-40

4 0

3 years ago

2(x + 4) − 1 = 2x + 7

Margarita [4]

Answer:

All real numbers are solutions to this problem.

Step-by-step explanation:

I) Simplify:

2(x + 4) - 1 = 2x + 7

x + 4 - 0.5 = x + 3.5 (Divide by 2)

x + 3.5 = x + 3.5 (Simplify x + 4 - 0.5)

x = x

II) Conclusion:

Any real number will work as a solution.

7 0

2 years ago

Read 2 more answers

In a shipment of 56 vials, only 13 do not have hairline cracks. If you randomly select 3 vials from the shipment, in how many wa

Elden [556K]

Answer: 27434

Step-by-step explanation:

Given : Total number of vials = 56

Number of vials that do not have hairline cracks = 13

Then, Number of vials that have hairline cracks =56-13=43

Since , order of selection is not mattering here , so we combinations to find the number of ways.

The number of combinations of m thing r things at a time is given by :-

$^nC_r=\dfrac{n!}{r!(n-r)!}$

Now, the number of ways to select at least one out of 3 vials have a hairline crack will be :-

$^{13}C_2\cdot ^{43}C_{1}+^{13}C_{1}\cdot ^{43}C_{2}+^{13}C_0\cdot ^{43}C_{3}\\\\=\dfrac{13!}{2!(13-2)!}\cdot\dfrac{43!}{1!(42)!}+\dfrac{13!}{1!(12)!}\cdot\dfrac{43!}{2!(41)!}+\dfrac{13!}{0!(13)!}\cdot\dfrac{43!}{3!(40)!}\\\\=\dfrac{13\times12\times11!}{2\times11!}\cdot (43)+(13)\cdot\dfrac{43\times42\times41!}{2\times41!}+(1)\dfrac{43\times42\times41\times40!}{6\times40!}\\\\=3354+11739+12341=27434$

Hence, the required number of ways =27434

5 0

2 years ago