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3 years ago

What is Y=2x+5 solve for x

Mathematics

1 answer:

Stells [14]3 years ago

4 0

Answer:

X=-5/2

Step-by-step explanation:

First for Y=2x+5 you change the y into 0

Then you replace the 0 with 2x but change the sign so it will be -2x so it will look like -2x=5

then you divide both sides by 2

Finally you get -5/2

Hopefully this helps

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What is 20-6n=-2(n+6)?

blsea [12.9K]

20n - 6n = -2(n + 6)

20 - 6n =-2n -12

6n + 2 n = -20 -12

8n = 32

n = 32 : 8

n = 8

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2 years ago

What is the GCF of 96x5 and 64x2? 32x 32x2 32x3 32x5

Lapatulllka [165]

The answer to your question would be 32x^2

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3 years ago

Read 2 more answers

How to differentiate y=x^n using the first principle. In this question, I cannot use the rule of differentiation. I have to do t

Zarrin [17]

By first principles, the derivative is

$\displaystyle\lim_{h\to0}\frac{(x+h)^n-x^n}h$

Use the binomial theorem to expand the numerator:

$(x+h)^n=\displaystyle\sum_{i=0}^n\binom nix^{n-i}h^i=\binom n0x^n+\binom n1x^{n-1}h+\cdots+\binom nnh^n$

$(x+h)^n=x^n+nx^{n-1}h+\dfrac{n(n-1)}2x^{n-2}h^2+\cdots+nxh^{n-1}+h^n$

where

$\dbinom nk=\dfrac{n!}{k!(n-k)!}$

The first term is eliminated, and the limit is

$\displaystyle\lim_{h\to0}\frac{nx^{n-1}h+\dfrac{n(n-1)}2x^{n-2}h^2+\cdots+nxh^{n-1}+h^n}h$

A power of $h$ in every term of the numerator cancels with $h$ in the denominator:

$\displaystyle\lim_{h\to0}\left(nx^{n-1}+\dfrac{n(n-1)}2x^{n-2}h+\cdots+nxh^{n-2}+h^{n-1}\right)$

Finally, each term containing $h$ approaches 0 as $h\to0$ , and the derivative is

$y=x^n\implies y'=nx^{n-1}$

4 0

3 years ago

6. Sammi spun a spinner with 5 equal parts labeled 1-5. If she spins the spinner 20 times, how many times should she expect to s

Tasya [4]

Answer:

Step-by-step explanation:

There is a 1/5 chance she spins a 1. (1/5 = 20% = 0.2)

She spins 20 times:

20 * 0.2 = 4

So she should expect getting 4 twos.

If there was anything that you didn't understand, just comment your question below this answer!

Have a good day! (●'◡'●)

6 0

2 years ago

An arithmetic sequence with a third term of 8 and a constant difference of 5.

larisa [96]

An arithmetic sequence (a_n) is as follows:

$a_1\\a_2=a_1+d\\a_3= a_1+2d\\a_4=a_1+3d,...$ where $a_1$ is the first term and d is the constant difference,

thus, we see that the n'th term of an arithmetic sequence is $a_n=a_1+(n-1)d$

in our particular case d=5, the third term is 8, so we have:

$a_3=8=a_1+2\cdot5\\\\8=a_1+10\\\\a_1=-2$

and the general term is $a_n=-2+5(n-1)$ ,

Answer: first term is -2, n'th term is -2+5(n-1)

8 0

2 years ago