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Sergeeva-Olga [200]
3 years ago
10

Please answer this and I’ll make you Brainly only if it’s right

Mathematics
2 answers:
alina1380 [7]3 years ago
7 0
The answer is d :) have a good day!
drek231 [11]3 years ago
3 0

Answer: D

hope it helps :)

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What is x2+5x=0? Please show all work!
Alexxandr [17]
X2+5x=0
7x+0
divide both sides by 7
7 divided by 0 is 0
x=0
0=0
7 0
3 years ago
Read 2 more answers
Differentiate with respect to X <br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B%20%5Cfrac%7Bcos2x%7D%7B1%20%2Bsin2x%20%7D%20
Mice21 [21]

Power and chain rule (where the power rule kicks in because \sqrt x=x^{1/2}):

\left(\sqrt{\dfrac{\cos(2x)}{1+\sin(2x)}}\right)'=\dfrac1{2\sqrt{\frac{\cos(2x)}{1+\sin(2x)}}}\left(\dfrac{\cos(2x)}{1+\sin(2x)}\right)'

Simplify the leading term as

\dfrac1{2\sqrt{\frac{\cos(2x)}{1+\sin(2x)}}}=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}

Quotient rule:

\left(\dfrac{\cos(2x)}{1+\sin(2x)}\right)'=\dfrac{(1+\sin(2x))(\cos(2x))'-\cos(2x)(1+\sin(2x))'}{(1+\sin(2x))^2}

Chain rule:

(\cos(2x))'=-\sin(2x)(2x)'=-2\sin(2x)

(1+\sin(2x))'=\cos(2x)(2x)'=2\cos(2x)

Put everything together and simplify:

\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{(1+\sin(2x))(-2\sin(2x))-\cos(2x)(2\cos(2x))}{(1+\sin(2x))^2}

=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{-2\sin(2x)-2\sin^2(2x)-2\cos^2(2x)}{(1+\sin(2x))^2}

=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{-2\sin(2x)-2}{(1+\sin(2x))^2}

=-\dfrac{\sqrt{1+\sin(2x)}}{\sqrt{\cos(2x)}}\dfrac{\sin(2x)+1}{(1+\sin(2x))^2}

=-\dfrac{\sqrt{1+\sin(2x)}}{\sqrt{\cos(2x)}}\dfrac1{1+\sin(2x)}

=-\dfrac1{\sqrt{\cos(2x)}}\dfrac1{\sqrt{1+\sin(2x)}}

=\boxed{-\dfrac1{\sqrt{\cos(2x)(1+\sin(2x))}}}

5 0
3 years ago
This was on my homework 5 X-5 = 5x+7
KengaRu [80]

Answer:

No solution

Step-by-step explanation:

Original Equation:

5x - 5 = 5x + 7

Add 5 to both sides

5x = 5x + 12

So we already know that this equation is not true, making it no solution

Hope this helped

6 0
3 years ago
What is 438,902 rounded to the nearest ten thousand
Ray Of Light [21]
440,000 is rounded to the nearest ten thousandth
7 0
3 years ago
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If x+1 and x-1 are the factors of the polynomial ax^3 + x^2 - 2x +b,find the values of a and b
Nadusha1986 [10]

The polynomial remainder theorem says that dividing a polynomial p(x) by x-c leaves a remainder of p(c)=0 if x-c is a factor of p(x). In this case, check c=-1 and c=1.

a(-1)^3+(-1)^2-2(-1)+b=-a+b+3=0

a(1)^3+(1)^2-2(1)+b=a+b-1=0

From the first equation,

-a+b+3=0\implies a=b+3

and substituting into the second gives

(b+3)+b-1=0\implies2b+2=0\implies b=-1\implies a=2

3 0
3 years ago
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