Ask question

Ask question

All categories

1 year ago

5

Solve for x. 9(x+1)=25+x x=2 0x=3 O x=4 O x=5

2 answers:

aleksandr82 [10.1K]1 year ago

3 0

Step-by-step explanation:

9(x+1) = 25+x

open the bracket

9x + 9 = 25+x

collect like terms

9x-x = 25-9

8x = 16

x = 2

ahrayia [7]1 year ago

3 0

Answer:

x = 2

Step-by-step explanation:

9(x + 1) = 25 + x ← distribute parenthesis on left side

9x + 9 = 25 + x ( subtract x from both sides )

8x + 9 = 25 ( subtract 9 from both sides )

8x = 16 ( divide both sides by 8 )

x = 2

You might be interested in

Mrs.Connners went to a shopping mall with $250 in her wallet. At the store, she bought a coat and spent 3/5 of the money she had

LenKa [72]

Step-by-step explanation:

step 1. 250(3/5) = 50(3) = 150.

step 2. She spent $150 for the coat.

3 0

3 years ago

Read 2 more answers

Can't anybody help me with this ik math sucks but help

Agata [3.3K]

Answer:

g

Step-by-step explanation:

6 0

2 years ago

Pleeease open the image and hellllp me

Verdich [7]

1. Rewrite the expression in terms of logarithms:

$y=x^x=e^{\ln x^x}=e^{x\ln x}$

Then differentiate with the chain rule (I'll use prime notation to save space; that is, the derivative of <em>y</em> is denoted <em>y' </em>)

$y'=e^{x\ln x}(x\ln x)'=x^x(x\ln x)'$

$y'=x^x(x'\ln x+x(\ln x)')$

$y'=x^x\left(\ln x+\dfrac xx\right)$

$y'=x^x(\ln x+1)$

2. Chain rule:

$y=\ln(\csc(3x))$

$y'=\dfrac1{\csc(3x)}(\csc(3x))'$

$y'=\sin(3x)\left(-\cot^2(3x)(3x)'\right)$

$y'=-3\sin(3x)\cot^2(3x)$

Since $\cot x=\frac{\cos x}{\sin x}$ , we can cancel one factor of sine:

$y'=-3\dfrac{\cos^2(3x)}{\sin(3x)}=-3\cos(3x)\cot(3x)$

3. Chain rule:

$y=e^{e^{\sin x}}$

$y'=e^{e^{\sin x}}\left(e^{\sin x}\right)'$

$y'=e^{e^{\sin x}}e^{\sin x}(\sin x)'$

$y'=e^{e^{\sin x}+\sin x}\cos x$

4. If you're like me and don't remember the rule for differentiating logarithms of bases not equal to <em>e</em>, you can use the change-of-base formula first:

$\log_2x=\dfrac{\ln x}{\ln2}$

Then

$(\log_2x)'=\left(\dfrac{\ln x}{\ln 2}\right)'=\dfrac1{\ln 2}$

So we have

$y=\cos^2(\log_2x)$

$y'=2\cos(\log_2x)\left(\cos(\log_2x)\right)'$

$y'=2\cos(\log_2x)(-\sin(\log_2x))(\log_2x)'$

$y'=-\dfrac2{\ln2}\cos(\log_2x)\sin(\log_2x)$

and we can use the double angle identity and logarithm properties to condense this result:

$y'=-\dfrac1{\ln2}\sin(2\log_2x)=-\dfrac1{\ln2}\sin(\log_2x^2)$

5. Differentiate both sides:

$\left(x^2-y^2+\sin x\,e^y+\ln y\,x\right)'=0'$

$2x-2yy'+\cos x\,e^y+\sin x\,e^yy'+\dfrac{xy'}y+\ln y=0$

$-\left(2y-\sin x\,e^y-\dfrac xy\right)y'=-\left(2x+\cos x\,e^y+\ln y\right)$

$y'=\dfrac{2x+\cos x\,e^y\ln y}{2y-\sin x\,e^y-\frac xy}$

$y'=\dfrac{2xy+\cos x\,ye^y\ln y}{2y^2-\sin x\,ye^y-x}$

6. Same as with (5):

$\left(\sin(x^2+\tan y)+e^{x^3\sec y}+2x-y+2\right)'=0'$

$\cos(x^2+\tan y)(x^2+\tan y)'+e^{x^3\sec y}(x^3\sec y)'+2-y'=0$

$\cos(x^2+\tan y)(2x+\sec^2y y')+e^{x^3\sec y}(3x^2\sec y+x^3\sec y\tan y\,y')+2-y'=0$

$\cos(x^2+\tan y)(2x+\sec^2y y')+e^{x^3\sec y}(3x^2\sec y+x^3\sec y\tan y\,y')+2-y'=0$

$\left(\cos(x^2+\tan y)\sec^2y+x^3\sec y\tan y\,e^{x^3\sec y}-1\right)y'=-\left(2x\cos(x^2+\tan y)+3x^2\sec y\,e^{x^3\sec y}+2\right)$

$y'=-\dfrac{2x\cos(x^2+\tan y)+3x^2\sec y\,e^{x^3\sec y}+2}{\cos(x^2+\tan y)\sec^2y+x^3\sec y\tan y\,e^{x^3\sec y}-1}$

7. Looks like

$y=x^2-e^{2x}$

Compute the second derivative:

$y'=2x-2e^{2x}$

$y''=2-4e^{2x}$

Set this equal to 0 and solve for <em>x</em> :

$2-4e^{2x}=0$

$4e^{2x}=2$

$e^{2x}=\dfrac12$

$2x=\ln\dfrac12=-\ln2$

$x=-\dfrac{\ln2}2$

7 0

2 years ago

Have an amazing day :) <br><br> Will give brainlst

Evgesh-ka [11]

Answer:

I got 16 but I dont know if that's the correct answer:(

3 0

2 years ago

50=5x−3−8x−7<br> what is X?

Black_prince [1.1K]

Answer:

x = -20

Step-by-step explanation:

Solve for x:

50 = 5 x - 8 x - 7 - 3

Grouping like terms, 5 x - 8 x - 7 - 3 = (5 x - 8 x) + (-3 - 7):

50 = (5 x - 8 x) + (-3 - 7)

5 x - 8 x = -3 x:

50 = -3 x + (-3 - 7)

-3 - 7 = -10:

50 = -10 - 3 x

50 = -3 x - 10 is equivalent to -3 x - 10 = 50:

-3 x - 10 = 50

Add 10 to both sides:

(10 - 10) - 3 x = 10 + 50

10 - 10 = 0:

-3 x = 50 + 10

50 + 10 = 60:

-3 x = 60

Divide both sides of -3 x = 60 by -3:

(-3 x)/(-3) = 60/(-3)

(-3)/(-3) = 1:

x = 60/(-3)

The gcd of 60 and -3 is 3, so 60/(-3) = (3×20)/(3 (-1)) = 3/3×20/(-1) = 20/(-1):

x = 20/(-1)

Multiply numerator and denominator of 20/(-1) by -1:

Answer: x = -20

4 0

2 years ago

Read 2 more answers