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KiRa [710]
4 years ago
8

Change the fraction 3/4 to a percent

Mathematics
1 answer:
Rom4ik [11]4 years ago
6 0
To convert 3/4 to a percent we need to convert 3/4 to a decimal and then multiply the decimal by 100. Lets do it:-

To convert a fraction to a decimal we have to divide the numerator by the denominator.

3/4  
3 ÷ 4 = 0.75

Decimal = 0.75

0.75 × 100 = 75
75%.

So, 3/4 in percent form is 75%.

Hope I helped ya!! xD 
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An electric generator can power 3550 watts if electricity. Write and solve an equation to find how many 75 watt light bulbs a ge
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7 0
4 years ago
Can you use the Associative Properties with subtraction and division? Use (14 - 8) - 2 and 24/(4/2 to explain.
larisa86 [58]
No, you cannot use the Associative properties with either subtraction and division, because it will not end to the same result. 14-8 is not the same as 8-14, and 24/4 is not the same as 4/24, but for example 3+2 is the same as 2+3. 
5 0
4 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
3 years ago
What is the slope of the line that passes through the points (-8, -3) and (-12, -3)
MA_775_DIABLO [31]

Step-by-step explanation:

slope=(y2-y1)/(x2-x1)

where (x1,y1)=(-8,-3)

(X2, y2)=(-12,-3)

slope=(-3+3)/(-12+8)=0

3 0
3 years ago
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