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Marat540 [252]
3 years ago
6

Let f(x)=3x2−5x+13 and g(x)=2x−7 . What is g(f(x)) ?

Mathematics
1 answer:
Debora [2.8K]3 years ago
6 0
G(f(x))=2(3x^2-5x+13)-7
G(f(x))=6x^2-10x+26-7
G(f(x))=6x^2-10x+19 =>A
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Carlos spent 1 1/4 hours doing his math homework. He spent 1/4 of this time practicing his multiplication facts.
Alex17521 [72]

Answer:

18.75 minutes or 0.3 (3/10 or almost 1/3)

Step-by-step explanation:

Carlos spent 1\frac{1}{4}hours doing math. If an hour is 60 minutes, this means he spent 60 minutes and 1/4 of 60 minutes which is 15 minutes. In total he spent 75 minutes on math.

We also know he spent 1/4 of that 75 minutes on multiplication facts. How long did he spend?

We will multiply by the two quantities to find the amount.

\frac{1}{4} (75)=\frac{75}{4} = 18.75 minutes

We will convert this back into hours by dividing it by 60 minutes since 60 minutes= 1 hour.

\frac{18.75}{60} =0.31

He spent 0.3 or 3/10 or about 1/3 of his time on multiplication facts.

3 0
3 years ago
What is 60+(-30)+40-80
klasskru [66]

<em>✽ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ✽</em>

<em>➷ </em>\boxed{-10}<em></em>

<em></em>60+(-30)+40-80<em></em>

<em></em>= 30 + 40 - 80<em></em>

<em></em>= 70 - 80<em></em>

<em></em>= -10<em></em>

<em>✽</em>

<em>➶ Hope This Helps You!</em>

<em>➶ Good Luck (:</em>

<em>➶ Have A Great Day ^-^</em>

<em>↬ May ♡</em>

5 0
3 years ago
Read 2 more answers
A Norman window is a rectangle with a semicircle on top. Suppose that the perimeter of a particular Norman window is to be 56 ft
harkovskaia [24]

Answer:

Dimensions of the window in order to allow maximum light is x=\dfrac{56}{4+\pi} and y=\dfrac{56}{4+\pi}

Step-by-step explanation:

Consider following Norman window, assuming ABCD as rectangle and arc AD as semicircle with center at E and radius r. (Refer attachment)

Given that perimeter of window is 56 ft. Therefore perimeter of window is given as,  

Perimerter = AB + BC + CD + arc\:AD

Calculate arc AD as follows,  

Let, x denote radius of semi circle. That is, r=x  

Since AD is the diameter of semi circle  

So AD = 2 r = 2 x.  

Now perimeter of semicircle is equal to circumference of semicircle, so calculate circumference of semicircle as follows  

Circumference of circle is C=2\pi r. So half of it will be  

C=\dfrac{2\pi r}{2}

C=\dfrac{2\pi x}{2}

C=\pi x

So, arc\:AD = \pi x

Calculate AB , BC and AD as follows,  

Consider rectangle ABCD,  

Since, AD is the diameter of semi circle which is also one of the side of the rectangle.  

So AD = 2 r = 2 x.  

Since AD is parallel to BC. Therefore, AD=BC=2x. Also length of rectangle be y  

\therefore BC=2x,AB=CD=y  

Substituting the value,  

Perimeter = AB + BC + CD + arc AD  

56=y+2x+y+\pi x  

56=2y+2x+\pi x  

To calculate value of y,subtracting 2x and \pi x on both sides,  

56-2x-\pi x=2y  

Dividing by 2,

28-x-\dfrac{\pi x}{2}=y  

Now calculate area of window.  

Area = Area of rectangle + Area of semicircle

From diagram,

Area = width\times length + \dfrac{1}{2}\times\pi r^{2}

Area = 2x\times y + \dfrac{1}{2}\times\pi x^{2}

Area = 2xy + \dfrac{\pi x^{2}}{2}

Substituting value of y in above equation,

Area = 2x\left (28-x-\dfrac{\pi x}{2} \right ) + \dfrac{\pi x^{2}}{2}

Simplifying,  

Area = 56x-2x^{2}-\pi x^{2}+ \dfrac{\pi x^{2}}{2}

Area = 56x-2x^{2}- \dfrac{\pi x^{2}}{2}

In order to find the maximum of area function, differentiate the equation with respect to x and find the critical points.  

Applying difference rule of derivative,  

\dfrac{dA}{dx}=\dfrac{d}{dx}\left(56x\right)-\dfrac{d}{dx}\left ( 2x^{2} \right )-\dfrac{d}{dx}\left ( \dfrac{\pi x^{2}}{2} \right )

Applying constant multiple rule of derivative,  

\dfrac{dA}{dx}=56\dfrac{d}{dx}\left(x\right)-2\dfrac{d}{dx}\left ( x^{2} \right )-\dfrac{\pi}{2}\dfrac{d}{dx}\left (x^{2}\right )

Applying power rule of derivative,

\dfrac{dA}{dx}=56\left(1x^{1-1}\right)-2\left(2x^{2-1}\right)- \dfrac{\pi}{2}\dfrac{d}{dx}\left (2x^{2-1}\right )

\dfrac{dA}{dx}=56\left(1\right)-2\left(2x\right)- \dfrac{\pi}{2}\dfrac{d}{dx}\left (2x\right )

\dfrac{dA}{dx}=56-4x-\pi x

Now find the critical number by solving as follows,

\dfrac{dA}{dx}=0

56-\left(4+\pi\right)x =0

56=\left(4+\pi \right)x

\dfrac{56}{4+\pi} =x

Since there is only one critical point, directly substitute the value of x into equation of A. If value of A is greater than 0, then the area is maximum at critical point.  

Area = 56\left (\dfrac{56}{4+\pi} \right )-2\left (\dfrac{56}{4+\pi} \right )^{2}- \dfrac{\pi \left (\dfrac{56}{4+\pi} \right )^{2}}{2}

Calculating the above expression,  

Area = \dfrac{1568}{4+\pi }

So area is greater than 0.  

Now calculate value of y,  

28-\dfrac{56}{4+\pi}-\dfrac{\pi}{2}\left ( \dfrac{56}{4+\pi} \right )=y

\dfrac{56}{4+\pi}=y

Hence dimensions are x=\dfrac{56}{4+\pi} and y=\dfrac{56}{4+\pi}

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4 years ago
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Solve for x in x²+8x+15=0 in factorisation method ​
Genrish500 [490]

Answer: x=-5, -3

Step-by-step explanation:

x^2 +8x+15=0\\\\(x+5)(x+3)=0\\\\x=-5, -3

7 0
2 years ago
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