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3 years ago

Suppose a van gets 22 mi/gal. The distance traveled D(g) is a function of the gallons of gas used

Mathematics

1 answer:

babymother [125]3 years ago

5 0

Answer:

a) Table and graph showed

b) The distance will be 231 miles

c) Yes

Step-by-step explanation:

We know the van gets 22 mi/gal, so the distance D in miles traveled by the van can be expressed as

D(g)=22g, being g the number of gallons of gas used

a) The graph of the function D and its corresponding table of values is shown below.

b) If the van used g=10.5 gallons of gas, the distance would be:

D(10.5)=22 x 10.5 = 231 miles

c) The values of g are real in nature because they represent the amount of gas consumed by the van and it can be any real positive number. Being D a linear function of g, it also happens to take positive real values. Then it makes sense to connect the points with lines.

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Differentiate x^2 e^x logx

zimovet [89]

Product rule:

$\dfrac{\mathrm d}{\mathrm dx}(x^2e^x\log x)$

$=\dfrac{\mathrm d(x^2)}{\mathrm dx}e^x\log x+x^2\dfrac{\mathrm d(e^x)}{\mathrm dx}\log x+x^2e^x\dfrac{\mathrm d(\log x)}{\mathrm dx}$

Power rule:

$\dfrac{\mathrm d(x^2)}{\mathrm dx}=2x$

The exponential function is its own derivative:

$\dfrac{\mathrm d(e^x)}{\mathrm dx}=e^x$

Assuming the base of $\log x$ is $e$ , its derivative is

$\dfrac{\mathrm d(\log x)}{\mathrm dx}=\dfrac1x$

But if you mean a logarithm of arbitrary base $b$ , we have

$y=\log_bx\implies x=b^y=e^{y\ln b}\implies1=e^{y\ln b}\ln b\dfrac{\mathrm dy}{\mathrm dx}$

$\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{e^{-y\ln b}}{\ln b}=\dfrac1{b^y\ln b}$

$\implies\dfrac{\mathrm d(\log_bx)}{\mathrm dx}=\dfrac1{x\ln b}$

So we end up with

$2xe^x\log x+x^2e^x\log x+\dfrac{x^2e^x}x$

$=xe^x(2\log x+x\log x+1)$

8 0

3 years ago

Please answer correctly !!!!! Will mark Brianliest !!!!!!!!!!!!!!

ELEN [110]

Answer:

$Volume~of~cylinder=\pi r^2h$

$\pi *(10)^2*5$

$500\pi$

$1570.8~m^3$

-------------------------

hope it helps..

have a great day!!

3 0

3 years ago

Read 2 more answers

I can’t find the the square centimeters of this

Temka [501]

Answer: 544?

Step-by-step explanation:

This is the answer if I'm supposed to solve the question in yellow

1/2*16(48+20) = 1/2*16*(68)

1/2*16=8

8*68 = 544

7 0

3 years ago

Can someone help me to find length CD

uysha [10]

Answer:

CD = 3.602019190339

Step-by-step explanation:

CD = DA - CA

DA = DB×Cos(29) = 18.7×cos(29) = 16.355388523507

BA = BA×cos(43) = 18.7×cos(43) = 13.676314220278

CA = BA÷tan(47) = 13.676314220278÷tan(47) = 12.753369333168

Then

CD = 16.355388523507 - 12.753369333168 = 3.602019190339

7 0

3 years ago

Two competitive neighbours build rectangular pools that cover the same area but are different shapes. Pool A has a width of (x +

GenaCL600 [577]

Answer:

a)Dimensions of pool A are length = 6.667m and width = 3.667 m and dimension of pool B are length = 7.333m and width = 3.333m.

b) Area of pool A is equal to area of pool B equal to 24.44 meters.

Solution:

Let’s first calculate area of pool A .

Given that width of the pool A = (x+3)

Length of the pool A is 3 meter longer than its width.

So length of pool A = (x+3) + 3 =(x + 6)

Area of rectangle = length x width

So area of pool A =(x+6) (x+3) ------(1)

Let’s calculate area of pool B

Given that length of pool B is double of width of pool A.

So length of pool B = 2(x+3) =(2x + 6) m

Width of pool B is 4 meter shorter than its length,

So width of pool B = (2x +6 ) – 4 = 2x + 2

Area of rectangle = length x width

So area of pool B =(2x+6)(2x+2) ------(2)

Since area of pool A is equal to area of pool B, so from equation (1) and (2)

(x+6) (x+3) =(2x+6) (2x+2)

On solving above equation for x

(x+6) (x+3) =2(x+3) (2x+2)

x+6 = 4x + 4

x-4x = 4 – 6

x = $\frac{2}{3}$

Dimension of pool A

Length = x+6 = ( $\frac{2}{3}$ ) +6 = 6.667m

Width = x +3 = ( $\frac{2}{3}$ ) +3 = 3.667m

Dimension of pool B

Length = 2x +6 = 2( $\frac{2}{3}$ ) + 6 = $\frac{22}{3}$ = 7.333m

Width = 2x + 2 = 2( $\frac{2}{3}$ ) + 2 = $\frac{10}{3}$ = 3.333m

Verifying the area:

Area of pool A = ( $\frac{20}{3}$ ) x ( $\frac{11}{3}$ ) = $\frac{220}{9}$ = 24.44 meter

Area of pool B = ( $\frac{22}{3}$ ) x ( $\frac{10}{3}$ ) = $\frac{220}{9}$ = 24.44 meter

Summarizing the results:

(a)Dimensions of pool A are length = 6.667m and width = 3.667 m and dimension of pool B are length = 7.333m and width = 3.333m.

(b)Area of pool A is equal to Area of pool B equal to 24.44 meters.

5 0

3 years ago