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3 years ago

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A construction crew is lengthening a road. The road started with a length of 54 miles, and the crew is adding 3 miles to the roa

d each day. Let L represent the total length of the road (in miles), and let D represent the number of days the crew has worked. Write an equation relating L to D . Then use this equation to find the total length of the road after the crew has worked 39 days.

1 answer:

patriot [66]3 years ago

6 0

53+3(D) = L

53 + 3(38) =L

53 +114 = L

167=L

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2. Draw the image of RST under the dilation with scale factor 2/3 and center of dilation (1,-1). Label the image RST .

professor190 [17]

Answer:

From the graph: we have the coordinates of RST i.e,

R = (2,1) , S = (2,-2) , T = (-1,-2)

Also, it is given the scale factor $\frac{2}{3}$ and center of dilation C (1,-1)

The mapping rule for the center of dilation applied for the triangle as shown below:

$(x, y) \rightarrow (\frac{2}{3}(x-1)+1, \frac{2}{3}(y+1)-1)$

or

$(x, y) \rightarrow (\frac{2}{3}x -\frac{2}{3}+1 , \frac{2}{3}y+\frac{2}{3}-1)$

or

$(x, y) \rightarrow (\frac{2}{3}x+\frac{1}{3} , \frac{2}{3}y-\frac{1}{3} )$

Now,

for R = (2,1)

the image R' = $(\frac{2}{3}(2)+\frac{1}{3} , \frac{2}{3}(1)-\frac{1}{3} )$ or

$(\frac{4}{3}+\frac{1}{3} , \frac{2}{3}-\frac{1}{3} )$

⇒ R' = $(\frac{5}{3} , \frac{1}{3})$

For S = (2, -2) ,

the image S'= $(\frac{2}{3}(2)+\frac{1}{3} , \frac{2}{3}(-2)-\frac{1}{3} )$ or

$(\frac{4}{3}+\frac{1}{3} , \frac{-4}{3}-\frac{1}{3} )$

⇒ S' = $(\frac{5}{3} , -\frac{5}{3})$

and For T = (-1, -2)

The image T' = $(\frac{2}{3}(-1)+\frac{1}{3} , \frac{2}{3}(-2)-\frac{1}{3} )$ or

$(\frac{-2}{3}+\frac{1}{3} , \frac{-4}{3}-\frac{1}{3} )$

⇒ T' = $(\frac{-1}{3} , \frac{-5}{3})$

Now, label the image of RST on the graph as shown below in the attachment:

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3 years ago

He charges $15 per hour to mow lawns and $20 per hour for gardening.

Reil [10]

Answer:

Step-by-step explanation:

he can make 300 dollars per week by doing either or. so if you're asking whether the statement is correct, it is

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2 years ago

Lou has an account with $10,000 which pays 6% interest compounded annually. If to that account, Lou deposits $5,000 at the begin

katovenus [111]

Answer:

Option d. $22154 is the right answer.

Step-by-step explanation:

To solve this question we will use the formula $A=P(1+\frac{r}{n})^{nt}$

In this formula A = amount after time t

P = principal amount

r = rate of interest

n = number of times interest gets compounded in a year

t = time

Now Lou has principal amount on the starting of first year = 10000+5000 = $15000

So for one year $A=15000(1+\frac{\frac{6}{100}}{1})^{1\times1}$

$= 15000(1+.06)^{1}$

$= 15000(1.06)$ = $15900

After one year Lou added $5000 in this amount and we have to calculate the final amount he got

Now principal amount becomes $15900 + $ 5000 = $20900

Then putting the values again in the formula

$A=20900(1+\frac{\frac{6}{100}}{1})^{1\times1}$

$= 20900(1+.06)^{1}$

$= 20900(1.06)=22154$

So the final amount will be $22154.

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3 years ago

You are ordering pencils from Amazon for $3.50 each. Shipping costs an additional $2.99. This situation can be represented with

DENIUS [597]

Answer:

Y=total cost of purchasing the pencils

Step-by-step explanation:

Y=3.5x+2.99

Where,

Y=total cost of purchasing the pencils

3.5=price of each pencil

X= number of pencils bought

2.99= shipping cost

For instance, if you order for 5 pencils on Amazon, the total cost of purchasing it is

Y=$3.50(5)+$2.99

Y=$17.50+$2.99

Y=$20.49

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3 years ago

Consider the function f(x)equalscosine left parenthesis x squared right parenthesis. a. Differentiate the Taylor series about 0

dybincka [34]

I suppose you mean

$f(x)=\cos(x^2)$

Recall that

$\cos x=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{x^{2n}}{(2n)!}$

which converges everywhere. Then by substitution,

$\cos(x^2)=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{(x^2)^{2n}}{(2n)!}=\sum_{n=0}^\infty(-1)^n\frac{x^{4n}}{(2n)!}$

which also converges everywhere (and we can confirm this via the ratio test, for instance).

a. Differentiating the Taylor series gives

$f'(x)=\displaystyle4\sum_{n=1}^\infty(-1)^n\frac{nx^{4n-1}}{(2n)!}$

(starting at $n=1$ because the summand is 0 when $n=0$ )

b. Naturally, the differentiated series represents

$f'(x)=-2x\sin(x^2)$

To see this, recalling the series for $\sin x$ , we know

$\sin(x^2)=\displaystyle\sum_{n=0}^\infty(-1)^{n-1}\frac{x^{4n+2}}{(2n+1)!}$

Multiplying by $-2x$ gives

$-x\sin(x^2)=\displaystyle2x\sum_{n=0}^\infty(-1)^n\frac{x^{4n}}{(2n+1)!}$

and from here,

$-2x\sin(x^2)=\displaystyle 2x\sum_{n=0}^\infty(-1)^n\frac{2nx^{4n}}{(2n)(2n+1)!}$

$-2x\sin(x^2)=\displaystyle 4x\sum_{n=0}^\infty(-1)^n\frac{nx^{4n}}{(2n)!}=f'(x)$

c. This series also converges everywhere. By the ratio test, the series converges if

$\displaystyle\lim_{n\to\infty}\left|\frac{(-1)^{n+1}\frac{(n+1)x^{4(n+1)}}{(2(n+1))!}}{(-1)^n\frac{nx^{4n}}{(2n)!}}\right|=|x|\lim_{n\to\infty}\frac{\frac{n+1}{(2n+2)!}}{\frac n{(2n)!}}=|x|\lim_{n\to\infty}\frac{n+1}{n(2n+2)(2n+1)}$

The limit is 0, so any choice of $x$ satisfies the convergence condition.

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3 years ago