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3 years ago

Total Cost (in dollars)

1 answer:

Crank3 years ago

6 0

This problem has been solved! See the answer Determine Um (mode ), average U, and Urms for a group of ten automobiles clocked by radar at speeds 38,44,48,50,55,55,57,58 and 60mi/h respectively

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PLEASE HELP ASAP NO LINKS

insens350 [35]

<u><em>function </em></u>: p(x) = 65 + 25(x)

for 3 sells

p(3) = 65 + 25(3)

p(3) = 140

for 10 sells

p(10) = 65 + 25(10)

p(10) = 315

So,

140 ≤ p(x) ≤ 315

8 0

2 years ago

Read 2 more answers

If a system of linear equations has infinitely many solutions, then the graph of the system is:

vichka [17]

I think the answer is a

6 0

3 years ago

50 POINTSSSS! Use trigonometric identities to verify that this expression is equal.

Llana [10]

Answer:

= cos

Step-by-step explanation:

4 0

3 years ago

1. Given the triangle below, answer the following questions

bogdanovich [222]

Answer:

x=81

y=99

z= 129

Step-by-step explanation:

the 3 angles of a triangle add to 180 degrees

51 + 48 + x = 180

99 + x = 180

subtract 99 from each side

99 + x -99 = 180 -99

x = 81

x+y = 180 they make a straight line

81+ y = 180

subtract 81 from each side

y = 180-81

y = 99

z+ 51 = 180 they make a straight line

subtract 51 from each side

z+51-51 = 180 -51

z = 129

4 0

3 years ago

Read 2 more answers

Find the mass of the triangular region with vertices (0, 0), (3, 0), and (0, 1), with density function ρ(x,y)=x2+y2.

ololo11 [35]

Since density is the ratio of mass to (in this case) area, we can find the mass of the triangular region $\mathcal T$ by computing the double integral of the density function over $\mathcal T$ :

$\mathrm{mass}=\displaystyle\iint_{\mathcal T}\rho(x,y)\,\mathrm dx\,\mathrm dy$

The boundary of $\mathcal T$ is determined by a set of lines in the $x,y$ plane. One way to describe the region $\mathcal T$ is by the set of points,

$\mathcal T=\left\{(x,y)\mid0\le x\le 3\,\land\,0\le y\le1-\dfrac x3\right\}$

So the mass is

$\mathrm{mass}=\displaystyle\int_{x=0}^{x=3}\int_{y=0}^{y=1-x/3}(x^2+y^2)\,\mathrm dy\,\mathrm dx$

$=\displaystyle\int_{x=0}^{x=3}\left(x^2y+\frac{y^3}3\right)\bigg|_{y=0}^{y=1-x/3}\,\mathrm dx$

$=\displaystyle\int_{x=0}^{x=3}\left(x^2\left(1-\frac x3\right)+\frac{\left(1-\frac x3\right)^3}3\right)\,\mathrm dx$

$=\displaystyle\frac1{81}\int_0^3(27-27x+90x^2-28x^3)\,\mathrm dx=\frac52$

6 0

3 years ago