HELP! If the area of a rectangle is 10x^2+3x-4 and the width is 2x-1, what is the side length of the rectangle?

Farmer bob's square plot ofland is slowly eroding away. Worried about the future of his farm. Farmer Bob measures the rate of er

kow [346]

Answer:

Option A.

Step-by-step explanation:

Area of a square is

$A=x^2$ .... (1)

where, x is side length.

The length of each side of his square plot is decreasing at the constant rate of 2 feet/year.

$\dfrac{dx}{dt}=2$

It is given that bob currently owns 250,000 square feet of land.

Fist find the length of each side.

$A=250000$

$x^2=250000$

Taking square root on both sides.

$x=500$

Differentiate with respect to t.

$\dfrac{dA}{dt}=2x\dfrac{dx}{dt}$

Substitute x=500 and $\frac{dx}{dt}=2$ in the above equation.

$\dfrac{dA}{dt}=2(500)(2)$

$\dfrac{dA}{dt}=2000$

Farmer bob is losing 2,000 square feet of land per year.

Therefore, the correct option is A.

7 0

3 years ago

A cone has a volume of 108 pi and a base diameter of 12. What is the height of the cone?

serious [3.7K]

Answer:

h = 9 units of length

Step-by-step explanation:

Volume of a circular right cone is:

V(c) = (1/3)*Area of the circular base * h

Where h is the height of the cone, then

V(c) = (1/3) * π * r² * h

And we have that V(c) = 108*π

Then:

108 * π = (1/3) * π * r² * h (1)

We know diameter of the circular base is 12 units, according to that

r = d/2 ⇒ r = 12/2 r = 6 units

Then:

108 * π [ cubic units ] = (1/3) * π * (6)² * h [square units] [ units of length]

When simplifying we have to take into account that in order to keep last equation h needs to have units of length

Therefore:

108 = (1/3) * 36 * h

h = ( 108* 3 )/ 36

h = 9 units of length

7 0

4 years ago

NEED HELP WITH A MATH QUESTION

Lady_Fox [76]

Answer:

$\dfrac{18}{5}$

Step-by-step explanation:

The picture shows three isosceles tirangles with the same legs. The base of each triangle is 12 units, 5x-3 units and 17 units.

Since the angles at vertex of each isosceles triangles are 27°, 28° and 29°, then the lengths of the bases satisfy the double inequality

15<5x-3<17

Add 3 to this inequality

15+3<5x-3+3<17+3

18<5x<20

Divide it by 5:

$\dfrac{18}{5}$

3 0

3 years ago

Find the slope of the line that passes through the points (−2/3, 3/4) and (5/6, −1/2).

EastWind [94]

Answer:

$-\frac{5}{6}$

Step-by-step explanation:

Use the slope formula : $\frac{y2-y1}{x2-x1}$

-1/2 is y2, 3/4 is y1, 5/6 is x2, and -2/3 is x1.

$\frac{-1/2-3/4}{5/6-(-2/3)}$

$\frac{-5/4}{5/6+2/3}$

$\frac{-5/4}{9/6}$

$\frac{-5}{4}$ ÷ $\frac{9}{6}$

$-\frac{5}{4}$ • $\frac{6}{9}$

$-\frac{5}{2}$ • $\frac{3}{9}$

$-\frac{15}{18}$

$-\frac{5}{6}$

Hope this helps! Let me know if I got it wrong.

5 0

3 years ago

The accompanying data represent the miles per gallon of a random sample of cars with a three-cylinder, 1.0 liter engine.a. Comp

Kobotan [32]

Answer:

1). Z score = 1.464733

Mean = 38.87917

Standard deviation = 3.609405

2). Quartiles:

Q1 = 37.025

Q2 = 38.450

Q3 = 40.800

3). IQR = 3.775

4).

CI (lower fence) = 37.4351

CI (upper fence) = 40.32323

5). There is outlier in the data set. Please see attached box plot for evidence.

Step-by-step explanation:

1). By Z score, we mean:

Z = $\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$ ,

where:

x-bar ==> mean(g) = 38.87917

$\mu$ = 37.80

standard deviation = 3.609405

sample size (n) = 24.

2). By quantile, we mean:

Q = L + (i*(n/4) - Cf)*c

Where L is the lower class limit of the quartile class

Cf is the cumulative frequency before the quartile class

c is the class size.

3) . IQR = Q3 - Q1

4). CI = $\mu \pm *Z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$ ,

Where $Z_{\alpha/2}$ ==> 1.96

In order to replicate and obtain the result, please use the R code below:

g = c(31.5, 36.0, 37.8, 38.5, 40.1, 42.2,34.2, 36.2, 38.1, 38.7, 40.6, 42.5,34.7, 37.3, 38.2, 39.5,

41.4, 43.4,35.6, 37.6, 38.4, 39.6, 41.7, 49.3)

boxplot(g)

Z = (mean(g) - 37.8)/(sd(g)/sqrt(length(g)))

mean(g)

quantile(g)

IQR(g)

CIl = mean(g) - 1.96*(sd(g)/sqrt(length(g)))

CIU = mean(g) + 1.96*(sd(g)/sqrt(length(g)))

4 0

3 years ago