All categories

3 years ago

Divide by long division: (3d^2+2d-29)/(d+3) help please thank you and show work

1 answer:

8 0

Ok so the answer i got was very complex, but i got the answers(if they r correct).
3d^2-27d-87.
i truly hope this helps.

You might be interested in

One half gallon is equivalent to 4 pints.how many gallons are the equivalent of 72 pints

malfutka [58]

Answer: 9 gallons

<u>Step-by-step explanation:</u>

Set up the proportion, then multiply to solve for the variable.

$\dfrac{\frac{1}{2}\ \text{gallon}}{4\ \text{pints}}=\dfrac{x}{72\ \text{pints}}$

$(72\ \text{pints})\dfrac{\frac{1}{2}\ \text{gallon}}{4\ \text{pints}}=(72\ \text{pints)}\dfrac{x}{72\ \text{pints}}$

$\dfrac{36\ \text{gallons}}{4}=x}$

9 gallons = x

5 0

3 years ago

HELP!!!!!!!!!!! What are two possible measures of the angle below?

Andreyy89

Answer: First Option -90° and 630°

Solution:

1) In clockwise the angle is negative, then is -90°

2) Counterclockwise the angle is positive and it could be:

270°

270°+360°=630°

Then, the answer is -90° and 630°

6 0

4 years ago

At 2:00 PM a car's speedometer reads 30 mi/h. At 2:20 PM it reads 50 mi/h. Show that at some time between 2:00 and 2:20 the acce

Bad White [126]

Answer:

Let v(t) be the velocity of the car t hours after 2:00 PM. Then $\frac{v(1/3)-v(0)}{1/3-0}=\frac{50 \:{\frac{mi}{h} }-30\:{\frac{mi}{h} }}{1/3\:h-0\:h} = 60 \:{\frac{mi}{h^2} }$ . By the Mean Value Theorem, there is a number c such that $0 < c$ with $v'(c)=60 \:{\frac{mi}{h^2}}$ . Since v'(t) is the acceleration at time t, the acceleration c hours after 2:00 PM is exactly $60 \:{\frac{mi}{h^2}}$ .

Step-by-step explanation:

The Mean Value Theorem says,

Let be a function that satisfies the following hypotheses:

f is continuous on the closed interval [a, b].
f is differentiable on the open interval (a, b).

Then there is a number c in (a, b) such that

$f'(c)=\frac{f(b)-f(a)}{b-a}$

Note that the Mean Value Theorem doesn’t tell us what c is. It only tells us that there is at least one number c that will satisfy the conclusion of the theorem.

By assumption, the car’s speed is continuous and differentiable everywhere. This means we can apply the Mean Value Theorem.

Let v(t) be the velocity of the car t hours after 2:00 PM. Then $v(0 \:h) = 30 \:{\frac{mi}{h} }$ and $v( \frac{1}{3} \:h) = 50 \:{\frac{mi}{h} }$ (note that 20 minutes is $20/60=1/3$ of an hour), so the average rate of change of v on the interval $[0 \:h, \frac{1}{3} \:h]$ is

$\frac{v(1/3)-v(0)}{1/3-0}=\frac{50 \:{\frac{mi}{h} }-30\:{\frac{mi}{h} }}{1/3\:h-0\:h} = 60 \:{\frac{mi}{h^2} }$

We know that acceleration is the derivative of speed. So, by the Mean Value Theorem, there is a time c in $(0 \:h, \frac{1}{3} \:h)$ at which $v'(c)=60 \:{\frac{mi}{h^2}}$ .

c is a time time between 2:00 and 2:20 at which the acceleration is $60 \:{\frac{mi}{h^2}}$ .

4 0

3 years ago

Math - Use unit conversion factor to convert the units.

nlexa [21]

Answer: 240 minutes, 20 cups, 4000 pounds

<u>Step-by-step explanation:</u>

Use the following conversions:

1 hour = 60 minutes

2 cups = 1 pint

2 pints = 1 quart

1 ton = 2000 pounds

$4\ hours\times \dfrac{60\ minutes}{1\ hour}=\large\boxed{240\ minutes}\\\\\\\\5\ quarts \times \dfrac{2\ pints}{1\ quart}\times \dfrac{2\ cups}{1\ pint}=\large\boxed{20\ cups}\\\\\\\\2\ tons\times \dfrac{2000\ pounds}{1\ ton}=\large\boxed{4000\ pounds}$

8 0

3 years ago

Equation of a line with slope 1/5 through the point (-5, -4)

Harrizon [31]

9514 1404 393

Answer:

y +4 = (1/5)(x +5)

Step-by-step explanation:

The point-slope form of the equation of a line is useful when you have a given point and slope.

y -k = m(x -h) . . . . . . line with slope m through point (h, k)

Filling in the given values, this is ...

y +4 = 1/5(x +5)

7 0

3 years ago