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2 years ago

Carlie biked 24 miles in four hours. What was Carlie's average speed in miles per hour?

Mathematics

2 answers:

garik1379 [7]2 years ago

8 0

Answer:

6 miles per hour

Step-by-step explanation:

love history [14]2 years ago

7 0

Answer:

6 miles per hour

Step-by-step explanation:

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Simplify 30! / 28! Please post the correct answer. Don’t send links

viva [34]

The answer is
3 divided by 2
Hope this helped

5 0

2 years ago

Can anyone help me integrate :

worty [1.4K]

Rewrite the second factor in the numerator as

$2x^2+6x+1=2(x+2)^2-2(x+2)-3$

Then in the entire integrand, set $x+2=\sqrt3\sec t$ , so that $\mathrm dx=\sqrt3\sec t\tan t\,\mathrm dt$ . The integral is then equivalent to

$\displaystyle\int\frac{(\sqrt3\sec t-2)(6\sec^2t-2\sqrt3\sec t-3)}{\sqrt{(\sqrt3\sec t)^2-3}}(\sqrt3\sec t)\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\sec^2t-1}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\tan^2t}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{|\tan t|}\,\mathrm dt$

Note that by letting $x+2=\sqrt3\sec t$ , we are enforcing an invertible substitution which would make it so that $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ requires $0\le t$ or $\dfrac\pi2$ . However, $\tan t$ is positive over this first interval and negative over the second, so we can't ignore the absolute value.

So let's just assume the integral is being taken over a domain on which $\tan t>0$ so that $|\tan t|=\tan t$ . This allows us to write

$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\tan t}\,\mathrm dt$
$=\displaystyle\int(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\csc t\,\mathrm dt$

We can show pretty easily that

$\displaystyle\int\csc t\,\mathrm dt=-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec t\csc t\,\mathrm dt=-\ln|\csc2t+\cot2t|+C$
$\displaystyle\int\sec^2t\csc t\,\mathrm dt=\sec t-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec^3t\csc t\,\mathrm dt=\frac12\sec^2t+\ln|\tan t|+C$

which means the integral above becomes

$=3\sqrt3\sec^2t+6\sqrt3\ln|\tan t|-18\sec t+18\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|-6\ln|\csc t+\cot t|+C$
$=3\sqrt3\sec^2t-18\sec t+6\sqrt3\ln|\tan t|+12\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|+C$

Back-substituting to get this in terms of $x$ is a bit of a nightmare, but you'll find that, since $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ , we get

$\sec t=\dfrac{x+2}{\sqrt3}$
$\sec^2t=\dfrac{(x+2)^2}3$
$\tan t=\sqrt{\dfrac{x^2+4x+1}3}$
$\cot t=\sqrt{\dfrac3{x^2+4x+1}}$
$\csc t=\dfrac{x+2}{\sqrt{x^2+4x+1}}$
$\csc2t=\dfrac{(x+2)^2}{2\sqrt3\sqrt{x^2+4x+1}}$

etc.

3 0

2 years ago

Charlie cannot remember how much he financed to buy his car. He does remember that his monthly payment is $200. His add-on inter

Mazyrski [523]

First find the total payments
Total paid
200×30=6,000 (this is the future value)

Second use the formula of the future value of annuity ordinary to find the monthly payment.
The formula is
Fv=pmt [(1+r/k)^(n)-1)÷(r/k)]
We need to solve for pmt
PMT=Fv÷[(1+r/k)^(n)-1)÷(r/k)]
PMT monthly payment?
Fv future value 6000
R interest rate 0.09
K compounded monthly 12
N=kt=12×(30months/12months)=30
PMT=6000÷(((1+0.09÷12)^(30)
−1)÷(0.09÷12))
=179.09 (this is the monthly payment)

Now use the formula of the present value of annuity ordinary to find the amount of his loan.
The formula is
Pv=pmt [(1-(1+r/k)^(-n))÷(r/k)]
Pv present value or the amount of his loan?
PMT monthly payment 179.09
R interest rate 0.09
N 30
K compounded monthly 12
Pv=179.09×((1−(1+0.09÷12)^(
−30))÷(0.09÷12))
=4,795.15

The answer is 4795.15

5 0

2 years ago

HELP WITH 4 QUESTIONS!!!!

m_a_m_a [10]

When dilation is about the origin, as it is here in every case, the image point coordinates are the original (pre-image) coordinates multiplied by the scale factor.

1. Multiply every coordinate value by 5:

... W' = (-5, 10), X' = (-15, -5), Y' = (25, -5), Z' = (15, 10)

2. Multiply every coordinate value by 1/3:

... A' = (-2, 5), B' = (0, 5/3), C' = (1, 10/3)

3. A' = (2, 8), B' = (6, 2), C' = (2, 2)

4. The image coordinates are 5 times the original coordinates, so ...

... the scale factor of the dilation is 5.

5 0

2 years ago

Betty measured an Italian restaurant and made a scale drawing. She used the scale 2 millimeters : 1 meter. If the actual width o

Karolina [17]

First convert all the units into millimetres.

1 metre = 1000 millimetres

10 metres = 10000 millimetres

2 : 1000 :: x : 10000

Remember this: Product of means is equal to the product of extremes.

2 × 10000 = 1000x

20000 = 1000x

∴ x = 20000 ÷ 1000

x = 20 millimetres

PLEASE MAKE THIS ANSWER THE BRAINLIEST

8 0

2 years ago

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