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Sindrei [870]
3 years ago
13

What is the length of the unknown leg in the right triangle? A right triangle has a side with length 20 meters, hypotenuse with

length 25 meters, and side labeled b. StartRoot 5 EndRoot m 5 m StartRoot 45 EndRoot m 15 m
Mathematics
2 answers:
galben [10]3 years ago
6 0

The length of the unknown leg of the triangle is 15 m.

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

Length of one leg = 20 m

Length of the hypotenuse= 25m

As it is a right angled triangle we can use pythogoras theorem.

Let the unknown length be y

(20) (20)  + y(y) = (25) (25)

400 + y(y) = 625

y(y) = 225

y = √225

y = 15

The length of the unknown leg is 15 m.

qwelly [4]3 years ago
6 0

Answer:

Y =15

Step-by-step explanation:

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3 trips

Step-by-step explanation:

We would divide the amount of riders the roller coaster could take in a single trip by the total number of riders to determine how many trips it can make.

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Hence, the rollercoaster made 3 trips.

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Demand for Tablet Computers The quantity demanded per month, x, of a certain make of tablet computer is related to the average u
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x = f ( p ) = \frac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } } \\\\ \qquad { p ( t ) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \quad ( 0 \leq t \leq 60 ) }

Answer:

12.0 tablet computers/month

Step-by-step explanation:

The average price of the tablet 25 months from now will be:

p ( 25) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { 25 } } + 200 \\= \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \times 5 } + 200\\\\=\dfrac { 400 } { 1 + \dfrac { 5 } { 8 } } + 200\\p(25)=\dfrac { 5800 } {13}

Next, we determine the rate at which the quantity demanded changes with respect to time.

Using Chain Rule (and a calculator)

\dfrac{dx}{dt}= \dfrac{dx}{dp}\dfrac{dp}{dt}

\dfrac{dx}{dp}= \dfrac{d}{dp}\left[{ \dfrac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } }\right] =-\dfrac{100}{9}p(810,000-p^2)^{-1/2}

\dfrac{dp}{dt}=\dfrac{d}{dt}\left[\dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \right]=-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}

Therefore:

\dfrac{dx}{dt}= \left[-\dfrac{100}{9}p(810,000-p^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}\right]

Recall that at t=25, p(25)=\dfrac { 5800 } {13} \approx 446.15

Therefore:

\dfrac{dx}{dt}(25)= \left[-\dfrac{100}{9}\times 446.15(810,000-446.15^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt {25} \right]^{-2}25^{-1/2}\right]\\=12.009

The quantity demanded per month of the tablet computers will be changing at a rate of 12 tablet computers/month correct to 1 decimal place.

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2, 11, 20, 29, 38, 47, 56,...
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Answer:

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Step-by-step explanation:

Hypothesis test on the population mean.

The claim is that the mean weight of one-year-old boys is greater than 25 pounds.

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The significance level is α=0.05.

The sample size is n=354. The sample mean is 25.8 pounds and the sample standard deviation is 5.3 pounds. As the population standard deviation is estimated from the sample standard deviation, we will use a t-statistic.

The degrees of freedom are:

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As the P-value is smaller than the significance level, the effect is significant and the null hypothesis is rejected.

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