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Svetllana [295]
2 years ago
9

In triangle △ABC, ∠ABC=90°, BH is an altitude. Find the missing lengths. AC=5 and BH=2, find AH and CH.

Mathematics
1 answer:
Yakvenalex [24]2 years ago
3 0

Answer:

  AH = 1 or 4

  CH = 4 or 1

Step-by-step explanation:

An altitude divides a right triangle into similar triangles. That means the sides are in proportion, so ...

  AH/BH = BH/CH

  AH·CH = BH²

The problem statement tells us AH + CH = AC = 5, so we can write

  AH·(5 -AH) = BH²

  AH·(5 -AH) = 2² = 4

This gives us the quadratic ...

  AH² -5AH +4 = 0 . . . . in standard form

  (AH -4)(AH -1) = 0 . . . . factored

This equation has solutions AH = 1 or 4, the values of AH that make the factors be zero. Then CH = 5-AH = 4 or 1.

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4(g - 1 )=24<br>what is the answer
babunello [35]
Hello There!

First Expand:
4g - 4 = 24
Then solve:
4g = 24 + 4
4g = 28
g = 28/4
g = 7.

Hope This Helps You!
Good Luck :) 
6 0
2 years ago
Read 2 more answers
The gas mileage for a certain vehicle can be approximated by m=−0.05x2+3.5x−49, where x is the speed of the vehicle in mph. Dete
Whitepunk [10]

Answer:

<h2>14mph</h2>

Step-by-step explanation:

Given the gas mileage for a certain vehicle modeled by the equation m=−0.05x²+3.5x−49 where x is the speed of the vehicle in mph. In order to determine the speed(s) at which the car gets 9 mpg, we will substitute the value of m = 9 into the modeled equation and calculate x as shown;

m = −0.05x²+3.5x−49

when m= 9

9 = −0.05x²+3.5x−49

−0.05x²+3.5x−49 = 9

0.05x²-3.5x+49 = -9

Multiplying through by 100

5x²+350x−4900 = 900

Dividing through by 5;

x²+70x−980 = 180

x²+70x−980 - 180 = 0

x²+70x−1160 = 0

Using the general formula to get x;

a = 1, b = 70, c = -1160

x = -70±√70²-4(1)(-1160)/2

x = -70±√4900+4640)/2

x = -70±(√4900+4640)/2

x = -70±√9540/2

x =  -70±97.7/2

x = -70+97.7/2

x = 27.7/2

x = 13.85mph

x ≈ 14 mph

Hence, the speed(s) at which the car gets 9 mpg to the nearest mph is 14mph

4 0
2 years ago
What is the quotient (2x4 – 3x3 – 3x2 + 7x – 3) ÷ (x2 – 2x + 1)?
Bess [88]
See attached picture for answer

5 0
2 years ago
The sample data below shows the number of hours spent by five students over the weekend to prepare for Monday's Business Statist
Levart [38]

Answer:

8.5 hrs

Step-by-step explanation:

-A 75th percentile mathematically means that  75% of the time data points are below that value and 25% of the time are above that value.

-We plot the given our data {3 12 2 3 5} to determine the 75th percentile

#We can use alcula.com to plot our boxplot.

-From the plot, we find our 75th percentile to be 8.5 hrs

Hence, 75% of the time the number of hours spent to prepare was less than 8.5hrs.

6 0
2 years ago
Consider a tank in the shape of an inverted right circular cone that is leaking water. The dimensions of the conical tank are a
Artyom0805 [142]

Answer:

\frac{dh}{dt} = \frac{216}{1875\pi}

Step-by-step explanation:

Given

Represent radius with r and height with h

h = 12ft

r = 10ft

Rate = \frac{8ft^3}{min} when h = 10ft

Express radius in terms of height

\frac{r}{h} = \frac{10}{12}

\frac{r}{h} = \frac{5}{6}

r = \frac{5}{6}h

First, we need to determine the volume of the cone in terms of height.

Volume = \frac{1}{3}\pi r^2h

Substitute r = \frac{5}{6}h

V = \frac{1}{3} * \pi * (\frac{5}{6}h)^2 * h

V = \frac{1}{3} * \pi * \frac{25}{36}h^2 * h

V = \frac{1}{3} * \pi * \frac{25}{36}h^3

V = \frac{25}{108} \pi h^3

Differentiate both sides w.r.t time (t)

\frac{dV}{dt} = 3 *  \frac{25}{108} \pi h^{3-1} * \frac{dh}{dt}

\frac{dV}{dt} = \frac{75}{108} \pi h^{2} \frac{dh}{dt}

Solve for \frac{dh}{dt}

\frac{dh}{dt} = \frac{dv}{dt} * \frac{108}{75\pi h^2}

At this point, h = 10 and \frac{dv}{dt} = \frac{8ft^3}{min}

So, we have:

\frac{dh}{dt} = 8 * \frac{108}{75\pi 10^2}

\frac{dh}{dt} = \frac{8 * 108}{75\pi 10^2}

\frac{dh}{dt} = \frac{864}{7500\pi}

\frac{dh}{dt} = \frac{216}{1875\pi}

Hence:

The rate is \frac{216}{1875\pi} ft/min

6 0
2 years ago
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