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2 years ago

An isosceles right triangle has legs of equal length if the hypotenuse is 10 cm long find the length of each leg

1 answer:

3 0

Since this is a right triangle, we know we have to use the Pythagorean Theorem: A² + B² = C² , where A and B = the legs of the triangle (which are shorter than the hypotenuse) and C = the hypotenuse.

In this isosceles triangle, A = B (because the legs are the same size) and C is 10cm longer than A and B.

Thus we get the following,

A = x

B = x

C = x + 10

We simply plug in these variables into the equation and solve for x:

x² + x² = (x+10)²

2x² = x² + 20x + 100

x² - 20x - 100 = 0

We solve for x by completing the square:

(x² - 20x + 100) - 100 = 100

(x-10)² -100 = 100

(x-10)² = 200

x - 10 = √(200)

x - 10 = 10 √2

x = 10 + 10√2 = 24

Finally,

A ≈ 24 (or 10 + 10√2)

B ≈ 24 (or 10 + 10√2)

C ≈ 24 + 10 ≈ 34 (or 10 + 10√2 + 10 = 20 + 10√2)

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Indefinite integral need help pleaseeeeeee

tatiyna

Integrate indefinite integral:
$I=\int\frac{dx}{e^{2x}+3e^x+2}dx$

Solution:
1. use substitution $u=e^x$
=>
$du=e^xdx$
=>
$dx=\frac{du}{e^x}$
=>
$dx=\frac{du}{u}$
=>
$I=\int\frac{du}{u(u^2+3u+2)}du$
$=\int\frac{du}{u(u+2)(u+1)}du$
2. decompose into partial fractions
$\frac{1}{u(u+2)(u+1)}$
$=\frac{A}{u}+\frac{B}{u+2}+\frac{C}{u+1}$
where A=1/2, B=1/2, C=-1
$=\frac{1}{2u}+\frac{1}{2(u+2)}-\frac{1}{u+1}$
3. Substitute partial fractions and continue
$I=\int\frac{du}{2u}+\int\frac{du}{2(u+2)}-\int\frac{du}{u+1}$
$=\frac{log(u)}{2}+\frac{log(u+2)}{2}-log(u+1)}$
4. back-substitute u=e^x
$=\frac{log(e^x)}{2}+\frac{log(e^x+2)}{2}-log(e^x+1)}$
$=\frac{x}{2}+\frac{log(e^x+2)}{2}-log(e^x+1)}$

Note: log(x) stands for natural log, and NOT log10(x)

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3 years ago

Alyssa walked 1.34 fewer miles than Emily. Alyssa walked 7.45 miles. How many miles did Emily walk?

Umnica [9.8K]

6.11 miles is Emily’s Distance amount

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2 years ago

One tank is filling at a rate of 5 gallon per 7 hour. A second tank is 8 10

seraphim [82]

Answer:

the second one

Step-by-step explanation:

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2 years ago

The data show the average monthly temperatures for two cities over a 6-month period. City 1: {20, 24, 40, 63, 76, 89} City 2: {4

Mrac [35]

Answer:

The MAD of city 2 is less than the MAD for city 1, which means the average monthly temperature of city 2 vary less than the average monthly temperatures for City 1.

Step-by-step explanation:

For comparing the mean absolute deviations of both data sets we have to calculate the mean absolute deviation for both data sets first,

So for city 1:

$Mean = x1 = \frac{20+24+40+63+76+89}{6}$

$x1 = \frac{312}{6}$

$x1 = 52$

Now to calculate the mean deviations mean will be subtracted from each data value. (Note: The minus sign is ignored as the deviation is the distance of value from the mean and it cannot be negative. For this purpose absolute is used)

$20-52 = -32=32\\24-52=-28=28\\40-52=-12=12\\63-52=11\\76-52=24\\89-52=37$

The deviations will be added then. $Mean Absolute Deviation = \frac{32+28+12+11+24+37}{6} \\=\frac{144}{6}\\=24$

So the mean absolute deviation for city 1 is 24 ..

For city 2:

$Mean = x2 = \frac{41+50+58+62+72+83}{6}$

$x2 = \frac{366}{6}$

$x2 = 61$

Now to calculate the mean deviations mean will be subtracted from each data value. (Note: The minus sign is ignored)

$41-61=-20=20\\50-61=-11=11\\58-61=-3=3\\62-61=1\\72-61=11\\83-61=22$

The deviations will be added then. $Mean Absolute Deviation = \frac{20+11+3+1+11+22}{6} \\=\frac{68}{6}\\=11.33$

So the MAD for city 2 is 11.33 ..

So,

The MAD of city 2 is less than the MAD for city 1, which means the average monthly temperature of city 2 vary less than the average monthly temperatures for City 1.

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3 years ago

Read 2 more answers

In the triangles, GK PN and HG 2MP.

dalvyx [7]

Answer:

G is smaller than P. Visually comparing G to P you are doing a rough comparison to a 90° (or right) angle. Angle G is a little bit tilted to the right making it just under 90°

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3 years ago