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3 years ago

15

) Find the integral of the following distributed loads over the length L, i.e., the total force. Show that it is equal to the ar

ea under the distributed load

1 answer:

serg [7]3 years ago

5 0

Jajahajaiahsvnsjajasnannanananannanakakqkehehsbnanakw

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Find the missing angle.

Advocard [28]

Answer:

35°+95°= 130°

a triangle equals 180° so,

180°-130°=50°

and to solve for the next triangle you would have to use 50° therefore,

50°+85°= 135°

since you are now solving for the next triangle you would do 180°-135°= 45°

so your missing angle is 45°

hope that made sense

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3 years ago

Caroline walked 40 laps on an indoor track in 20 minutes. What was Caroline's average speed in laps per minute?

kakasveta [241]

The answer would be 2 laps per minute

Explanation

====================================================================

We can set this problem up as $\frac{40}{20}$ . We simply divide and we get 2 laps per minute.

<em>I hope this helps : )</em>

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3 years ago

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3 9/12 + (4 7/12 + 5/12) = 3 9/12 + ____?

Maru [420]

Answer:

5

Step-by-step explanation:

4 7/12 + 5/12 = 4 12/12 which equals 5

8 0

3 years ago

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Solve the equation algebraically x^2=192

lapo4ka [179]

The square root of 192 is 13.856. so the answer is more than likely 14 or 13.85

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4 years ago

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What is the length of BC , rounded to the nearest tenth?

Arte-miy333 [17]

Step $1$

In the right triangle ADB

<u>Find the length of the segment AB</u>

Applying the Pythagorean Theorem

$AB^{2} =AD^{2}+BD^{2}$

we have

$AD=5\ units\\BD=12\ units$

substitute the values

$AB^{2}=5^{2}+12^{2}$

$AB^{2}=169$

$AB=13\ units$

Step $2$

In the right triangle ADB

<u>Find the cosine of the angle BAD</u>

we know that

$cos(BAD)=\frac{adjacent\ side }{hypotenuse}=\frac{AD}{AB}=\frac{5}{13}$

Step $3$

In the right triangle ABC

<u>Find the length of the segment AC</u>

we know that

$cos(BAC)=cos (BAD)=\frac{5}{13}$

$cos(BAC)=\frac{adjacent\ side }{hypotenuse}=\frac{AB}{AC}$

$\frac{5}{13}=\frac{AB}{AC}$

$\frac{5}{13}=\frac{13}{AC}$

solve for AC

$AC=(13*13)/5=33.8\ units$

Step $4$

<u>Find the length of the segment DC</u>

we know that

$DC=AC-AD$

we have

$AC=33.8\ units$

$AD=5\ units$

substitute the values

$DC=33.8\ units-5\ units$

$DC=28.8\ units$

Step $5$

<u>Find the length of the segment BC</u>

In the right triangle BDC

Applying the Pythagorean Theorem

$BC^{2} =BD^{2}+DC^{2}$

we have

$BD=12\ units\\DC=28.8\ units$

substitute the values

$BC^{2}=12^{2}+28.8^{2}$

$BC^{2}=973.44$

$BC=31.2\ units$

therefore

<u>the answer is</u>

$BC=31.2\ units$

8 0

3 years ago

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