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gtnhenbr [62]
3 years ago
5

The figure shows part of a freeway interchange. The raised freeway is supported by vertical, parallel pillars. Set up a ratio an

d solve to find the length of PQ to the nearest tenth of a yard.

Mathematics
1 answer:
Ugo [173]3 years ago
7 0
We will use ratio and proportion between the inclined sides with its corresponding horizontal bases. Then we could determine for the value of segment AQ.

PA/SB = AQ/BR
32/30 = AQ/24
AQ = 25.6 yd

Since what is asked is segment PQ,
PQ = PA + AQ
PQ = 32 + 25.6
PQ = 57.6 yards
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-6.8-c divided 1.2=-2.9
Diano4ka-milaya [45]

Answer:

3.44

Step-by-step explanation:

(-6.8 - c)  /  1.2  =  -2.8

(multiply both sides by 1.2)

-6.8 - c = -3.36

(add 6.8 to both sides)

-c = 3.44

c = 3.44

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What does 5/6 ÷ 8.5 equal, in fractions and division form
nydimaria [60]

Answer:

5/51 or 0.098

Step-by-step explanation:

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Why resultant magnitude by using Pythagorean theorem is different than using ( x,y )components addition vectors ???
Andreyy89

9514 1404 393

Explanation:

Your different answers came about as a result of an error you made in the Pythagorean theorem calculation.

  √(100 +100) = √200 = 14.142 . . . . same as vector calculation

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2 years ago
Which type of parent function is f(x) = |x|?
mariarad [96]
The absolute value parent function.
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2 years ago
cylinder shaped can needs to be constructed to hold 200 cubic centimeters of soup. The material for the sides of the can costs 0
Dafna11 [192]

Answer:

Radius=2.09 cm

Height,h=14.57 cm

Step-by-step explanation:

We are given that

Volume of cylinderical shaped can=200 cubic cm.

Cost of sides of can=0.02 cents per square cm

Cost of top and bottom of the can =0.07 cents per square cm

Curved surface area of cylinder=2\pi rh

Area of circular base=Area of circular top=\pi r^2

Total cost,C(r)=0.02\times 2\pi rh+2\pi r^2\times 0.07

Volume of cylinder,V=\pi r^2 h

200=\pi r^2 h

h=\frac{200}{\pi r^2}

Substitute the value of h

C(r)=0.02\times 2\pi r\times \frac{200}{\pi r^2}+2\pi r^2\times 0.07

C(r)=\frac{8}{r}+0.14\pi r^2

Differentiate w.r.t r

C'(r)=-\frac{8}{r^2}+0.28\pi r

C'(r)=0

-\frac{8}{r^2}+0.28\pi r=0

0.28\pi r=\frac{8}{r^2}

r^3=\frac{8}{0.28\pi}=9.095

r=(9.095)^{\frac{1}{3}}=2.09

Again, differentiate w.r.t r

C''(r)=\frac{16}{r^3}+0.28\pi

Substitute the value of r

C''(2.09)=\frac{16}{(2.09)^3}+0.28\pi=2.63>0

Therefore,the product cost is minimum at r=2.09

h=\frac{200}{\pi (2.09)^2}=14.57

Radius of can,r=2.09 cm

Height of cone,h=14.57 cm

4 0
3 years ago
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