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2 years ago

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To help racecar drivers safely negotiate turns, racetracks are banked at the curves. At one location, a 58 ft track is banked su

ch that the highest point on the track is 23.6 ft above ground level. The cross section of the track is in the shape of a right triangle, as shown below. What is the width of the base of this cross section? A. 34.4 ft B. 53.0 ft C. 62.6 ft D. 81.6 ft

1 answer:

zmey [24]2 years ago

7 0

The correct answer is Choice B: 53.0 feet.

If you draw a picture of the cross section, you will have a right triangle. The hypotenuse is 58 and one of the legs is 23.6. All you have to do is use the Pythagorean Theorem and solve for the missing side.

x^2 + 23.6^2 = 58^2

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2 years ago

What is the equvilent fraction of 7•4 OVER 1002•9

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2 years ago

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<1 and <2 are supplementary. m<1=4x and m<2 =92 what is the value of x

monitta

Answer:

x=22

Step-by-step explanation:

Supplementary means that the two angles that are supplementary equal 180 degrees. 180-m<2=m<1

180-92=88 (m<2=92)

m<1=88

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2 years ago

PLS HELP ASAP IW ILL GIVE BRIANLEST IF RIGHT DO NOT ANSWER UNUSEFUL ANSWERS

Roman55 [17]

1 equals 0.46 (make a line over the 6 to show it is repeated)

2 equals 0.4 (make a line over the 4 to show it is repeated)

3 equals -0.6 (make a line over the 6 to show it is repeated)

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5 equals 3.3409 (make a line over the 09 at the end to show it is repeated)

6 equals -1.7727 (make a line over the 27 at the end to show it is repeated)

7 equals 66.5%

8 equals 3.16 (make a line over the 6 to show it is repeated)

9 equals -9 over 10

10 equals -17 over 20

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2 years ago

Read 2 more answers

Please answer correctly !!!!!!! Will mark brainliest !!!!!!!!!!!!

LUCKY_DIMON [66]

Answer:

$=8\sqrt{15}b^{\frac{7}{2}}$

Step-by-step explanation:

$\sqrt{24b^3}\sqrt{40b^2}\sqrt{b^2}$

$=\sqrt{40}\sqrt{b^2}\sqrt{b^2}\sqrt{24b^3}$

$=\sqrt{40}b^2\sqrt{24b^3}$

$\sqrt{24b^3}$

$=\sqrt{24}\sqrt{b^3}$

$=\sqrt{24}b^{\frac{3}{2}}$

$=\sqrt{24}b^{\frac{3}{2}}\sqrt{40}b^2$

$=\sqrt{2^3\cdot \:3}b^{\frac{3}{2}}\sqrt{40}b^2$

$=\sqrt{2^3}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2$

$\sqrt{2^3}$

$=2^{3\cdot \frac{1}{2}$

$=2^{3\cdot \frac{1}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2$

$=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2$

$=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{2^3\cdot \:5}b^2$

$=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{2^3}\sqrt{5}b^2$

$=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\cdot \:2^{\frac{3}{2}}\sqrt{5}b^2$

$=\sqrt{3}b^{\frac{3}{2}}\cdot \:2^{\frac{3}{2}+\frac{3}{2}}\sqrt{5}b^2$

$=\sqrt{3}\cdot \:2^{\frac{3}{2}+\frac{3}{2}}\sqrt{5}b^{\frac{3}{2}+2}$

$2^{\frac{3}{2}+\frac{3}{2}}$

$=2^3$

$=2^3\sqrt{3}\sqrt{5}b^{\frac{3}{2}+2}$

$b^{\frac{3}{2}+2}$

$=b^{\frac{7}{2}}$

$=2^3\sqrt{3}\sqrt{5}b^{\frac{7}{2}}$

$=2^3\sqrt{3\cdot \:5}b^{\frac{7}{2}}$

$=8\sqrt{15}b^{\frac{7}{2}}$

4 0

2 years ago