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BabaBlast [244]

3 years ago

12

The time required to drive a fixed distance varies inversely as the speed. It takes 40 hr at a speed of 10 km/h to drive a fixed

distance. How long will it take to drive the same distance at a speed of 8 km/h?

1 answer:

patriot [66]3 years ago

3 0

Answer:

50 hours

Step-by-step explanation:

Given data

Time=40hr

Speed=10km/hr

Let us find the fixed distance first

Speed= distance/Time

10= distance/40

distance= 10*40

distance= 400miles

Now our speed= 8km/h

hence the time is

8= 400/time

time= 400/8

time=50 hours

Hence the time is 50 hours

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8 0

4 years ago

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Sanders found 6/7 of a pie in the fridge. he eats 1/3 and gives 1/2 to sister. How much is left

Nadya [2.5K]

Answer:

1/42

Step-by-step explanation:

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3 0

3 years ago

Solve using law of sines or law of cosines!

malfutka [58]

Answer:

Part 5) The length of the ski lift is $1.15\ miles$

Part 6) The height of the tree is 18.12 m

Step-by-step explanation:

Part 5)

Let

A -----> Beginning of the ski lift

B -----> Top of the mountain

C -----> Base of mountain

we have

$b=0.75\ miles$

$A=20\°$

$C=180\°-50\°=130\°$ ----> by supplementary angles

Find the measure of angle B

Remember that the sum of the interior angles must be equal to 180 degrees

$B=180\°-A-C$

substitute

$B=180\°-20\°-130\°=30\°$

Applying the law of sines

$\frac{b}{sin(B)}=\frac{c}{sin(C)}$

substitute

$\frac{0.75}{sin(30\°)}=\frac{c}{sin(130\°)}$

$c=\frac{0.75}{sin(30\°)}(sin(130\°))$

$c=1.15\ miles$

Par 6)

see the attached figure with letters to better understand the problem

<u><em>Applying the law of sines in the right triangle BDC</em></u>

In the right triangle BDC 20 degrees is the complement of 70 degrees

$\frac{BC}{sin(70\°)}=\frac{x}{sin(20\°)}$

$BC=(sin(70\°))\frac{x}{sin(20\°)}$ -----> equation A

<u><em>Applying the law of sines in the right triangle ABC</em></u>

In the right triangle ABC 50 degrees is the complement of 40 degrees

$\frac{BC}{sin(40\°)}=\frac{x+15}{sin(50\°)}$

$BC=(sin(40\°))\frac{x+15}{sin(50\°)}$ -----> equation B

Equate equation A and equation B and solve for x

$(sin(70\°))\frac{x}{sin(20\°)}=(sin(40\°))\frac{x+15}{sin(50\°)}\\\\2.7475x=0.8391(x+15)\\\\2.7475x=0.8391x+12.5865\\\\2.7475x-0.8391x=12.5865\\\\x=6.60\ m$

<u><em>Find the value of BC</em></u>

$BC=(sin(70\°))\frac{6.6}{sin(20\°)}$

$BC=18.12\ m$

therefore

The height of the tree is 18.12 m

5 0

4 years ago

Multiply. √3 ⋅ √5 ⋅ √7 Enter your answer, in radical form, in the box.

Galina-37 [17]

Answer

<h2> $\sqrt{105}$ </h2>

Step by step explanation

$\sqrt{3} \times \sqrt{5} \times \sqrt{7}$

The product of roots with the same index is equal to root of the product

$= \sqrt{3 \times 5 \times 7}$

Calculate the product

$= \sqrt{105}$

Hope this helps...

Best Regards!

3 0

3 years ago