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2 years ago

Anyone please could help

Mathematics

1 answer:

Shalnov [3]2 years ago

5 0

The answer would be c.127

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Joe needs to push his car 20 feet up a hill. If

Sever21 [200]

Answer:

What if you have to push a heavy object up a ramp? Say, for example, you have to move a refrigerator. You want to go camping, and because you expect to catch plenty of fish, you decide to take your 100-kilogram refrigerator with you. The only catch is getting the refrigerator into your vehicle (see the figure). The refrigerator has to go up a 30-degree ramp that happens to have a static coefficient of friction with the refrigerator of 0.20 and a kinetic coefficient of friction of 0.15. The good news is that you have two friends to help you move the fridge. The bad news is that you can supply only 350 newtons of force each, so your friends panic.

The minimum force needed to push that refrigerator up the ramp has a magnitude Fpush, and it has to counter the component of the weight of the refrigerator acting along the ramp and the force due to friction.

The first step in this problem is to resolve the weight of the refrigerator into components parallel and perpendicular to the ramp. Take a look at the figure, which shows the refrigerator and the forces acting on it. The component of the weight of the refrigerator along the ramp is

Step-by-step explanation:

please rate thank and follow

5 0

2 years ago

A fox leaves its den and travels 725 yards northeast, then 950 yards west. How far, in yards is the fox from its den

Alenkinab [10]

Answer:

1,675

Step-by-step explanation:

add it

7 0

2 years ago

Present a picture of an object (TV, Ipad Screen, Book Cover) that contains a right angle (it could be something in nature or tha

myrzilka [38]

The measure of hypotenuse using the ruler is 25.5 cm and using the formula is 23.43 cm, they are not same, and formula can be used to in the real life problems as well.

<h3>What is a right-angle triangle?</h3>

It is a triangle in which one of the angles is 90 degrees and the other two are sharp angles. The sides of a right-angled triangle are known as the hypotenuse, perpendicular, and base.

The question is incomplete:

The complete question is:

Present a picture of an object (TV, Ipad Screen, Book Cover) that contains a right angle (it could be something in nature or that is man-made). Use a ruler or measuring tape to measure the two sides that make the right angle and measure the distance from the end of one side to the end of the other side (hypotenuse). Draw a diagram of the object including the measurements. Use your leg measurements on your diagram to calculate a theoretical hypotenuse. (Show all steps).

Does the hypotenuse that measured with a ruler/measuring tape equal the hypotenuse you calculated?
Why do you think they are or are not exactly the same?
What did you learn from developing this presentation?

We are assuming the length of one side x:

x = 18 cm

Length of other side y:

y = 15 cm

And length of the hypotenuse z:

z = 23.5 cm

From the Pythagoras theorem:

z² = x² + y²

z² = 18² + 15²

z² = 549

z = √54 = 23.43 cm

The measure of hypotenuse is:

Using ruler: x = 25.5 cm

Using formula: x = 23.43 cm

They are not equal due to the reading in the ruler.

The formula can be used to in the real life problems as well.

Thus, the measure of hypotenuse using the ruler is 25.5 cm and using the formula is 23.43 cm, they are not same, and formula can be used to in the real life problems as well.

Learn more about the right angle triangle here:

brainly.com/question/3770177

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8 0

1 year ago

If you know that 1% of a population has a certain disease. A test was made. This test is positive in 95% of the people with the

LenKa [72]

Answer:

ali probley has a 50% chance

Step-by-step explanation:

6 0

2 years ago

Find all possible values of α+

const2013 [10]

Answer:

$\rm\displaystyle 0,\pm\pi$

Step-by-step explanation:

please note that to find but α+β+γ in other words the sum of α,β and γ not α,β and γ individually so it's not an equation

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

we want to find all possible values of α+β+γ when tanα+tanβ+tanγ = tanαtanβtanγ to do so we can use algebra and trigonometric skills first

cancel tanγ from both sides which yields:

$\rm\displaystyle \tan( \alpha ) + \tan( \beta ) = \tan( \alpha ) \tan( \beta ) \tan( \gamma ) - \tan( \gamma )$

factor out tanγ:

$\rm\displaystyle \tan( \alpha ) + \tan( \beta ) = \tan( \gamma ) (\tan( \alpha ) \tan( \beta ) - 1)$

divide both sides by tanαtanβ-1 and that yields:

$\rm\displaystyle \tan( \gamma ) = \frac{ \tan( \alpha ) + \tan( \beta ) }{ \tan( \alpha ) \tan( \beta ) - 1}$

multiply both numerator and denominator by-1 which yields:

$\rm\displaystyle \tan( \gamma ) = - \bigg(\frac{ \tan( \alpha ) + \tan( \beta ) }{ 1 - \tan( \alpha ) \tan( \beta ) } \bigg)$

recall angle sum indentity of tan:

$\rm\displaystyle \tan( \gamma ) = - \tan( \alpha + \beta )$

let α+β be t and transform:

$\rm\displaystyle \tan( \gamma ) = - \tan( t)$

remember that tan(t)=tan(t±kπ) so

$\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta\pm k\pi )$

therefore when k is 1 we obtain:

$\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta\pm \pi )$

remember Opposite Angle identity of tan function i.e -tan(x)=tan(-x) thus

$\rm\displaystyle \tan( \gamma ) = \tan( -\alpha -\beta\pm \pi )$

recall that if we have common trigonometric function in both sides then the angle must equal which yields:

$\rm\displaystyle \gamma = - \alpha - \beta \pm \pi$

isolate -α-β to left hand side and change its sign:

$\rm\displaystyle \alpha + \beta + \gamma = \boxed{ \pm \pi }$

when is 0:

$\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta \pm 0 )$

likewise by Opposite Angle Identity we obtain:

$\rm\displaystyle \tan( \gamma ) = \tan( -\alpha -\beta\pm 0 )$

recall that if we have common trigonometric function in both sides then the angle must equal therefore:

$\rm\displaystyle \gamma = - \alpha - \beta \pm 0$

isolate -α-β to left hand side and change its sign:

$\rm\displaystyle \alpha + \beta + \gamma = \boxed{ 0 }$

and we're done!

8 0

2 years ago

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