Ask question

Ask question

All categories

3 years ago

5

A six foot man is standing 10 feet away from a 20 foot lamppost. what is the length of his shadow?

2 answers:

timofeeve [1]3 years ago

7 0

Answer:

4.29 feet ( approx )

Step-by-step explanation:

Let x be the length of the shadow of man,

Given,

The height of the man = 6 feet,

The height of the lamppost = 20 foot,

Distance of the man from the lampost = 10 feet,

By making the diagram,

We found two similar triangles in which first triangle having sides 6 and x,

Second triangle having sides 20 and x + 10,

∵ The corresponding sides of similar triangle are in same proportion,

$\implies \frac{x}{x+10}=\frac{6}{20}$

$20x = 6x + 60$

$20x - 6x = 60$

$14x = 60$

$\implies x = \frac{60}{14}=4.28571428571\approx 4.29$

Hence, the length of the shadow is approximately 4.29 feet.

elena55 [62]3 years ago

4 0

There are two congruent triangles that are being formed from the lamppost, the light, the ground and the man. Always try to draw a picture for these types of problems. It helps a lot when solving. Your answer should be about 4.3ft

You might be interested in

Which answer is right?????

cricket20 [7]

Answer:

See the attachment

Step-by-step explanation:

The point of the dashed line y=x in the problem statement graph is that the inverse function is a reflection of the function across that line. (y and x are interchanged) The graph of selection C has the appropriate pair of curves.

8 0

3 years ago

Find the set of values of k for which the line y=kx-4 intersects the curve y=x²-2x at 2 distinct points?

Sonbull [250]

Answer:

$-6 < k < 2$

Step-by-step explanation:

Given

$y = x^2 - 2x$

$y =kx -4$

Required

Possible values of k

The general quadratic equation is:

$ax^2 + bx + c = 0$

Subtract $y = x^2 - 2x$ and $y =kx -4$

$y - y = x^2 - 2x - kx +4$

$0 = x^2 - 2x - kx +4$

Factorize:

$0 = x^2 +x(-2 - k) +4$

Rewrite as:

$x^2 +x(-2 - k) +4=0$

Compare the above equation to: $ax^2 + bx + c = 0$

$a = 1$

$b= -2-k$

$c =4$

For the equation to have two distinct solution, the following must be true:

$b^2 - 4ac > 0$

So, we have:

$(-2-k)^2 -4*1*4>0$

$(-2-k)^2 -16>0$

Expand

$4 +4k+k^2-16>0$

Rewrite as:

$k^2 + 4k - 16 + 4 >0$

$k^2 + 4k - 12 >0$

Expand

$k^2 + 6k-2k - 12 >0$

Factorize

$k(k + 6)-2(k + 6) >0$

Factor out k + 6

$(k -2)(k + 6) >0$

Split:

$k -2 > 0$ or $k + 6> 0$

So:

$k > 2$ or k $> -6$

To make the above inequality true, we set:

$k < 2$ or $k >-6$

So, the set of values of k is:

$-6 < k < 2$

3 0

3 years ago

Find b in triangle TRG. Please help, I'm not sure how to solve for b.

Marina CMI [18]

Answer:

b=4.9

C. is the correct option.

Step-by-step explanation:

In triangle TGN,

Let <RTG be x

cosx= b/h

or, cosx = 2/a

Again,

In triangle TRG,

cosx = b/h

(for bigger right angled triangle)

cosx = a/(4+2)

cosx = a/6

Now

cosx = 2/a = a/6

or, 2/a = a/6

or, 12 = a²

so, a² = 12

Now,

For TRG,

h² = p²+b²

or, 6² = b² + a²(a²=12)

or, 6² = b² + 12

or, 36 = b² + 12

or, 36 -12 = b²

or, b² = 24

or, b = square root 24

so, b = 4.8989

so, b = 4.9

7 0

2 years ago

Daron purchased a cell phone case from the corner store. Daron paid with a $20 bill. The picture to the left shows his change. H

svlad2 [7]

Do you have a picture of the question

3 0

3 years ago

What is the value of x

Archy [21]

I think it might be a 10

8 0

4 years ago