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3 years ago

6

HEEEEELP The height of a street light is 50 feet. It casts a 40−foot shadow. At the same time, a man standing next to the street

light casts a 5−foot shadow. How tall is the man?

1 answer:

Ganezh [65]3 years ago

5 0

Answer:

5 foot

Step-by-step explanation:

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Find the measure of each side indicated. Round to the nearest tenth.

mihalych1998 [28]

Answer:

D. 25.3

Step-by-step explanation:

tan∅ = opposite over adjacent

Step 1: Write equation

tan66.5° = x/11

Step 2: Multiply both sides by 11

11tan66.5° = x

Step 3: Evaluate

x = 25.2983

x ≈ 25.3

4 0

3 years ago

Read 2 more answers

Which polynomial equation of least degree has -2, -2, 3, and 3 as four of its roots? (x + 2)(x - 3) = 0 (x - 2)-2(x + 3)3 = 0 (x

azamat

Answer:

(x + 2)²(x - 3)² = 0

Step-by-step explanation:

Since we have a degree of 2 and double of the same roots, we know that each root would have a multiplicity of 2. Therefore, our answer is(x + 2)²(x - 3)² = 0

6 0

4 years ago

10 , 12 , 6 ,1 ,9 ,16 , 20, 17,8<br><br> Median ???

allochka39001 [22]

Answer:

its 10

Step-by-step explanation:

You have to put them from least to most and see the middle number.

5 0

3 years ago

Read 2 more answers

Find the value of the variable y, where the sum of the fraction 2/y-3 and 6/y+3 is equal to the quotient.

NISA [10]

Answer:

Here we need to solve:

$\frac{2}{y - 3} + \frac{6}{y + 3 } = \frac{\frac{2}{y-3}}{\frac{6}{y + 3} }$

The sum of the fractions is equal to the quotient between the fractions.

Notice that the two values:

y = 3

y = -3

make the denominator equal to zero, so those values are restricted.

We can simplify the right side to get:

$\frac{2}{y - 3} + \frac{6}{y + 3 } = \frac{\frac{2}{y-3}}{\frac{6}{y + 3} } = \frac{2*(y + 3)}{6*(y - 3)} = 3*\frac{y + 3}{y - 3}$

Now we can multiply both sides by (y - 3)

$(y - 3)*(\frac{2}{y - 3} + \frac{6}{y + 3 }) = 3*(y + 3)\\2 + 6*\frac{y -3}{y + 3} = 3*(y + 3)$

Now we can multiply both sides by (y + 3)

$(2 + 6*\frac{y -3}{y + 3})*(y + 3) = 3*(y + 3)*(y + 3)$

$2*(y + 3) + 6*(y - 3) = 3*(y + 3)*(y + 3)\\\\2*y + 6 + 6*y - 18 = 3*(y^2 + 2*y*3 + 9)\\\\8*y - 12 = 3*y^2 + 6*y + 33\\\\0 = 3*y^2 + 6*y + 33 - 8*y + 12\\\\0 = 3*y^2 - 2*y + 45$

First, let's see the determinant of that quadratic equation:

$D = (-2)^2 - 4*3*45 = -536$

We can see that it is negative, thus, there are no real solutions of the equation.

Thus, there is no value of y such that the origina equation is true,

6 0

3 years ago

What are the length and width of a rectangle if the length exceeds the width by 55 inches and the perimeter is 270 inches?

astra-53 [7]

Here, P = 2L + 2W, and L=W+55. Subbing W+55 for L in P = 2L + 2W, we get:

P = 2(W+55) + 2W = 270 in
2W + 110 + 2W = 270 in, or 4W = 160 in. Then W=40 in, and L=(40+55)in.

3 0

4 years ago