Find the values that will make the equations true.

lbvjy [14]

Angle of elevation L = angle JKL (alternate interior) 55°

tan (angle JKL) = OPPOSITE SIDE / ADJACENT SIDE

tan(55°) = JL/26

1.4281 = JL/26 and JL = 26 x 1.4281 =37.13 or JL = 37.1

8 0

3 years ago

The sum of 8 and a number h is 12 into an equation

Mademuasel [1]

Answer: 4

Step-by-step explanation:

The sum of 12 and 8 is 12 add 8 which is 20. The difference between 12 and 8 is 12 minus 8 which is 4.

I don't know if this is right but if it is not tell me plz.

:) Thx

6 0

3 years ago

A 4-foot tall child walks directly away from a 12-foot tall lamppost at 2 mph. How quickly is the length of her shadow increasin

xz_007 [3.2K]

Answer:

The length of the shadow is increasing with the rate of 1.5 feet per sec

Step-by-step explanation:

Let AB and CD represents the height of the lamppost and child respectively ( shown below )

Also, let E be a point represents the position of child.

In triangles ABE and CDE,

$\angle ABE\cong \angle CDE$ ( right angles )

$\angle AEB\cong \angle CED$ ( common angles )

By AA similarity postulate,

$\triangle ABE\sim \triangle CDE$

∵ Corresponding sides of similar triangles are in same proportion,

$\implies \frac{AB}{CD}=\frac{BE}{DE}$

We have, AB = 12 ft, CD = 4 ft, BE = BD + DE = 6 + DE,

$\implies \frac{12}{4}=\frac{6+DE}{DE}$

$12DE = 24 + 4DE$

$8DE = 24$

$DE=3$

Now, the speed of walking = 2 mph = $\frac{2\times 5280}{3600}\approx 2.933\text{ ft per sec}$

Note: 1 mile = 5280 ft, 1 hour = 3600 sec

Thus, the time taken by child to reach at E

$= \frac{\text{Walked distance}}{\text{Walking speed}}$

$=\frac{6}{2.933}$

= 2.045 hours

Hence, the change rate in the length of shadow

$= \frac{\text{Length of shadow}}{\text{Time taken}}$

$=\frac{3}{2.045}$

= 1.5 ft per sec.

3 0

3 years ago

I don't understand how to do this problem

Naya [18.7K]

It is 270 i believe hope it helps

8 0

3 years ago

I can’t figure out how to use the zeros in the polynomial. Please explain

Temka [501]

Answer:

a) $P (x) = (x + 3) (x-1) (x-4)$

b) $P (x) = (2x + 5) (5x - 4) (x-6)$

c) $P (x) = (x-3) (x-1) (x-4) (x + 1) ^ 2$

Step-by-step explanation:

For the question a * you need to find a polynomial of degree 3 with zeros in -3, 1 and 4.

This means that the polynomial P(x) must be zero when x = -3, x = 1 and x = 4.

Then write the polynomial in factored form.

$P (x) = (x + 3) (x-1) (x-4)$

Note that this polynomial has degree 3 and is zero at x = -3, x = 1 and x = 4.

For question b, do the same procedure.

Degree: 3

Zeros: -5/2, 4/5, 6.

The factors are

$x = -\frac{5}{2}\\\\x +\frac{5}{2} = 0\\\\(2x +5) = 0$

---------------------------------------

$x =\frac{4}{5}\\\\x-\frac{4}{5} = 0\\\\(5x-4) = 0$

--------------------------------------

$x = 6\\\\(x-6) = 0$

--------------------------------------

$P (x) = (2x + 5) (5x - 4) (x-6)$

Finally for the question c we have

Degree: 5

Zeros: -3, 1, 4, -1

Multiplicity 2 in -1

$x = -3\\\\(x-3) = 0$

--------------------------------------

$x = 1\\\\(x-1) = 0$

--------------------------------------

$x = 4\\\\(x-4) = 0$

----------------------------------------

$x = -1\\\\(x + 1) = 0$

-----------------------------------------

$P (x) = (x-3) (x-1) (x-4) (x + 1) ^ 2$

8 0

3 years ago