G(t)= t^2 and f(x)= 1 + x f(2g(1)) A. 1 B. 3 C. 30 D. -1

The product of -5 and the number is 35 what is the number a. 40 b. 30 c. -7 d. 7

skad [1K]

Step-by-step explanation:

-5y=35

=35÷(-5)

= -7

=-5×-7

=35

6 0

2 years ago

What is the size of angle c?

lesantik [10]

Answer:

55°

Step-by-step explanation:

d = 180 - 80 = 100

100 + 130 + 90 + (180 - 85) + (180 - c)

= 540

595 - c = 540

c = 595 - 540

c = 55°

6 0

3 years ago

The common factor to 18xy and 24 y

vladimir1956 [14]

18xy + 24y = 6y(3x + 8)

The common factor is 6y

6 0

3 years ago

A street light is mounted at the top of a 15-ft-tall pole. A man 6 ft tall walks away from the pole with a speed of 4 ft/s along

mina [271]

Answer:

The tip of the man shadow moves at the rate of $\frac{20}{3} ft.sec$

Step-by-step explanation:

Let's draw a figure that describes the given situation.

Let "x" be the distance between the man and the pole and "y" be distance between the pole and man's shadows tip point.

Here it forms two similar triangles.

Let's find the distance "y" using proportion.

From the figure, we can form a proportion.

$\frac{y - x}{y} = \frac{6}{15}$

Cross multiplying, we get

15(y -x) = 6y

15y - 15x = 6y

15y - 6y = 15x

9y = 15x

y = $\frac{15x}{9\\} y = \frac{5x}{3}$

We need to find rate of change of the shadow. So we need to differentiate y with respect to the time (t).

$\frac{dy}{t} = \frac{5}{3} \frac{dx}{dt}$ ----(1)

We are given $\frac{dx}{dt} = 4 ft/sec$ . Plug in the equation (1), we get

$\frac{dy}{dt} = \frac{5}{3} *4 ft/sect\\= \frac{20}{3} ft/sec$

Here the distance between the man and the pole 45 ft does not need because we asked to find the how fast the shadow of the man moves.

7 0

3 years ago

Express as a difference: a+b, a+5

Tcecarenko [31]

Answer:

a - (-b), a - (-5)

Step-by-step explanation:

A double negative negative is a positive, so a - (-b) = a + b, and a - (-5) is a + 5. However, a double negative is still a difference, so this answer works.

6 0

3 years ago