Simplify.

M_6 is (2x – 5)º and m_8 is (x + 5)º.

HACTEHA [7]

m_6 + m_8 = 180º because they form a straight line.

So, (2x-5) + (x+5) = 180

3x = 180

x = 60

So, m_115º (2•60 - 5 = 115)

Also, because the two lines are parallel m_6 = m_3 by alternate interior angles.

So, m_3 = 115º

4 0

2 years ago

The mathematics department of a college has 6 male professors, 9 female professors, 5 male teaching assistants, and 6 female tea

Hitman42 [59]

Given:

6 male professores

9 female professores

5 male teaching assistants

6 female teaching assistants

Sol:.

$N(A\text{ or B)=N(A)+N(B)-N(A and B)}$

N (professors)

$\begin{gathered} =6+9 \\ =15 \end{gathered}$

N(Male)

$\begin{gathered} =6+5 \\ =11 \end{gathered}$

N(professors and male)

$=6$

N(professors OR male) = N(professors) + N(males) -N(professor OR male)

$\begin{gathered} N(\text{ Professors OR male)=15+11-6} \\ =20 \end{gathered}$

N(People to choose from)

$\begin{gathered} =6+9+5+6 \\ =26 \end{gathered}$

Then probablitiy is:

$\begin{gathered} =\frac{20}{26} \\ =\frac{10}{13} \end{gathered}$

Then the probability is 10/13

6 0

1 year ago

45% of what number is 27?

gtnhenbr [62]

45% of what number is 27

0.45x = 27
x = 27 / 0.45
x = 60 <== 45% of 60 = 27

5 0

3 years ago

The Fifth Grade classes are going on a field trip. Mrs. Lafont fills an ice chest with drinks. The ice chest holds 122 cubic inc

forsale [732]

122+118=240 because you add it.

5 0

3 years ago

Read 2 more answers

An object is launched from a launching pad 144 ft. above the ground at a velocity of 128ft/sec. what is the maximum height reach

ollegr [7]

Answer:

18) a. h(x) = -16x² + vx + h(0) ⇒ h(x) = -16x² + 128x + 144

b. The maximum height = 400 feet

c. Attached graph

19) The rocket will reach the maximum height after 4 seconds

20) The rocket hits the ground after 9 seconds

Step-by-step explanation:

* Lets study the rule of motion for an object with constant acceleration

# The distance S = ut ± 1/2 at², where u is the initial velocity, t is the time

and a is the acceleration of gravity

# The vertical distances h in x second is h(x) - h(0), where h(0)

is the initial height of the object above the ground

∵ h(x) = vx + 1/2 ax², where h is the vrtical distance, v is the initial

velocity, a is the acceleration of gravity (32 feet/second²) and x

is the time

18)a.

∵ The value of a = -32 ft/sec² ⇒ negative because the direction

of the motion

is upward

∴ h(x) - h(0) = vx - (1/2)(32)(x²) ⇒ (1/2)(32) = 16

∴ h(x) = vx - 16x² + h(0)

∴ h(x) = -16x² + vx + h(0) ⇒ proved

* Find the height of the object after x seconds from the ground

∵ h(0) = 144 and v = 128 ft/sec

∴ h(x) = -16x² + 128x + 144

b.

* At the maximum height h'(x) = 0

∵ h'(x) = -32x + 128

∴ -32x + 128 = 0 ⇒ subtract 128 from both sides

∴ -32x = -128 ⇒ ÷ -32

∴ x = 4 seconds

- The time for the maximum height = 4 seconds

- Substitute this value of x in the equation of h(x)

∴ The maximum height = -16(4)² + 128(4) + 144 = 400 feet

c. Attached graph

19)

- The object will reach the maximum height after 4 seconds

20)

- When the rocket hits the ground h(x) = 0

∵ h(x) = -16x² + 128x + 144

∴ 0 = -16x² + 128x + 144 ⇒ divide the two sides by -16

∴ x² - 8x - 9 = 0 ⇒ use the factorization to find the value of x

∵ x² - 8x - 9 = 0

∴ (x - 9)( x + 1) = 0

∴ x - 9 = 0 OR x + 1 = 0

∴ x = 9 OR x = -1

- We will rejected -1 because there is no -ve value for the time

* The time for the object to hit the ground is 9 seconds

8 0

3 years ago