Please help

Practice exercise 1-5 number line

jolli1 [7]

Answer:

Wait there is no exercise 1-5 number line maybe its like this? 1-2-3-4-5

^2

Step-by-step explanation:

5 0

3 years ago

HELP WITH THE CORRECT ANSWER

castortr0y [4]

Take 30% of all the numbers from colors and add up

4 0

3 years ago

Please help me really quick pleasee answer choices: a)2 b)11 c)20 d)29

mihalych1998 [28]

Answer:

C. $x=20^{\circ}$

Step-by-step explanation:

Angle 6 and angle with measure 45° are vertical angles (opposite angles whan two lines intersect). By Vertical Angles Theorem, vertical angles are congruent, so

$m\angle 6=45^{\circ}$

Angle 6 and angle with measure $5x+35^{\circ}$ are the same side interior angles when two parallel lines are cut by transversal. The same side interior angles add up to 180°, then

$m\angle 6+5x+35^{\circ}=180^{\circ}\\ \\45^{\circ}+5x+35^{\circ}=180^{\circ}\\ \\5x=180^{\circ}-45^{\circ}-35^{\circ}\\ \\5x=100^{\circ}\\ \\x=20^{\circ}$

7 0

3 years ago

Someone has a Visa Card with an annual percentage rate of 16.8%. The unpaid balance for his June billing cycle is $1009.49. Du

iragen [17]

69420694206942069420

4 0

3 years ago

A ball is thrown vertically in the air with a velocity of 95 ft/s. The ball is at a height of 120 ft.

sattari [20]

Answer:

The ball is at a height of 120 feet after 1.8 and 4.1 seconds.

Step-by-step explanation:

The statement is incomplete. The complete statement is perhaps the following "A ball is thrown vertically in the air with a velocity of 95 ft/s. What time in seconds is the ball at a height of 120ft. Round to the nearest tenth of a second."

Since the ball is launched upwards, gravity decelerates it up to rest and moves downwards. The position of the ball can be determined as a function of time by using this expression:

$y = y_{o} + v_{o}\cdot t +\frac{1}{2}\cdot g \cdot t^{2}$

Where:

$y_{o}$ - Initial height of the ball, measured in feet.

$v_{o}$ - Initial speed of the ball, measured in feet per second.

$g$ - Gravitational constant, equal to $-32.174\,\frac{ft}{s^{2}}$ .

$t$ - Time, measured in seconds.

Given that $y_{o} = 0\,ft$ , $v_{o} = 95\,\frac{ft}{s}$ , $g = -32.174\,\frac{ft}{s^{2}}$ and $y = 120\,ft$ , the following second-order polynomial is found:

$120\,ft = 0\,ft + \left(95\,\frac{ft}{s} \right)\cdot t +\frac{1}{2}\cdot \left(-32.174\,\frac{ft}{s^{2}} \right) \cdot t^{2}$

$-16.087\cdot t^{2} + 95\cdot t -120 =0$

The roots of this polynomial are, respectively:

$t_{1} \approx 4.075\,s$ and $t_{2} \approx 1.831\,s$ .

Both roots solutions are physically reasonable, since $t_{1}$ represents the instant when the ball reaches a height of 120 ft before reaching maximum height, whereas $t_{2}$ represents the instant when the ball the same height after reaching maximum height.

In nutshell, the ball is at a height of 120 feet after 1.8 and 4.1 seconds.

8 0

4 years ago