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3 years ago

What is the area of the triangle?

Mathematics

1 answer:

Vika [28.1K]3 years ago

4 0

Answer:

9 units²

Step-by-step explanation:

Area of triangle= ½ × base × height

The base here is 3 units and the height is 6 units.

Area of traingle

= ½(3)(6)

= 9 units²

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Where do the parentheses go and solve 3×14=3×2×7=3×2×7=×7

grin007 [14]

Answer:

I don't think this is a real math equation.

Step-by-step explanation:

8 0

3 years ago

If f ( x ) = 2 x - 5, then f (4) is _____. 3 5 4 17

ICE Princess25 [194]

The answer is 3

if you substitute x as 4 then it would be 2 x 4 = 8

8-5=3

7 0

2 years ago

Read 2 more answers

Your plan was to be on the road by 9 A.M. but you did not leave the garage until 10 A.M. You then drove with the cruise control

Wittaler [7]

Answer:

The average speed over the time interval from 9 A.M. to noon was of 33.33 mph.

Step-by-step explanation:

The average speed is given by the following equation:

$v = \frac{d}{t}$

From 10A.M. to noon.

You traveled at 50mph during 2 hours. So how long you traveled?

$50 = \frac{d}{2}$

$d = 50*2$

$d = 100$

You traveled 100 miles.

What was your average speed over the time interval from 9 A.M. to noon

From 9 A.M. to 10 A.M, you did not travel.

From 10 P.M. to noon, 100 miles.

So 100 miles during 3 hours.

$v = \frac{d}{t} = \frac{100}{3} = 33.33$

The average speed over the time interval from 9 A.M. to noon was of 33.33 mph.

8 0

3 years ago

Show your work −126 = 14k

Yuliya22 [10]

Answer: k = -9

Step-by-step explanation:

-126 = 14k

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

14 14

-126 divided by 14 = -9

k = -9

4 0

3 years ago

The height H of an ball that is thrown straight upward from an initial position 3 feet off the ground with initial velocity of 9

mariarad [96]

Answer:

The ball will be 84 feet above the ground 1.125 seconds and 4.5 seconds after launch.

Step-by-step explanation:

Statement is incorrect. Correct form is presented below:

The height $h(t)$ of an ball that is thrown straight upward from an initial position 3 feet off the ground with initial velocity of 90 feet per second is given by equation $h(t) = 3 +90\cdot t -16\cdot t^{2}$ , where $t$ is time in seconds. After how many seconds will the ball be 84 feet above the ground.

We equalize the kinematic formula to 84 feet and solve the resulting second-order polynomial by Quadratic Formula to determine the instants associated with such height:

$3+90\cdot t -16\cdot t^{2} = 84$

$16\cdot t^{2}-90\cdot t +81 = 0$ (1)

By Quadratic Formula:

$t_{1,2} = \frac{90\pm \sqrt{(-90)^{2}-4\cdot (16)\cdot (81)}}{2\cdot (16)}$

$t_{1} = 4.5\,s$ , $t_{2} = 1.125\,s$

The ball will be 84 feet above the ground 1.125 seconds and 4.5 seconds after launch.

3 0

3 years ago