Which graph shows the inequality n>2

The base of a prism is a right triangle with a base and height measurement of 12cm and a 5cm. If the volume of the prism is 210c

bezimeni [28]

Answer:

The height is 7 cm

Step-by-step explanation:

210 × 2 = 420

12 × 5 = 60

420 ÷ 60 = 7

6 0

3 years ago

Read 2 more answers

Someone please help me. ASAP

gizmo_the_mogwai [7]

Answer: x=10

Step-by-step explanation:

1/5(x-2)=1/10(x+6)

1/5x -2/5=1/10x + 6/10

+2/5 +2/5

1/5x =1/10x + 1

-1/10 -1/10

1/10x= 1 divide both sides by 1/10

x= 10

3 0

2 years ago

A triangle has side lengths measuring 3x cm, 7x cm, and hcm Which expression describes the possible values of h, in cm?

GREYUIT [131]

Answer:

OLX is your answer if 10 x

4 0

2 years ago

A field in the shape of a right triangle is bordered on one side by a stream, on another side by a road, and on a third side by

trasher [3.6K]

Answer:

42°

Step-by-step explanation:

For this problem, it may help you to draw it out to visualize it. You want to find the angle between the stream and road. Based on the description the triangle looks like this.

/ |

Stream(27) / | Road( 20)

/__|

Fence

You know the length of the road is 20, and the stream is 27. The angle can be found by using Cos∅=a/h. Adjacent to where the road and stream meet is the road, 20, and hypotenuse is the stream, 27.

So 20/27 = .74 = Cos∅

To find the angle ∅ use inverse cosine to get 42°.

Hope that helps and let me know if I did anything incorrectly!

5 0

2 years ago

you drop a ball from a height of 98 feet. at the same time, your friend throws a ball upward. the polynomials represent the heig

Ostrovityanka [42]

a) $h_0 -u_y t$

b) See interpretation below

Step-by-step explanation:

a)

The motion of both balls is a free-fall motion: it means that the ball is acted upon the force of gravity only.

Therefore, this means that the motion of the ball is a uniformly accelerated motion, with constant acceleration equal to the acceleration of gravity:

$g=32 ft/s^2$

in the downward direction.

For the ball dropped from the initial height of $h_0 = 98 ft$ , the height at time t is given by

$h(t) = h_0 -\frac{1}{2}gt^2$ (1)

The ball which is thrown upward from the ground instead is fired with an initial vertical velocity $u_y$ , and its starting height is zero, so its position at time t is given by

$h'(t)=u_y t - \frac{1}{2}gt^2$ (2)

Therefore, the polynomial that represents the distance between the two balls is:

$h(t)-h'(t)=h_0 - \frac{1}{2}gt^2 - (u_y t - \frac{1}{2}gt^2) = h_0 -u_y t$

b)

Now we interpret this polynomial, which is:

$\Delta h(t) = h_0 -u_y t$

which represents the distance between the two balls at time t.

The interpretation of the two terms is the following:

- The constant term, $h_0$ , is the initial distance between the two balls, at time t=0 (in fact, the first ball is still at the top of the building, while the second ball is on the ground). For this problem, $h_0 = 98 ft$

- The coefficient of the linear term, $u_y$ , is the initial velocity of the second ball; this terms tells us that the distance between the two balls decreases every second by $u_y$ feet.

5 0

3 years ago