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pantera1 [17]
3 years ago
13

What angle relationship describes ∠UQR and ∠WRT?

Mathematics
2 answers:
sashaice [31]3 years ago
5 0

Answer:

corresponding angles because it is in a F pattern shape

always corresponding angle is denoted from F pattern and its sum is always equal

damaskus [11]3 years ago
5 0

Answer:

corresponding angles

Step-by-step explanation:

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A cuboid, with dimensions 28cm by 16cm by 15cm, has a volume of 6720cm cube. The cuboid is melted to form smaller cubes of lengt
Dovator [93]

Answer:

Vol of the cube =Vol of cubiod

Vol of each smaller cube =s×s×s= 4×4×4=64

Vol of cuboid /Vol of each smaller cube = no of cubes that can be obtained

6720/64= 105

Hence 105 cubes can be obtained

3 0
3 years ago
In parallelogram ABCD above, AE =2x+4 and AC=6x. What is the length of CE?
lawyer [7]

Answer:

4-4x

Step-by-step explanation:

From the parallelogram

CE = CA+AE

Since CA = -AC

CE = -AC+AE

GIVEN

AC = 6x

AE= 2x+4

Substitute

CE = -6x+2x+4.

CE = -4x+4

Gene the length of CE is 4-4x

4 0
2 years ago
Identify the property demonstrated by the equation.
Dima020 [189]

option B is the correct answer friend.

it is distributive property

4 0
3 years ago
Find ratio equal to 12:7
Dmitry [639]


12:7
12(2):7(2)
24:14

12:7 = 24:14
I hoped I helped(:
6 0
3 years ago
Read 2 more answers
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
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