All categories

3 years ago

Which expressions are equivalent to 12r-5?

Mathematics

1 answer:

Yanka [14]3 years ago

7 0

Answer:

Step-by-step explanation: 12r-6+1= 12r-5

-8+12r+3= 12r-5

A and B

You might be interested in

Please help me!!! Will mark Brainliest!!!

Kruka [31]

Answer:

Mean = 28$ MAD = 3.6$

Step-by-step explanation:

Mean-

Every day added= 140/5 = 28

MAD-

It is hard to explain

6 0

3 years ago

Classify each angle.

serg [7]

Answer:

∠MDP = 90° = Right angle

∠MDP = 180° = Straight angle

∠ODM = 90° = Right angle

∠OND = Not an angle

∠NDO = 90° = Right angle

∠NDM = 180° = Straight angle

∠PMN = Not an angle

∠PDO = 180° = Straight angle

∠PDN = 90° = Right angle

If my answer helped, please mark me as the brainliest!

Thank You !

7 0

3 years ago

The position function of a particle in rectilinear motion is given by s(t) = 2t3 – 21t2 + 60t + 3 for t ≥ 0 with t measured in s

Norma-Jean [14]

The positions when the particle reverses direction are:

$s(t_1)=55ft\\\\s(t_2)=28ft$

The acceleraton of the paticle when reverses direction is:

$a(t_1)=-18\frac{ft}{s^{2}}\\ \\a(t_2)=a(5s)=18\frac{ft}{s^{2}}$

Why?

To solve the problem, we need to remember that if we derivate the position function, we will get the velocity function, and if we derivate the velocity function, we will get the acceleration function. So, we will need to derivate two times.

Also, when the particle reverses its direction, the velocity is equal to 0.

We are given the following function:

$s(t)=2t^{3}-21t^{2}+60t+3$

So,

- Derivating to get the velocity function, we have:

$v(t)=\frac{ds}{dt}=(2t^{3}-21t^{2}+60t+3)\\\\v(t)=3*2t^{2}-2*21t+60*1+0\\\\v(t)=6t^{2}-42t+60$

Now, making the function equal to 0, to find the times when the particle reversed its direction, we have:

$v(t)=6t^{2}-42t+60\\\\0=6t^{2}-42t+60\\\\0=t^{2}-7t+10\\(t-5)*(t-2)=0\\\\t_{1}=5s\\t_{2}=2s$

We know that the particle reversed its direction two times.

- Derivating the velocity function to find the acceleration function, we have:

$a(t)=\frac{dv}{dt}=6t^{2}-42t+60\\\\a(t)=12t-42$

Now, substituting the times to calculate the accelerations, we have:

$a(t_1)=a(2s)=12*2-42=-18\frac{ft}{s^{2}}\\ \\a(t_2)=a(5s)=12*5-42=18\frac{ft}{s^{2}}$

Now, substitutitng the times to calculate the positions, we have:

$s(t_1)=2*(2)^{3}-21*(2)^{2}+60*2+3=16-84+120+3=55ft\\\\s(t_2)=2*(5)^{3}-21*(5)^{2}+60*5+3=250-525+300+3=28ft$

Have a nice day!

3 0

3 years ago

You plant a garden in the shape of a triangle as shown in the figure. What is the perimeter of the garden? Find the values of x

Vadim26 [7]

Answer:

x = 42°

y = 69°

Step-by-step explanation:

From the picture attached,

Given triangle is an isosceles triangle.

Therefore, two sides will be equal and measure 8 yd.

Measure of 3rd side of the triangle = 6 yd

Perimeter of the garden = 8 + 8 + 6

= 22 yd

In an isosceles triangle, opposite angles of the equal sides will be equal.

Therefore, y = 69°

By the property of interior angles of a triangle,

x° + y° + 69° = 180°

x° + 69° + 69° = 180°

x = 180 - 138

x = 42°

7 0

3 years ago

Which relationship describes angles 1 and 2?

Alexxx [7]

Conplementary angles

4 0

3 years ago

Read 2 more answers