All categories

3 years ago

Im sorry im spamming but i really need help with this question, if someone could help please

Mathematics

1 answer:

Anni [7]3 years ago

5 0

Answer:

AB, BC and AC.

Step-by-step explanation:

The length of a side of a triangle is proportional to the angle across it.

Here the order of angles are, from smallest to largest, C, A and B.

The corresponding order of lengths (of the opposite sides) are therefore:

AB, BC and AC.

You might be interested in

A point located at (6, -4) is reflected over the x -axis. What are the coordinates of the image?

Vika [28.1K]

Ur answer is A or number 1

7 0

3 years ago

In Jerome's physics class, he had to solve the equation 40 = 12mv2 for m. Each step of his work is shown below.

Aleksandr-060686 [28]

Answer:

Step-by-step explanation:

I think you mean 1/2 for 12. I will assume so

40 = (1/2) m v^2 Multiply both sides by 2

40*2 = (1/2)*2 *m * v^2 Combine the 2s cancel.

80 = m * v^2 Divide by v^2

80/v^2 = m * v^2/v^2 Cancel

80/v^2 = m

Since I don't have the drop down menus, you'll have do the matching.

6 0

3 years ago

Read 2 more answers

What are the answers for the multiplying decimals

AleksAgata [21]

Starting on the right, multiply each digit in the top number by each digit in the bottom number, just as with whole numbers

8 0

4 years ago

You weigh a gallon of 2% milk in science class and learn that is approximately 8.4 pounds. You pass the milk to the next group,

iris [78.8K]

1 pound is 0.453592 kilograms
so 8.4 pounds is 3.81018 kilograms
You would cross multiply and make sure the units are lined up with each other

1 $\frac{1 lb}{0.453592 kg} = \frac{8.4 lb}{x}$
then you would cross multiply and get 3.81018

6 0

4 years ago

Read 2 more answers

assume x and y are both differentiable functions of t. find dx/dt given x=-1 and dy/dt=8 for the relation: 4x^2+3y^3=28

melomori [17]

Given:

x and y are both differentiable functions of t.

$4x^2+3y^3=28$

$x=-1\text{ and }\dfrac{dy}{dt}=8$

To find:

The value of $\dfrac{dx}{dt}$ .

Solution:

We have,

$4x^2+3y^3=28$ ...(i)

At x=-1,

$4(-1)^2+3y^3=28$

$4+3y^3=28$

$3y^3=28-4$

$3y^3=24$

Divide both sides by 3.

$y^3=8$

Taking cube root on both sides.

$y=2$

So, y=2 at x=-1.

Differentiate (i) with respect to t.

$4(2x\dfrac{dx}{dt})+3(3y^2\dfrac{dy}{dt})=0$

Putting x=-1, y=2 and $\dfrac{dy}{dt}=8$ , we get

$4(2(-1)\dfrac{dx}{dt})+3(3(2)^28)=0$

$-8\dfrac{dx}{dt}+9(4)(8)=0$

$-8(\dfrac{dx}{dt}-9(4))=0$

Divide both sides by -8.

$\dfrac{dx}{dt}-36=0$

$\dfrac{dx}{dt}=36$

Therefore, the value of $4x^2+3y^3=28$ is 36.

6 0

3 years ago