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3 years ago

10

Help me pleaseeee, im having trouble solving.

2 answers:

Goryan [66]3 years ago

8 0

Answer:

-4

Step-by-step explanation:

4+3+5/ 3-6

12/-3

-4

hope this helps

shutvik [7]3 years ago

6 0

You might want to try and use photomath buddy

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What is 12.062 rounded to the nearest tenth?

babunello [35]

Well, the answer to this problem is 12.1

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3 years ago

Evaluate 7y - 2<br> when y = 9

kotykmax [81]

Answer:

The answer is 61.

Step-by-step explanation:

7(9)-2

63-2

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3 years ago

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Someone please help me with this.

notka56 [123]

Forty_five point five

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3 years ago

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

Find the length of the unknown side. Round your answer to the nearest tenth. (4 points) Image of a right triangle with a leg lab

Anika [276]

using pythagorean theorem

a^2 + b^2 =c^2

a^2 +8^2 =14^2

a^2 +64 = 196

subtract 64 from each side

a^2 = 132

take the square root of each side

a^2 = sqrt (132)

a=11.48912529

a =11.5

5 0

4 years ago