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3 years ago

6

WILL GIVE BRAINLIEST AND 50 POINTS!

1 answer:

Cerrena [4.2K]3 years ago

4 0

Answer: The function M if derived from the integral of the derivative of the function. This is valid by the fundamental theorem of calculus.

M(x)=43x32−7x+C

Step-by-step explanation: not 100 about this but it's what I got

You might be interested in

Solve the equation for<br> x 3/4(x+21) = 10.5 <br> x =

MrRa [10]

<h3>✍️ Answer:</h3><h3> $x=-\frac{294}{13}$ </h3>

✽✽✽✽

$\frac{x\cdot \:3}{4\left(x+21\right)}=10.5$

$\mathrm{Multiply\:both\:sides\:by\:}4\left(x+21\right)$

$\frac{x\cdot \:3}{4\left(x+21\right)}\cdot \:4\left(x+21\right)=10.5\cdot \:4\left(x+21\right)$

$3x=42\left(x+21\right)$

$3x=42x+882$

$\mathrm{Subtract\:}42x\mathrm{\:from\:both\:sides}$

$3x-42x=42x+882-42x$

$-39x=882$

$\mathrm{Divide\:both\:sides\:by\:}-39$

$\frac{-39x}{-39}=\frac{882}{-39}$

$x=-\frac{294}{13}$

❅❅❅❅❅❅❅❅❅❅

<h3>hope it helps...</h3><h3>have a great day!!</h3>

4 0

3 years ago

The sum of three consecutive integers is 141. What is the smallest integer?

Sonja [21]

Answer:

The Consecutive number are 46 , 47 and 48 having sum as 141

The smallest number is 46

Step-by-step explanation:

• A consecutive numbers means number are in a sequence and are one after the another !

e.g. n , (n+1). Or 1 , 2 are consecutive numbers

• A consecutive even numbers differ by 2 because all even numbers are divisible by 2 !

• If the number is divisible by x we must had taken number as n , ( n + x ) ,

( n + 2x )

• Let the numbers be ( n - 1 ) , n and

( n + 1 )

Given ,the sum of numbers is 141

n - 1 + n + n + 1 = 141

3n = 141

n = 141/3

n = 47

So, the numbers are

n- 1 = 47 - 1 = 46

n = 47

n + 1 = 48

And , the smallest number is 46

• Verification :-

46 + 48 + 47 = 141

Which satisfies the given condition !!

Hence , verified !!

8 0

3 years ago

What is the value of the next term of the sequence?

user100 [1]

Would it be (-3)^7? (-2187)

4 0

3 years ago

Find the mass of the lamina that occupies the region D = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1} with the density function ρ(x, y) = xye

Alona [7]

Answer:

The mass of the lamina is 1

Step-by-step explanation:

Let $\rho(x,y)$ be a continuous density function of a lamina in the plane region D,then the mass of the lamina is given by:

$m=\int\limits \int\limits_D \rho(x,y) \, dA$ .

From the question, the given density function is $\rho (x,y)=xye^{x+y}$ .

Again, the lamina occupies a rectangular region: D={(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1}.

The mass of the lamina can be found by evaluating the double integral:

$I=\int\limits^1_0\int\limits^1_0xye^{x+y}dydx$ .

Since D is a rectangular region, we can apply Fubini's Theorem to get:

$I=\int\limits^1_0(\int\limits^1_0xye^{x+y}dy)dx$ .

Let the inner integral be: $I_0=\int\limits^1_0xye^{x+y}dy$ , then

$I=\int\limits^1_0(I_0)dx$ .

The inner integral is evaluated using integration by parts.

Let $u=xy$ , the partial derivative of u wrt y is

$\implies du=xdy$

and

$dv=\int\limits e^{x+y} dy$ , integrating wrt y, we obtain

$v=\int\limits e^{x+y}$

Recall the integration by parts formula: $\int\limits udv=uv- \int\limits vdu$

This implies that:

$\int\limits xye^{x+y}dy=xye^{x+y}-\int\limits e^{x+y}\cdot xdy$

$\int\limits xye^{x+y}dy=xye^{x+y}-xe^{x+y}$

$I_0=\int\limits^1_0 xye^{x+y}dy$

We substitute the limits of integration and evaluate to get:

$I_0=xe^x$

This implies that:

$I=\int\limits^1_0(xe^x)dx$ .

Or

$I=\int\limits^1_0xe^xdx$ .

We again apply integration by parts formula to get:

$\int\limits xe^xdx=e^x(x-1)$ .

$I=\int\limits^1_0xe^xdx=e^1(1-1)-e^0(0-1)$ .

$I=\int\limits^1_0xe^xdx=0-1(0-1)$ .

$I=\int\limits^1_0xe^xdx=0-1(-1)=1$ .

No unit is given, therefore the mass of the lamina is 1.

3 0

3 years ago

Point R on the coordinate grid below shows the location of a car in a parking lot. A second car needs to park 7 units above R to

VLD [36.1K]

If this helps I got 2,-3 when I plug it into the graph though it says 2,3

6 0

3 years ago

Read 2 more answers