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Scorpion4ik [409]
3 years ago
12

Please help ASAP! Thank you<3

Mathematics
2 answers:
Alex73 [517]3 years ago
6 0
Answer: 2

Explanation:

(4^2/2 + 6)/4 + 3
= (16/2 + 6)/7
= (8 + 6)/7
= 14/7
= 2

Therefore, the answer is 2
aliina [53]3 years ago
3 0

Answer: I believe its 2

Step-by-step explanation:

Start on the top, 4 squared is 16, then divided by two is 8, plus 6 is 14, then on the bottom 4+3 equals 7 so then you get, 14/7 and that equals 2.

hope this helps

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Solve for the value of x.
alexandr402 [8]

Answer:

x = 15

Step-by-step explanation:

Similar triangles and proportional side lengths.

10/4 = x/6

5/2 = x/6

2x = 5 * 6

2x = 30

x = 15

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Estimate $0.89 ×14 find each product
viva [34]
The product of $0.89 x 14 equals 12.46
5 0
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What is the least common denominator of the fraction 3 x over x plus 1 plus the fraction x plus 1 over 2 x plus the fraction 5 o
Hoochie [10]
3x/(x+1) + (x+1)/(2x) + (5/x)

To find the least common denominator, we multiply the denominators that cannot factor into eachother. x, the denominator in 5/x, fits into 2x, so we do not need to multiply this number.
Therefore, the least common denominator is:
2x(x+1) 

The first term will be multiplied by 2x/2x
The second term will be multiplied by (x+1)/(x+1)
The third term will be multiplied by 2(x+1)
3 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
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