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2 years ago

I really don't get it can you give me the answer then explain why it is that one please

Mathematics

2 answers:

Ronch [10]2 years ago

7 0

Answer:

$\frac{100}{9}$ is the answer.

Step-by-step explanation:

Well the first thing we would do is simplify all the exponents first inside the parenthesis. So 2^-3 is equal to 1/8. To solve negative exponents you would just divide the number instead of multiply, so 2 divided by 2 divided by 2 is equal to ⅛ or .125

Next would be to do the 2^-9. This would equal 1/512

Now would be to simplify 5^0. Anything to the power of 0 is always one. So now we simplify all the products meaning 7*5*2 and 7*3.

Now we would simplify both fractions.

The picture are pretty much self explanatory. If you have any questions feel free to ask in the comments - Mark

Also when you have the chance please mark me brainliest.

avanturin [10]2 years ago

4 0

Answer:

e (i said this so the other dude could be mark brainlist)

Step-by-step explanation:

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3 years ago

When two fractions are between 0 & 1 half how do you know which fraction is greater explaina

oksian1 [2.3K]

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3 years ago

For Problems 11-12, expand the expressions using the Distributive Property. Then simplify the expressions. 12. 9(3y-5)

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2 years ago

I need help please.

Lina20 [59]

The 1st choice, the third and the fifth are correct
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2 years ago

Approximate the integral integral integral integral f(x, y) dA by dividing the rectangle R with vertices (0, 0), (4, 0), (4, 2),

amm1812

Answer:

Step-by-step explanation:

Approximate the integral $\int\int\limits_R {f(x,y)} \, dA$ by dividing the region $R$ with vertices (0,0),(4,0),(4,2) and (0,2) into eight equal squares.

Find the sum $\sum\limits^8_{i=1}f(x_i,y_i)\delta A_i$

Since all are equal squares, so $\delta A_i=1$ for every $i$

$\sum\limits^8_{i=1}f(x_i,y_i)\delta A_i=f(x_1,y_1)\delta A_1+f(x_2,y_2)\delta A_2+f(x_3,y_3)\delta A_3+f(x_4,y_4)\delta A_4+f(x_5,y_5)\delta A_5+f(x_6,y_6)\delta A_6+f(x_7,y_7)\delta A_7+f(x_8,y_8)\delta A_8\\\\=f(0.5,0.5)(1)+f(1.5,0.5)(1)+f(2.5,0.5)(1)+f(3.5,0.5)(1)+f(0.5,1.5)(1)+f(1.5,1.5)(1)+f(2.5,1.5)(1)+f(3.5,1.5)(1)\\\\=0.5+0.5+1.5+0.5+2.5+0.5+3.5+0.5+0.5+1.5+1.5+1.5+2.5+1.5+3.5+1.5\\\\=24$

Thus, $\sum\limits^8_{i=1}f(x_i,y_i)\delta A_i=24$

Evaluating the iterate integral $\int\limits^4_0 \int\limits^2_0 {(x+y)} \, dydx=\int\limits^4_0 {[xy+\frac{y^2}{2} ]}\limits^2_0 \, dx =\int\limits^4_0 {[2x+2]}dx\\\\=[x^2+2x]\limits^4_0=24.$

Thus, $\int\limits^4_0 \int\limits^2_0 {(x+y)} \, dydx=24$

7 0

2 years ago

I really don't get it can you give me the answer then explain why it is that one please ​

I really don't get it can you give me the answer then explain why it is that one please