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3 years ago

Cecilia simplified an expression. Her work is shown below.

Mathematics

2 answers:

Alika [10]3 years ago

8 0

Answer:

Step 2

Step-by-step explanation:

Given work on simplification

(6 and one-half + 2 and three-fourths) minus 1.5 times (4.5 divided by 0.5)

(6 1/2 + 2 3/4) - 1.5 (4.5 ÷ 0.5)

Step 1 9 and one-fourth minus 1.5 times (4.5 divided by 0.5)

9 1/4 - 1.5 (4.5 ÷ 0.5)
This is ok as first step is solving parenthesis

Step 2 7.75 times (4.5 divided by 0.5)

7.75 (4.5 ÷ 0.5)
Incorrect, subtraction should be the last step

Citrus2011 [14]3 years ago

4 0

Answer: second step.

Step-by-step explanation:

you then have to distribute the answer not subtract 9 1/4 from 1.5

you have to do the parenthesis next

4.5/0.5=45/5=9

then

1.5*9= 13.5

then

9.25-13.5 which I think you are capable of solving

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3 years ago

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3 years ago

Read 2 more answers

a) Estimate the volume of the solid that lies below the surface z = 7x + 5y2 and above the rectangle R = [0, 2]⨯[0, 4]. Use a Ri

SSSSS [86.1K]

In the $x$ direction we consider the $m=2$ subintervals [0, 1] and [1, 2] (each with length 1), while in the $y$ direction we consider the $n=2$ subintervals [0, 2] and [2, 4] (with length 2). Then the lower right corners of the cells in the partition of $R$ are (1, 0), (2, 0), (1, 2), (2, 2).

Let $f(x,y)=7x+5y^2$ . The volume of the solid is approximately

$\displaystyle\iint_Rf(x,y)\,\mathrm dx\,\mathrm dy\approx f(1,0)\cdot1\cdot2+f(2,0)\cdot1\cdot2+f(1,2)\cdot1\cdot2+f(2,2)\cdot1\cdot2=\boxed{164}$

###

More generally, the lower-right-corner Riemann sum over $m=\mu$ and $n=\nu$ subintervals would be

$\displaystyle\sum_{m=1}^\mu\sum_{n=1}^\nu\left(7\frac{2m}\mu+5\left(\frac{4n-4}\nu\right)^2\right)\frac{2-0}\mu\frac{4-0}\nu=\frac83\left(101+\frac{21}\mu+\frac{40}{\nu^2}-\frac{120}\nu\right)$

Then taking the limits as $\mu\to\infty$ and $\nu\to\infty$ leaves us with an exact volume of $\dfrac{808}3$ .

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3 years ago