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Rama09 [41]
3 years ago
13

What is the coefficient of x2 in the expansion of (x + 2)^3

Mathematics
1 answer:
galina1969 [7]3 years ago
6 0

Answer:6

Step-by-step explanation:

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One cup of​ grated, fresh Parmesan cheese weighs 3 ounces. How many cups of grated cheese can you expect to get from a​ 4.5-poun
Ksenya-84 [330]

Answer:

9 cups

Step-by-step explanation:

If one pound is equal to 2 cups, then you would multiply 2 by 4, which gives you 8. If one pound equals 2 cups, then half a pound equals 1 cup. So since we already have 8 cups, you just add one more cup, which gives you 9 cups total

SUMMARY: You can expect to get 9 cups of grated parmesan cheese from a 4.5-pound block.

5 0
2 years ago
Money of Rp. 24,000.00 can buy 3 kg of apples. If Anto buys apples with money
ss7ja [257]

Answer:

5 Kg of apples

Step-by-step explanation:

cost of apples per kg = 24000.00/3

=8000.00

amount of apples bought = 40000.00/8000.00

=5 Kg

He got 5kg of apples with Rp.40,000.00

8 0
2 years ago
Use lagrange multiplier techniques to find the local extreme values of f(x, y) = x2 − y2 − 2 subject to the constraint x2 + y2 =
dexar [7]
Given f(x,\ y)=x^2-y^2-2 subject to the constraint x^2+y^2=16

Let g(x,\ y)=x^2+y^2.

The gradient vectors of f and g are:

\nabla f(x,\ y)=\langle2x,-2y\rangle and \nabla g(x,\ y)=\langle2x,2y\rangle

By Lagrange's theorem, there is a number \lambda, such that

\langle2x,-2y\rangle=\lambda\langle2x,2y\rangle=\langle2\lambda x,2\lambda y\rangle

\lambda=\pm1

It can be seen that f(x,\ y)=x^2-y^2-2 has local extreme values at the given region.
6 0
3 years ago
PLSSS HELP IMMEDIATELY!!!!!!! only answer if u know. i’ll be giving brainiest if u don’t leave a link.
podryga [215]

Answer:

1. isosceles because two of the sides are equal

2. right because there is a right angle

Step-by-step explanation:

7 0
2 years ago
Use the Midpoint Rule with n = 5 to estimate the volume V obtained by rotating about the y-axis the region under the curve y = 1
velikii [3]

Using the shell method, the volume is given exactly by the definite integral,

2\pi\displaystyle\int_0^1x(1+9x^3)\,\mathrm dx

Splitting up the interval [0, 1] into 5 subintervals gives the partition,

[0, 1/5], [1/5, 2/5], [2/5, 3/5], [3/5, 4/5], [4/5, 5]

with left and right endpoints, respectively, for the i-th subinterval

\ell_i=\dfrac{i-1}5

r_i=\dfrac i5

where 1\le i\le5. The midpoint of each subinterval is

m_i=\dfrac{\ell_i+r_i}2=\dfrac{2i-1}{10}

Then the Riemann sum approximating the integral above is

2\pi\displaystyle\sum_{i=1}^5m_i(1+9{m_i}^3)\frac{1-0}5

\dfrac{2\pi}5\displaystyle\sum_{i=1}^5\left(\frac{2i-1}{10}+9\left(\frac{2i-1}{10}\right)^4\right)

\dfrac{2\pi}{5\cdot10^4}\displaystyle\sum_{i=1}^5\left(16i^4-32i^3+24i^2+1992i-999\right)=\frac{112,021\pi}{25,000}\approx\boxed{14.08}

(compare to the actual value of the integral of about 14.45)

3 0
3 years ago
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