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Sergio [31]
3 years ago
7

Noor has 2 cups of hot chocolate in his thermos. He offers 3/8 cup servings of hot chocolate to some friends. What is the greate

st number of friends noor can offer got chocolate too?
Mathematics
1 answer:
mezya [45]3 years ago
6 0
The greatest number of friends is 5. once he gives 5 friends, he'll have given away 15/8 cups. he has 2 (16/8) cups so he can't give any more.
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The polygons in each pair are similar find the scale factor smaller figure to the larger
GrogVix [38]

Answer:

Smaller Figure 3

Larger Figure 5

Scale factor = 15/25 = 3/5

3 0
2 years ago
18÷6 solve it<br><br><br>answer:​
Oduvanchick [21]

Answer:3 is your answer

Step-by-step explanation:

3 x 6 = 18

8 0
3 years ago
Find the area of the kite !!!! PLEASE HELP!!!!
Andrew [12]

ANSWER

D. 18 m^2

EXPLANATION

The area of a kite is half the product of the diagonals.

The diagonals of the kite are

3+3=6m

and

2+4=6m

The area of the kite

=  \frac{1}{2}  \times 6 \times 6

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The correct answer is D.

7 0
3 years ago
-X + 2y = 10<br> -3x+ 6y = 11
Tanzania [10]

Answer:

1. y = 1 /2X +5

Step-by-step explanation:

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7 0
2 years ago
Which definite integral approximation formula is this: the integral from a to b of f(x)dx ≈ (b-a)/n * [<img src="https://tex.z-d
Stella [2.4K]

The answer is most likely A.

The integration interval [<em>a</em>, <em>b</em>] is split up into <em>n</em> subintervals of equal length (so each subinterval has width (<em>b</em> - <em>a</em>)/<em>n</em>, same as the coefficient of the sum of <em>y</em> terms) and approximated by the area of <em>n</em> rectangles with base (<em>b</em> - <em>a</em>)/<em>n</em> and height <em>y</em>.

<em>n</em> subintervals require <em>n</em> + 1 points, with

<em>x</em>₀ = <em>a</em>

<em>x</em>₁ = <em>a</em> + (<em>b</em> - <em>a</em>)/<em>n</em>

<em>x</em>₂ = <em>a</em> + 2(<em>b</em> - <em>a</em>)/<em>n</em>

and so on up to the last point <em>x</em> = <em>b</em>. The right endpoints are <em>x</em>₁, <em>x</em>₂, … etc. and the height of each rectangle are the corresponding <em>y </em>'s at these endpoints. Then you get the formula as given in the photo.

• "Average rate of change" isn't really relevant here. The AROC of a function <em>G(x)</em> continuous* over an interval [<em>a</em>, <em>b</em>] is equal to the slope of the secant line through <em>x</em> = <em>a</em> and <em>x</em> = <em>b</em>, i.e. the value of the difference quotient

(<em>G(b)</em> - <em>G(a)</em> ) / (<em>b</em> - <em>a</em>)

If <em>G(x)</em> happens to be the antiderivative of a function <em>g(x)</em>, then this is the same as the average value of <em>g(x)</em> on the same interval,

g_{\rm ave}=\dfrac{G(b)-G(a)}{b-a}=\dfrac1{b-a}\displaystyle\int_a^b g(x)\,\mathrm dx

(* I'm actually not totally sure that continuity is necessary for the AROC to exist; I've asked this question before without getting a particularly satisfying answer.)

• "Trapezoidal rule" doesn't apply here. Split up [<em>a</em>, <em>b</em>] into <em>n</em> subintervals of equal width (<em>b</em> - <em>a</em>)/<em>n</em>. Over the first subinterval, the area of a trapezoid with "bases" <em>y</em>₀ and <em>y</em>₁ and "height" (<em>b</em> - <em>a</em>)/<em>n</em> is

(<em>y</em>₀ + <em>y</em>₁) (<em>b</em> - <em>a</em>)/<em>n</em>

but <em>y</em>₀ is clearly missing in the sum, and also the next term in the sum would be

(<em>y</em>₁ + <em>y</em>₂) (<em>b</em> - <em>a</em>)/<em>n</em>

the sum of these two areas would reduce to

(<em>b</em> - <em>a</em>)/<em>n</em> = (<em>y</em>₀ + <u>2</u> <em>y</em>₁ + <em>y</em>₂)

which would mean all the terms in-between would need to be doubled as well to get

\displaystyle\int_a^b f(x)\,\mathrm dx\approx\frac{b-a}n\left(y_0+2y_1+2y_2+\cdots+2y_{n-1}+y_n\right)

7 0
3 years ago
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