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kolbaska11 [484]
3 years ago
13

A basketball team scored 78 points in their championship win over their rival. The team scored 25 baskets worth 2 points and mad

e seven free throws. How many 3 point baskets did the team make?
Mathematics
1 answer:
solong [7]3 years ago
3 0
25(2) + 7(1) + 3x = 78....x represents the number of 3 point shots
50 + 7 + 3x = 78
57 + 3x = 78
3x = 78 - 57
3x = 21
x = 21/3
x = 7....so there were 7 three-point shots made
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Evgen [1.6K]
10 and 1. Because 10 times 1 equals 10 and 10 plus 1 equals 11.
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Mrs kin picked some tomatoes from her garden. She use 5/9 of her tomatoes to make pasta sauce. Then she used 5/8 of the remainde
IgorC [24]

Answer:

1/6 of the tomatoes

Step-by-step explanation:

So first, after she uses 5/9 she has 1 - 5/9 = 4/9 of her tomatoes left. Then, she uses 5/8 of that 4/9 to make a salad, so she used 5/8 * 4/9 = 5/18 of her tomatoes overall.  That means that she has 4/9 - 5/18 = 8/18 - 5/18 = 3/18 = 1/6 of her tomatoes saved.

7 0
3 years ago
Read 2 more answers
37. Verify Green's theorem in the plane for f (3x2- 8y2) dx + (4y - 6xy) dy, where C is the boundary of the
Nastasia [14]

I'll only look at (37) here, since

• (38) was addressed in 24438105

• (39) was addressed in 24434477

• (40) and (41) were both addressed in 24434541

In both parts, we're considering the line integral

\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dy

and I assume <em>C</em> has a positive orientation in both cases

(a) It looks like the region has the curves <em>y</em> = <em>x</em> and <em>y</em> = <em>x</em> ² as its boundary***, so that the interior of <em>C</em> is the set <em>D</em> given by

D = \left\{(x,y) \mid 0\le x\le1 \text{ and }x^2\le y\le x\right\}

• Compute the line integral directly by splitting up <em>C</em> into two component curves,

<em>C₁ </em>: <em>x</em> = <em>t</em> and <em>y</em> = <em>t</em> ² with 0 ≤ <em>t</em> ≤ 1

<em>C₂</em> : <em>x</em> = 1 - <em>t</em> and <em>y</em> = 1 - <em>t</em> with 0 ≤ <em>t</em> ≤ 1

Then

\displaystyle \int_C = \int_{C_1} + \int_{C_2} \\\\ = \int_0^1 \left((3t^2-8t^4)+(4t^2-6t^3)(2t))\right)\,\mathrm dt \\+ \int_0^1 \left((-5(1-t)^2)(-1)+(4(1-t)-6(1-t)^2)(-1)\right)\,\mathrm dt \\\\ = \int_0^1 (7-18t+14t^2+8t^3-20t^4)\,\mathrm dt = \boxed{\frac23}

*** Obviously this interpretation is incorrect if the solution is supposed to be 3/2, so make the appropriate adjustment when you work this out for yourself.

• Compute the same integral using Green's theorem:

\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dy = \iint_D \frac{\partial(4y-6xy)}{\partial x} - \frac{\partial(3x^2-8y^2)}{\partial y}\,\mathrm dx\,\mathrm dy \\\\ = \int_0^1\int_{x^2}^x 10y\,\mathrm dy\,\mathrm dx = \boxed{\frac23}

(b) <em>C</em> is the boundary of the region

D = \left\{(x,y) \mid 0\le x\le 1\text{ and }0\le y\le1-x\right\}

• Compute the line integral directly, splitting up <em>C</em> into 3 components,

<em>C₁</em> : <em>x</em> = <em>t</em> and <em>y</em> = 0 with 0 ≤ <em>t</em> ≤ 1

<em>C₂</em> : <em>x</em> = 1 - <em>t</em> and <em>y</em> = <em>t</em> with 0 ≤ <em>t</em> ≤ 1

<em>C₃</em> : <em>x</em> = 0 and <em>y</em> = 1 - <em>t</em> with 0 ≤ <em>t</em> ≤ 1

Then

\displaystyle \int_C = \int_{C_1} + \int_{C_2} + \int_{C_3} \\\\ = \int_0^1 3t^2\,\mathrm dt + \int_0^1 (11t^2+4t-3)\,\mathrm dt + \int_0^1(4t-4)\,\mathrm dt \\\\ = \int_0^1 (14t^2+8t-7)\,\mathrm dt = \boxed{\frac53}

• Using Green's theorem:

\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dx = \int_0^1\int_0^{1-x}10y\,\mathrm dy\,\mathrm dx = \boxed{\frac53}

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3 years ago
A leaky faucet wastes 2,400 gallons of water per year. How many quarts of water are wasted per month? [1 gallon = 4 quarts]
bixtya [17]

2,400÷4=600

hope this is good :)

o_________________o

7 0
4 years ago
Read 2 more answers
Quien me puede Ayudar Con esto Por Favor (con procesos) ES URGENTE
MrMuchimi
Adonde es tu pregunta
6 0
3 years ago
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