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yKpoI14uk [10]
3 years ago
9

(8x + 9) + (15 - 8x)

Mathematics
1 answer:
serious [3.7K]3 years ago
5 0

Answer:

24

Step-by-step explanation:

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HELP I WILL GIVE 50 POINTS AND BRAINLIEST TO WHOEVER ANSWERS FIRST Graph ​y>1−3x.
AleksandrR [38]

Answer:

The last graph

Step-by-step explanation:

This is because when there is a greater than sign, the line on the graph is dotted and the shaded area is facing upwards. I try to remember that when there is a greater than sign or a greater than or equal to sign, the shaded area faces up towards the sky.

Hope this helps. Please inform me if my answer is wrong.

7 0
3 years ago
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Lucy rented a movie. She started the movie at 9:48 a.m. and it was 3 hours 39 minutes long. When did the movie end
iren2701 [21]

Think through this one...

9:48 + 9 minutes = 9:57

9:57 + 30 minutes = 10:27

10:27 + 3 hours = 1:27 p.m.

6 0
3 years ago
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If 300 words or typed in 6 minutes how many words will be typed in 17 minutes
Ierofanga [76]

Answer:

850 words can be typed in 17 minutes

Step-by-step explanation:

Divide 300 by 6 and multiply it by 17 and that is your answer, 850.

5 0
2 years ago
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How do you turn 0.65 decimal into a fraction or mixed number into simplest form
Yakvenalex [24]
Put 65 over 100 so it's a fraction and simplify from there i think
5 0
3 years ago
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Use Green's Theorem to evaluate the line integral along the given positively oriented curve. ∫C (3y +5e√x)dx + (10x + 3 cos
elena-s [515]

By Green's theorem, the line integral

\displaystyle \int_C f(x,y)\,\mathrm dx + g(x,y)\,\mathrm dy

is equivalent to the double integral

\displaystyle \iint_D \frac{\partial g}{\partial x} - \frac{\partial f}{\partial y} \,\mathrm dx\,\mathrm dy

where <em>D</em> is the region bounded by the curve <em>C</em>, provided that this integrand has no singularities anywhere within <em>D</em> or on its boundary.

It's a bit difficult to make out what your integral should say, but I'd hazard a guess of

\displaystyle \int_C \left(3y+5e^{-x}\right)\,\mathrm dx + \left(10x+3\cos\left(y^2\right)\right)\,\mathrm dy

Then the region <em>D</em> is

<em>D</em> = {(<em>x</em>, <em>y</em>) : 0 ≤ <em>x</em> ≤ 1 and <em>x</em> ² ≤ <em>y</em> ≤ √<em>x</em>}

so the line integral is equal to

\displaystyle \int_0^1\int_{x^2}^{\sqrt x} \frac{\partial\left(10x+3\cos\left(y^2\right)\right)}{\partial x} - \frac{\partial\left(3y+5e^{-x}\right)}{\partial y}\,\mathrm dy\,\mathrm dx \\\\ = \int_0^1 \int_{x^2}^{\sqrt x} (10-3)\,\mathrm dy\,\mathrm dx \\\\ = 7\int_0^1 \int_{x^2}^{\sqrt x} \mathrm dy\,\mathrm dx

which in this case is 7 times the area of <em>D</em>.

The remaining integral is trivial:

\displaystyle 7\int_0^1\int_{x^2}^{\sqrt x}\mathrm dy\,\mathrm dx = 7\int_0^1y\bigg|_{y=x^2}^{y=\sqrt x}\,\mathrm dx \\\\ = 7 \int_0^1\left(\sqrt x-x^2\right)\,\mathrm dx \\\\ = 7 \left(\frac23x^{3/2}-\frac13x^3\right)\bigg|_{x=0}^{x=1} = 7\left(\frac23-\frac13\right) = \boxed{\frac73}

6 0
3 years ago
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