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4 years ago

4x-3 (x -2y) + 1/2 (6x - 8y)

Mathematics

1 answer:

nikitadnepr [17]4 years ago

3 0

Answer:

4x+2y

Step-by-step explanation:

4x-3(x-2y)+1/2x(6x-8y)

remove parentheses

Factor

4x-3x+6y+ 1/2*2(3x - 4y)

Reduce the numbers

4x-3x+6y+(3x - 4y)

remove parentheses

4x-3x+6y+3x - 4y

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Write these numbers in standard form.

atroni [7]

Answer:

a)250,000

2.5×10^5

b)81,000

8.1×10^4

c)906,000,000

=9.06×10^8

d)10340,000,000

=1.034×10^10

i hope this will help you :)

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3 years ago

Find the measure of the arc indicated.<br> A) 144° <br> B) 150° <br> C) 130° <br> D) 131°

mariarad [96]

Answer:

Option B. $150\°$

Step-by-step explanation:

we know that

In this problem

$(15+15x)=(17x-3)$

solve for x

$17x-15x=15+3$

$2x=18$

$x=9$

Find the measure of arc BC

$arc\ BC=17x-3=17(9)-3=150\°$

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3 years ago

Are these two triangles congruent why or why not!!!!!!!!!???????? WORTH 24 POINTS AND I WILL GIVE YOU BRAINIEST!!!!!!1111

Natalija [7]

Answer:

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Step-by-step explanation:

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5 0

3 years ago

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Alona [7]

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Step-by-step explanation:

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4 years ago

Find the counterclockwise circulation and outward flux of the field Fequals7 xy Bold i plus 2 y squared Bold j around and over t

77julia77 [94]

Answer:

The counterclockwise circulation is $\frac{7}{12}$ and the outward flux is $\frac{11}{15}$

Step-by-step explanation:

We are given the field $F(x,y) = (7xy,2y^2)$ . A picture of the region and the path we are considering is attached. Recalll the following theorems.

Given a field of the form F(x,y)=(f(x,y),g(x,y) with f,g having continous partial derivates, C is a closed path counterclockwise oriented, R is the region enclosed by C and n is the normal vector pointing outwards of the path C. Then

$\oint_C F\cdot dr =\iint_R \frac{\partial f}{dy}- \frac{\partial g}{dx} dA$ (this one calculates the counterclockwise circulation)

$\oint_C F\cdot n ds =\iint_R (\frac{\partial f}{dx}+ \frac{\partial g}{dy} dA$ (This one calculates the outward flux)

Then, recall that in our case f(x,y) = 7xy, g(x,y)=2y^2[/tex]. Then

$\frac{\partial f}{dx} = 7y,\frac{\partial f}{dy} = 7x$

$\frac{\partial g}{dx}=0, \frac{\partial g}{dy} = 4y$ .

Note that we just need to describe our region R. The region R lies between the parabola y=x^2 and the line y=x. Thus, one way to describe the region is as follows $0\leq x \leq 1, x^2\leq y \leq x$ . Then, using the previous results, we get that

$\oint_C F\cdot dr =\int_{0}^{1}\int_{x^2}^{x}7x-0dydx = 7\int_{0}^1x(x-x^2)dx = 7 \left.(\frac{x^3}{3}-\frac{x^4}{4})\right|_{0}^1 = 7(\frac{1}{3}-\frac{1}{4}) = \frac{7}{12}$ (circulation)

$\oint_C F\cdot n ds=\int_{0}^{1}\int_{x^2}^{x}7y+4ydydx = \frac{11}{2}\int_{0}^1\left.y^2\right_{x^2}^{x}dx = \frac{11}{2}\int_{0}^{1}x^2-x^4 dx = \frac{11}{2}\left(\frac{x^3}{3}-\frac{x^5}{5})\right|_{0}^{1}=\frac{11}{2}(\frac{1}{3}-\frac{1}{5})=\frac{11}{15}$ (flux)

5 0

3 years ago