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3 years ago

Simplify the following expressions. Cross out the letter that matches your answer.

Mathematics

1 answer:

OverLord2011 [107]3 years ago

8 0

1. 3x + 5x + 14x = 22x

2. 4x + 7 + 11x = 15x+7

3. 10x + 8 - 3x= 7x+8

4. 6x - 5 + X - 11 = 7x-16

5. -8x + 19 + 2 - 2x = -10x+21

6. X + 23 + 4x - 34 - X = 4x-11

7. 105 + 27x - 68 - 19x - 27 + x = 9x +10

8. 7x + 16 - 3x + 3 - 2x - 2x = 19

9. -18 + 6 - 4x + 10 - 11x + 2 = -15x

10. 16 + 4x - 12x - 28 - 6x + 17 + x - 1 = -13x+4

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Walter takes 7/10 of an hour to mow 2/5 of an acre of lawn. At this rate, how many acres will he mow in an hour?

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hours&\stackrel{lawn}{acre}\\
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\end{array}\implies \cfrac{\frac{7}{10}}{1}=\cfrac{\frac{2}{5}}{a}\implies \cfrac{\frac{7}{10}}{\frac{1}{1}}=\cfrac{\frac{2}{5}}{\frac{a}{1}}\implies \cfrac{7}{10}\cdot \cfrac{1}{1}=\cfrac{2}{5}\cdot \cfrac{1}{a}
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4 years ago

Write an equation of the line that passes through the point (6, -9) with the given slope m = 3.

adelina 88 [10]

Answer:

y = 3x - 27

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Here m = 3 , then

y = 3x + c ← is the partial equation

To find c substitute (6, - 9) into the partial equation

- 9 = 18 + c ⇒ c = - 9 - 18 = - 27

y = 3x - 27 ← equation of line

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3 years ago

Read 2 more answers

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3 years ago

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Use the Green's Theorem to calculate the work done by the field [ F (x, y) = -3y^5 i + 5y^2x^3 j ] to move a particle along the

lianna [129]

$\displaystyle\int_C\mathbf F\cdot\mathrm d\mathbf r=\iint_R\left(\frac{\partial(5y^2x^3)}{\partial x}-\frac{\partial(-3y^5)}{\partial y}\right)\,\mathrm dx\,\mathrm dy$

$\dfrac{\partial(5y^2x^3)}{\partial x}=15y^2x^2$
$\dfrac{\partial(-3y^5)}{\partial y}=-15y^2$

$=\displaystyle\int_{x=-2}^{x=2}\int_{y=-\sqrt{4-x^2}}^{y=\sqrt{4-x^2}}15y^2(x^2+1)\,\mathrm dy\,\mathrm dx$

Converting to polar coordinates, the integral is equivalent to

$=\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=2}15r^3\sin^2\theta(r^2\cos^2\theta+1)\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle15\int_0^{2\pi}\int_0^2\left(\frac{r^5}4\sin^22\theta+r^3\sin^2\theta\right)\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle\frac{15}4\left(\int_0^2r^5\,\mathrm dr\right)\left(\int_0^{2\pi}\sin^22\theta\,\mathrm d\theta\right)+15\left(\int_0^2r^3\,\mathrm dr\right)\left(\int_0^{2\pi}\sin^2\theta\,\mathrm d\theta\right)$
$=100\pi$

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3 years ago