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lana [24]
3 years ago
15

Help asap this is ELA

Mathematics
2 answers:
almond37 [142]3 years ago
7 0

Answer:

Step-by-step explanation:

7.personification

6.hyperbole

Burka [1]3 years ago
3 0
7) personification
6) hyperbole
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Find r if (r, 6) and (5, -4) are two points on a line with a slope of 5/7 (Hint: use the slope formula)
andreyandreev [35.5K]

The formula of a slope:

m=\dfrac{y_2-y_1}{x_2-x_1}

We have

(r,\ 6),\ (5,\ -4)\\\\m=\dfrac{5}{7}

Substitute

\dfrac{-4-6}{5-r}=\dfrac{5}{7}\\\\\dfrac{-10}{5-r}=\dfrac{5}{7}\qquad|\text{cross multiply}\\\\5(5-r)=(-10)(7)\qquad|\text{use distributive property}\\\\25-5r=-70\qquad|-25\\\\-5r=-95\qquad|:(-5)\\\\\boxed{r=19}

4 0
3 years ago
Prove that the diagonals of a parallelogram bisect each other.<br> The midpoint of AC is
iren2701 [21]

Answer:

Theorem: The diagonals of a parallelogram bisect each other. Proof: Given ABCD, let the diagonals AC and BD intersect at E, we must prove that AE ∼ = CE and BE ∼ = DE. The converse is also true: If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram.

Step-by-step explanation:

3 0
3 years ago
Frederick works in a grocery store and also mows lawns . He earns $ 10 per hour at the grocery store and $ 8.50 per hour mowing
Mariana [72]

Answer:

just multiply 35 by x and then 10 by however many lawns he mowed (if that makes sense)

Step-by-step explanation:

8 0
2 years ago
Which construction of parallel lines is justified by the theorem "when two lines are intersected by a transversal and the corres
Ugo [173]

Answer:

c

Step-by-step explanation:

I think you missed attaching the photo, please see my attachment.

And the correct answer is C,

When you look at where the arc meets the parallel lines, if you create a seam between two points, you get a straight line parallel to the horizontal lines  so it makes the corresponding angles are congruent.

5 0
3 years ago
1. Express <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bx%282x%2B3%29%20%7D" id="TexFormula1" title="\frac{1}{x(2x+3) }" a
katovenus [111]

1. Let a and b be coefficients such that

\dfrac1{x(2x+3)} = \dfrac ax + \dfrac b{2x+3}

Combining the fractions on the right gives

\dfrac1{x(2x+3)} = \dfrac{a(2x+3) + bx}{x(2x+3)}

\implies 1 = (2a+b)x + 3a

\implies \begin{cases}3a=1 \\ 2a+b=0\end{cases} \implies a=\dfrac13, b = -\dfrac23

so that

\dfrac1{x(2x+3)} = \boxed{\dfrac13 \left(\dfrac1x - \dfrac2{2x+3}\right)}

2. a. The given ODE is separable as

x(2x+3) \dfrac{dy}dx} = y \implies \dfrac{dy}y = \dfrac{dx}{x(2x+3)}

Using the result of part (1), integrating both sides gives

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + C

Given that y = 1 when x = 1, we find

\ln|1| = \dfrac13 \left(\ln|1| - \ln|5|\right) + C \implies C = \dfrac13\ln(5)

so the particular solution to the ODE is

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + \dfrac13\ln(5)

We can solve this explicitly for y :

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3| + \ln(5)\right)

\ln|y| = \dfrac13 \ln\left|\dfrac{5x}{2x+3}\right|

\ln|y| = \ln\left|\sqrt[3]{\dfrac{5x}{2x+3}}\right|

\boxed{y = \sqrt[3]{\dfrac{5x}{2x+3}}}

2. b. When x = 9, we get

y = \sqrt[3]{\dfrac{45}{21}} = \sqrt[3]{\dfrac{15}7} \approx \boxed{1.29}

8 0
2 years ago
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