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IRISSAK [1]
2 years ago
8

PLZ HELP IWILL GIVE BRAINLIESTTT

Mathematics
1 answer:
liq [111]2 years ago
8 0

Answer:

Choice B

A = b x h

Step-by-step explanation:

In other words the area of a parallelogram is the base times height

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3 0
2 years ago
The following integral requires a preliminary step such as long division or a change of variables before using the method of par
shtirl [24]

Division yields

\dfrac{x^4+7}{x^3+2x} = x-\dfrac{2x^2-7}{x^3+2x}

Now for partial fractions: you're looking for constants <em>a</em>, <em>b</em>, and <em>c</em> such that

\dfrac{2x^2-7}{x(x^2+2)} = \dfrac ax + \dfrac{bx+c}{x^2+2}

\implies 2x^2 - 7 = a(x^2+2) + (bx+c)x = (a+b)x^2+cx + 2a

which gives <em>a</em> + <em>b</em> = 2, <em>c</em> = 0, and 2<em>a</em> = -7, so that <em>a</em> = -7/2 and <em>b</em> = 11/2. Then

\dfrac{2x^2-7}{x(x^2+2)} = -\dfrac7{2x} + \dfrac{11x}{2(x^2+2)}

Now, in the integral we get

\displaystyle\int\frac{x^4+7}{x^3+2x}\,\mathrm dx = \int\left(x+\frac7{2x} - \frac{11x}{2(x^2+2)}\right)\,\mathrm dx

The first two terms are trivial to integrate. For the third, substitute <em>y</em> = <em>x</em> ² + 2 and d<em>y</em> = 2<em>x</em> d<em>x</em> to get

\displaystyle \int x\,\mathrm dx + \frac72\int\frac{\mathrm dx}x - \frac{11}4 \int\frac{\mathrm dy}y \\\\ =\displaystyle \frac{x^2}2+\frac72\ln|x|-\frac{11}4\ln|y| + C \\\\ =\displaystyle \boxed{\frac{x^2}2 + \frac72\ln|x| - \frac{11}4 \ln(x^2+2) + C}

7 0
2 years ago
Find the coordinate of K' after a Glide reflection of the triangle: translation 4 units up and 4 units right then a reflection a
monitta

Answer:

  a.  (-1, 3)

Step-by-step explanation:

The "glide" 4 units right and up is the transformation ...

  (x, y) ⇒ (x+4, y+4)

The reflection across y=1 is the transformation ...

  (x, y) ⇒ (x, 2-y)

Combined, you have the transformation ...

  (x, y) ⇒ (x +4, 2-(y +4)) = (x +4, -2-y)

Then, ...

  K(-5, -5) ⇒ K'(-5 +4, -2-(-5)) = K'(-1, 3)

4 0
3 years ago
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