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Luda [366]
3 years ago
13

1. Jan spends part of her year as a member of a

Mathematics
1 answer:
antiseptic1488 [7]3 years ago
3 0

Answer:

13 months in a half......

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Write the equation of the line parallel to y=4 that passes through (1,6)​
nadya68 [22]

Answer:

y=0x+6

Step-by-step explanation:

6 0
3 years ago
Which equation shows how to determine the volume of the dollhouse?
777dan777 [17]

Answer:

first choice

Step-by-step explanation:

The volume of a rectangular prism is L*H*B.

So the volume of the rectangular prism pictured here is 24*10*8

The volume of a triangle prism is 1/2 *H*L*B.

So the volume of the triangle prism pictured here is 1/2 *6*24*8

The volume of the whole figure is the sum of the volume of the rectangular prism and volume of the triangular prism which is in this case:

24*10*8 + 1/2 *6*24*8.

Answer is the first choice.

(It wouldn't be third because 10 has nothing to with the triangular prism part)

7 0
3 years ago
Find the product of (x + 3)(x - 3)
Roman55 [17]

Answer:

(x+3)(x-3) is x²-9

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Someone help please want it asap anyone??
Gre4nikov [31]

Answer:

Pretty sure it's the last one at the bottom.

8 0
2 years ago
Read 2 more answers
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
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