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3 years ago

Estimate, then add. Write each sum in the simplest form.

Mathematics

1 answer:

Likurg_2 [28]3 years ago

5 0

Answer:

add the fractions and then divide

Step-by-step explanation:

so you add the top two and then the bottem two and then after you get those two anwsers divide them

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Find and compare the median and

Over [174]

Answer:

Median= $2.45/2

=$1.225

Mode= $2.87

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3 years ago

an equilateral triangle has one angle measureme of 60. What is the measurement of the two remaining angles?

Anna [14]

The equilateral triangle has one angle measurement of 60 what is the measurement of two remaining angles divide 60 by 2 with sugar 30 that your answer

6 0

3 years ago

Resssspooond quickkk

Doss [256]

The slope of the given line is -1/3. The perpendicular line's slope is the opposite reciprocal of the given line's slope. So, the perpendicular line's slope is 3. Then we have slope intercept form y=mx+b

y=3x+b
We don't know "b" which is the y-intercept for this equation, but we have the coordinates (6, -1). We can use these to find the slope by plugging them into the equation.

-1=3(6)+b
-1=18+b
-1-18=18+b-18
-19=b
So, the resulting y-intercept is -19.

The final perpendicular equation would be y=3x-19

7 0

3 years ago

Read 2 more answers

Write out the form of the partial fraction decomposition of the function. Do not determine the numerical values of the coefficie

Dvinal [7]

For part (a), you have

$\dfrac x{x^2+x-6}=\dfrac x{(x+3)(x-2)}=\dfrac a{x+3}+\dfrac b{x-2}$
$x=a(x-2)+b(x+3)$

If $x=2$ , then $2=b(2-3)\implies b=-2$ .

If $x=-3$ , then $-3=a(-3-2)\implies a=\dfrac35$ .

So,

$\dfrac x{x^2+x-6}=\dfrac 3{5(x+3)}-\dfrac 2{x-2}$

For part (b), since the degrees of the numerator and denominator are the same, you first need to find the quotient and remainder upon division.

$\dfrac{x^2}{x^2+x+2}=\dfrac{x^2+x+2-x-2}{x^2+x+2}=1-\dfrac{x+2}{x^2+x+2}$

In the remainder term, the denominator $x^2+x+2$ can't be factorized into linear components with real coefficients, since the discriminant is negative $(1-4\times1\times2=-7)$ . However, you can still factorized over the complex numbers, so a partial fraction decomposition in terms of complexes does exist.

$x^2+x+2=0\implies x=-\dfrac12\pm\dfrac{\sqrt7}2i$
$\implies x^2+x+2=\left(x-\left(-\dfrac12+\dfrac{\sqrt7}2i\right)\right)\left(x-\left(-\dfrac12-\dfrac{\sqrt7}2i\right)\right)$
$\implies x^2+x+2=\left(x+\dfrac12-\dfrac{\sqrt7}2i\right)\left(x+\dfrac12+\dfrac{\sqrt7}2i\right)$

Then you have

$\dfrac{x+2}{x^2+x+2}=\dfrac a{x+\dfrac12-\dfrac{\sqrt7}2i}+\dfrac b{x+\dfrac12+\dfrac{\sqrt7}2i}$
$x+2=a\left(x+\dfrac12+\dfrac{\sqrt7}2i\right)+b\left(x+\dfrac12-\dfrac{\sqrt7}2i\right)$

When $x=-\dfrac12-\dfrac{\sqrt7}2i$ , you have

$-\dfrac12-\dfrac{\sqrt7}2i+2=b\left(-\dfrac12-\dfrac{\sqrt7}2i+\dfrac12-\dfrac{\sqrt7}2i\right)$
$\dfrac32-\dfrac{\sqrt7}2i=-\sqrt7ib$
$b=\dfrac12+\dfrac3{2\sqrt7}i=\dfrac1{14}(7+3\sqrt7i)$

When $x=-\dfrac12+\dfrac{\sqrt7}2i$ , you have

$-\dfrac12+\dfrac{\sqrt7}2i+2=a\left(-\dfrac12+\dfrac{\sqrt7}2i+\dfrac12+\dfrac{\sqrt7}2i\right)$
$\dfrac32+\dfrac{\sqrt7}2i=\sqrt7ia$
$a=\dfrac12-\dfrac3{2\sqrt7}i=\dfrac1{14}(7-3\sqrt7i)$

So, you could write

$\dfrac{x^2}{x^2+x+2}=1-\dfrac{x+2}{x^2+x+2}=1-\dfrac {7-3\sqrt7i}{14\left(x+\dfrac12-\dfrac{\sqrt7}2i\right)}-\dfrac {7+3\sqrt7i}{14\left(x+\dfrac12+\dfrac{\sqrt7}2i\right)}$

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5 0

3 years ago

Read 2 more answers

Perform the indicated operation <br><br> 3/4 divided by 1/3

SVETLANKA909090 [29]

Answer:

Step-by-step explanation:

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3 years ago

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