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2 years ago

I’m in dyer need of the answer to this problem please help!!

Mathematics

1 answer:

ElenaW [278]2 years ago

5 0

Answer:

the x intercepts are -4 and -2

the y intercept are -2

Step-by-step explanation:

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Which is bigger 5/12 or 7/24

7nadin3 [17]

I would say 7/24 because when you multiply it you should be able to get a large number.

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2 years ago

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What is the power of 15<br> ------------------

miv72 [106K]

Answer:

If something is "to the power of 15" then it means it is multiplied by itself 14 times (one less than 15 as the original number is already to the power of 1).

An example:

So 2 to the power of 15, which we write as 215 = 2 * 2 * 2 * 2 etc... (so we have 15 written 2's in total meaning we will make 14 multiplications) is equal to 32,768.

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2 years ago

Solve the equation 7(w +13) = 35.<br> ) What is the value of w?

Nina [5.8K]

Answer: -8

Step-by-step explanation:

7(w+13)=35 [given]
w+13=5 [divide both sides by 7]
w = -8 [subtract 13 from both sides]

3 0

11 months ago

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What is the length of the longest side of a triangle that has vertices at (-2,6) (-4,6) and (-4,4)?

zalisa [80]

Length (2, 6) to (-4, 6) is sqrt((x2 - x1))^2 + (y2 - y1)^2) = sqrt((-4 -2)^2 + (6 - 6)^2) = sqrt((-6)^2 + 0) = 6

Length (2, 6) to (-4, 4) is sqrt((-4 - 2)^2 + (4 - 6)^2) = sqrt((-6)^2 + (-2)^2) = sqrt(36 + 4) = sqrt(40) = 2sqrt(10) units

Length (-4, 6) to (-4, 4) is sqrt((-4 - (-4))^2 + (4 - 6)^2) = sqrt(0^2 + (-2)^2) = 2

So the length of the longest side is 2sqrt(10) units

6 0

2 years ago

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

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

but that may or may not be considered acceptable by that webpage.

5 0

2 years ago

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