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2 years ago

I'm giving 50 POINTS pls answer ASAP

Mathematics

2 answers:

Ray Of Light [21]2 years ago

7 0

Answer:

Step-by-step explanation:

arlik [135]2 years ago

4 0

Answer:

Step-by-step explanation:

No, in order for a set of numbers to be a function, it has to pass the vertical line test. Draw a vertical line to see if it intersects the line more than once. Another way to think about it, is that one singular x value cannot give two different y values. That is impossible.

For example at x = 0, there are two y values that relate to that point. I'm going to estimate the y value for simplicity and the purpose of explanation.

(0,2) and (0,-4)

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mamaluj [8]

Answer:

16 petunias

Step-by-step explanation:

28/7 = x/4

7x=28*4

7x= 112

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x=16

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2 years ago

Judson collected information about the name length and the population of a random sample of 296 American cities. Here are the re

Natasha2012 [34]

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2 years ago

Kara was baking a cake. The recipe called for 8.5 cups of flour and she put in 3.2 cups so far. How many more cups of flour does

alina1380 [7]

Answer:

5.3 cups of flour

Step-by-step explanation:

8.5 - 3.2 = 5.3 cups of flour

5 0

2 years ago

Help pls …..———<br> :….—-

LenaWriter [7]

Answer:

36s ~ 360º

1s ~ ?

so 1 * 360 / 36 = 10º

36*3/4=27s (closest)

36*1/4=9s (farthest)

20ft

4 0

2 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

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

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

5 0

2 years ago

Read 2 more answers