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3 years ago

14

≧﹏≦ PLEASE HELP PLEASE HELP PLEASE HELP ≧﹏≦

1 answer:

liq [111]3 years ago

8 0

Answer:

the answer is j=x and k=y

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Fill in the blank and which goes where, drag and drop.. <br><br> 80%<br> 95%<br> 33 1/3%<br> 15%

scoundrel [369]

1. = 80 2 = 95 3 = 33 4 = 15

4 0

2 years ago

• (4xy^4)(– 2x^2y^5)

Eva8 [605]

Answer:

This will be -8x^3y^9.

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3 years ago

What is the area of this figure drag and drop the appropriate number into the box

lyudmila [28]

Answer:

142ft squared

Step-by-step explanation:

It was actually very easy, the working may not be clear and hard to understand but it's on the attached image. Hope I helped! :)

4 0

3 years ago

<img src="https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cfrac%7B%282x%2B1%29%28x-5%29%7D%7B%28x-5%29%28x%2B4%29%5E%7B2%7D%20%7D" id

dybincka [34]

i) The given function is

$f(x)=\frac{(2x+1)(x-5)}{(x-5)(x+4)^2}$

The domain is

$(x-5)(x+4)^2\ne 0$

$(x-5)\ne0,(x+4)^2\ne 0$

$x\ne5,x\ne -4$

ii) For vertical asymptotes, we simplify the function to get;

$f(x)=\frac{(2x+1)}{(x+4)^2}$

The vertical asymptote occurs at

$(x+4)^2=0$

$x=-4$

iii) The roots are the x-intercepts of the reduced fraction.

Equate the numerator of the reduced fraction to zero.

$2x+1=0$

$2x=-1$

$x=-\frac{1}{2}$

iv) To find the y-intercept, we substitute $x=0$ into the reduced fraction.

$f(0)=\frac{(2(0)+1)}{(0+4)^2}$

$f(0)=\frac{(1)}{(4)^2}$

$f(0)=\frac{1}{16}$

v) The horizontal asymptote is given by;

$lim_{x\to \infty}\frac{(2x+1)}{(x+4)^2}=0$

The horizontal asymptote is $y=0$ .

vi) The function has a hole at $x-5=0$ .

Thus at $x=5$ .

This is the factor common to both the numerator and the denominator.

vii) The function is a proper rational function.

Proper rational functions do not have oblique asymptotes.

6 0

3 years ago

The sum of three consecutive integers if the first is z

valentinak56 [21]

The answer should 1x

3 0

3 years ago