All categories

2 years ago

I need help finding functions

Mathematics

1 answer:

velikii [3]2 years ago

3 0

Answer:

y= 7
y= -3
y= -4
y= -8

Step-by-step explanation:

y= -5x+1

=-5.-1+1

Now you can do this method

You might be interested in

Explain how bacteria are both helpful and harmful to out bodies

Pavel [41]

Answer:

Step-by-step explanation:

Bacteria is helpful because it helps protect the cells in your intestines. Its, harmful because itI can cause a disease in your lungs.

Hope this helps!

8 0

2 years ago

Suppose a person has the following pairs of pants: white, black, khaki, and navy; the following shirts: red, blue,

marissa [1.9K]

Answer:

The answer is A, 72

8 0

2 years ago

The value of the exponent in (22)−5 is

lyudmila [28]

Answer:

Let's calculate:

22-5

• Remove minus in the exponent and invert the base:

= (1/22)5

• Expand the multiplication:

= (1/22) * (1/22) * (1/22) * (1/22) * (1/22)

• Find your result:

= 1/5153632 ≈ 0.000000194038

3 0

2 years ago

Cathy earns $32 after babysitting for 5 hours. How much money does she earn per hour?

telo118 [61]

The answer would be $6.40

4 0

2 years ago

Read 2 more answers

Pls help this is pretty urgent

KIM [24]

Answer:

(a) 0

(b) f(x) = g(x)

Step-by-step explanation:

Given rational function:

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

Part (a)

Factor the numerator and denominator of the given rational function:

$\begin{aligned} \implies f(x) & = \dfrac{x^2+2x+1}{x^2-1} \\\\& = \dfrac{(x+1)^2}{(x+1)(x-1)}\\\\& = \dfrac{x+1}{x-1}\end{aligned}$

Substitute x = -1 to find the limit:

$\displaystyle \lim_{x \to -1}f(x)=\dfrac{-1+1}{-1-1}=\dfrac{0}{-2}=0$

Therefore:

$\displaystyle \lim_{x \to -1}f(x)=0$

Part (b)

From part (a), we can see that the simplified function f(x) is the same as the given function g(x). Therefore, f(x) = g(x).

Part (c)

As x = 1 is approached from the right side of 1, the numerator of the function is positive and approaches 2 whilst the denominator of the function is positive and gets smaller and smaller (approaching zero). Therefore, the quotient approaches infinity.

$\displaystyle \lim_{x \to 1^+} f(x)=\dfrac{\to 2^+}{\to 0^+}=\infty$

5 0

9 months ago

Read 2 more answers

I need help finding functions​

I need help finding functions