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2 years ago

Does the graph represent a function and if so, why?

Mathematics

1 answer:

Dmitriy789 [7]2 years ago

7 0

Answer:

A. yes

Step-by-step explanation:

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PLEASE HELP ME I HAVING TROUBLE

viktelen [127]

Answer: f(-6) = $-\frac{12}{11}$ , f(-4) = $\frac{8}{9}$ , f(4) = $-\frac{8}{9}$ , f(6) = $\frac{12}{11}$

<u>Step-by-step explanation:</u>

(-6, f(-6)) is an x,y coordinate. They are asking what the y-value is when you plug in -6 for x.

f(x) = $\frac{2x}{x^{2}-25}$

f(-6) = $\frac{2(-6)}{(-6)^{2}-25}$

= $\frac{-12}{36-25}$

= $-\frac{12}{11}$

f(-4) = $\frac{2(-4)}{(-4)^{2}-25}$

= $\frac{-8}{16-25}$

= $\frac{-8}{-9}$

= $\frac{8}{9}$

f(4) = $\frac{2(4)}{(4)^{2}-25}$

= $\frac{8}{16-25}$

= $\frac{8}{-9}$

= $-\frac{8}{9}$

f(6) = $\frac{2(6)}{(6)^{2}-25}$

= $\frac{12}{36-25}$

= $\frac{12}{11}$

5 0

2 years ago

In the scale drawing, what will the length of the west side of the backyard?

Shtirlitz [24]

The diagram shows a lawn with the north side measuring 50 units and the west side measuring 28 units. The deck is scaled down by a factor of 1/5.

This is because the ratio of the length of the lawn to that of the deck is 1 : 5.

Hence we can now ca

7 0

1 year ago

the first term of an arithmetic sequence is 4 and the tenth term is 67. what is the common difference?

Tatiana [17]

Plss see the attachment

3 0

2 years ago

Statistics??? Construct a 95% confidence interval for the population mean. Assume the population has a normal distribution. A sa

antiseptic1488 [7]

Given:
μ = $3120, population mean
σ = $677, population standard deviation.
The population is normally distributed.
n = 20, sample size.

At the 95% confidence level, the confidence interval is
$( \mu - 1.96\frac{\sigma}{ \sqrt{n} } ,\, \mu + 1.96\frac{\sigma}{ \sqrt{n} })$

1.96(σ/√n) = 1.96(677/√20) = 296.71
Th confidence interval for the mean is
($2823.29, $3416.71)

Answer: ($2823.29, $3416.71)

6 0

2 years ago

A certain restaurant offers 8 different salads, 5 different main courses, 6 different desserts. If customers choose one salad, o

Naddika [18.5K]

Answer: E. 1200

Step-by-step explanation:

Given : A certain restaurant offers 8 different salads, 5 different main courses, 6 different desserts.

i.e. No. of choices for different salads= 8 -----(1)

No. of choices for different main courses = 5 ----(2)

No. of choices for different desserts = 6

When we choose two different desserts then we use permutation(repeatition not allowed) :

$^6P_2=\dfrac{6!}{(6-2)!}=\dfrac{6\times5\times4!}{4!}=6\times5=30$ ----(3)

[∵ No. of ways to choose r things out of n = $^nP_r=\dfrac{n!}{(n-r)!}$ ]

If customers choose one salad, one main course and two different desserts for their meal , then By Fundamental principle of counting (Multiply (1) , (2) and (3)), the number of different meals are possible :-

$8\times5\times30\\\\=1200$

Hence, the correct answer is E. 1200 .

8 0

2 years ago