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3 years ago

What’s the domain of (3,7),(4,1),(-1,-8),(2,2),(4,-1)

Mathematics

1 answer:

mafiozo [28]3 years ago

6 0

Answer:

{-1,2,3,4}

Step-by-step explanation:

The domain is the input values

(3,7),(4,1),(-1,-8),(2,2),(4,-1)

(3,4,-1,2,4)

We do not list the repeats

(3,4,-1,2)

(-1,2,3,4)

You might be interested in

Distributive property to solve -9(2-p)

Nikitich [7]

Answer:

9p−18

Step-by-step explanation:

Expand by distributing terms

(−9×2 ) (−9×−p)

Simplify
−18−9×−p

Regroup terms

−18−(−9p)

Remove parentheses

9p−18

4 0

2 years ago

What is the fraction and the percent of the shaded part in each square?

elena-14-01-66 [18.8K]

Answer:

Part a) The fraction is $\frac{5}{9}$

Part b) The percent is $55\frac{5}{9}\%$

Step-by-step explanation:

Part a)

Find the fraction

we know that

To find the fraction of the shaded part in the square divide the number of shaded squares by the total number of squares

Let

x-------> the number of shaded squares

y-------> the total number of squares

f-------> the fraction of the shaded part

$f=\frac{x}{y}$

In this problem we have

$x=20\ shaded\ squares$

$y=36\ total\ squares$

substitutes

$f=\frac{20}{36}$

Simplify

$\frac{20}{36}=\frac{5}{9}$

Part b) Find the percent

To find the percent multiply the fraction by 100

$\frac{5}{9}*100=\frac{500}{9}\%$

convert to mixed number

$\frac{500}{9}=\frac{495}{9}+\frac{5}{9}=55\frac{5}{9}\%$

7 0

3 years ago

Read 2 more answers

If t a n squared θ = 3 /8 , what is the value of s e c θ

TEA [102]

We can use 3 facts:

$\tan{\theta}=\frac{y}{x}$

$\sec{\theta}=\frac{r}{x}$

$x^2+y^2=r^2$

So, to find our x and our y we can do:

$\tan^2{\theta}=\frac{3}{8} \implies \\ \tan{\theta}=\pm{\sqrt{\frac{3}{8}}=\pm\frac{\sqrt{3}}{\sqrt{8}}$ So, now that we know our x and our y we can find our r.

$(\sqrt{3})^2+(\sqrt{8})^2=11 \implies r=\sqrt{11}$

You'll notice, though, that there were two solutions for $\tan{\theta}$ , which means we have a positive and a negative for our x. So:

$\sec{\theta}=\frac{\sqrt{11}}{\sqrt{8}} \\ \sec{\theta}=-\frac{\sqrt{11}}{\sqrt{8}}}\\ \text{Rationalizing both of them:}\\ \sec{\theta}=\frac{\sqrt{88}}{8} \\ \sec{\theta}=-\frac{\sqrt{88}}{8}$