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2 years ago

If the domain and range have the same number is it still a function??

Mathematics

1 answer:

Delicious77 [7]2 years ago

6 0

Yes ..................

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Consider the expression: (x + 3)•(y + 1)•(x + 2)

emmainna [20.7K]

Answer:

Step-by-step explanation:

First we multiply x through (y+1). This equals xy +x.

Second we take 3 through (y+1). This equals 3y+3.

Next we take xy + x through (x+2). This equals 2x squared, y + 2x squared.

Fourth we take 3y+3 through (x+2). This equals 6xy + 6x.

5 0

2 years ago

12 belongs to which set(s) of numbers?o Natural Numbers, Whole Numbers, and Integers onlyNatural Numbers and Whole Numbers onlyo

erastova [34]

Integers are written without no fractional components so 12 is an integer .

Natural numbers are positive numbers just like the number we count. 12 is a natural number.

Rational numbers

3 0

4 months ago

Trigonometry

saul85 [17]

Answer:

9.7

Step-by-step explanation:

tan=opposite/adjacent

tan=9/adjacent

tan 43=9/x

x tan 43=9

0.9325x=9

0.9325x/0.9325=9/0.9325

=9.65

hence;

x=9.7

7 0

2 years ago

To solve the equation w/8 - 4 = 20, you would first add 4 to each side and then divide each side by 8.

bearhunter [10]

False because, you DO add 4 to the sides but you would multiply the 8 because you to the opposite to cancel it out

4 0

2 years ago

Read 2 more answers

Sinx = 1/2, cosy = sqrt2/2, and angle x and angle y are both in the first quadrant.

Leviafan [203]

Answer:

Option D. 3.73

Step-by-step explanation:

we know that

$tan(x+y)=\frac{tan(x)+tan(y)}{1-tan(x)tan(y)}$

and

$sin^{2}(\alpha)+cos^{2}(\alpha)=1$

step 1

Find cos(X)

we have

$sin(x)=\frac{1}{2}$

we know that

$sin^{2}(x)+cos^{2}(x)=1$

substitute

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

$cos^{2}(x)=1-\frac{1}{4}$

$cos^{2}(x)=\frac{3}{4}$

$cos(x)=\frac{\sqrt{3}}{2}$

step 2

Find tan(x)

$tan(x)=sin(x)/cos(x)$

substitute

$tan(x)=1/\sqrt{3}$

step 3

Find sin(y)

we have

$cos(y)=\frac{\sqrt{2}}{2}$

we know that

$sin^{2}(y)+cos^{2}(y)=1$

substitute

$sin^{2}(y)+(\frac{\sqrt{2}}{2})^{2}=1$

$sin^{2}(y)=1-\frac{2}{4}$

$sin^{2}(y)=\frac{2}{4}$

$sin(y)=\frac{\sqrt{2}}{2}$

step 4

Find tan(y)

$tan(y)=sin(y)/cos(y)$

substitute

$tan(y)=1$

step 5

Find tan(x+y)

$tan(x+y)=\frac{tan(x)+tan(y)}{1-tan(x)tan(y)}$

substitute

$tan(x+y)=[1/\sqrt{3}+1}]/[{1-1/\sqrt{3}}]=3.73$

7 0

2 years ago