#19: I need to find out if the equation is even, odd or neither and I'm stuck

Use the graph of f below. Assume the entire function is graphed below. Find the domain

enot [183]

ANSWER

The domain of the function is

$- 3 \leqslant x \leqslant 6$

or

[-3,6]

EXPLANATION

The domain of a function refers to the values of x for which the function is defined.

From the graph, we can see that the function begins at (-3,6) and ends at (6,-4).

Therefore the function is defined for x=-3 to x=6.

Hence the domain of the function is

$- 3 \leqslant x \leqslant 6$

or

[-3,6]

5 0

3 years ago

At an animal shelter, there are 15 dogs and 12 cats.

Bond [772]

Answer:

Dogs: 44.44 Cats: 55.556

Step-by-step explanation:

*this is only if you are allowed to use a calculator*

1. You have to add the total of the cats and dogs

Why? Here, let me put it this way. You have a test with two sections: math and science. If there are 2 Maths Questions and 2 Science questions, what percent of the test are math questions? You would not say 2/2 and say 100. You would say 2/4 because it is obvious. You do the same in this case but it is not as noticeable because you are dealing with bigger numbers.

15 + 12 = 27.

2. Solve

a. There are 15 dogs and 12 cats that makes up 27 pets total.

So you have to do 12/27 for cats,

15/27 for dogs. The fractions equal one and It shows you how much of one is there compared to the total.

b. Simplify

You can simplify these fractions:

for 12/27, divide the numerator and the denominator by 3.

You get 4/9

Divide 15/27 for dogs by 3 as well to get 5/9.

5+4 is 9, which means that I am doing the math correctly.

According to the calculator, 5/9 is 55.556, and 4/9 is 44.444

They should be accurate because there are more dogs than cats and the percentage for dogs are higher. Hence, this is correct.

Cats = 4/9 = 44.444%

Dogs = 5/9 = 55.556%

Hope this helps :)

-jp524

7 0

2 years ago

You are given a number line with a left endpoint R, a point in between the two endpoints (not a midpoint though) of S, and a rig

Nina [5.8K]

You just add the lengths of RS and ST together. In your case RT= 24

3 0

2 years ago

Can you help me solve this please

Lostsunrise [7]

9514 1404 393

Answer:

180 days

Step-by-step explanation:

The formulas for the volume of a cone and a cylinder are ...

V = πr²h . . . . . volume of a cone

V = (1/3)πr²h . . . . . volume of a cylinder

Comparing these formulas, you can see that the 6' high cone at the top will have the same volume as a cylinder of the same radius that is 2' high. So, we can simplify the problem to finding the volume of a cylinder that is 66+2 = 68' high.

V = π(12 ft)²(68 ft) = 9792π ft³ ≈ 30752.5 ft³

When the corn is used at the rate of 170 ft³/day, it will last ...

(30752.5 ft³)/(170 ft³/day) ≈ 180.96 days

The corn will last 180 days.

_____

Additional comment

In scenarios like this, it is inappropriate to round up. There will be about 162 ft³ of corn left after 180 days, which is not enough for an additional day.

7 0

2 years ago

Find an n^th degree polynomial with real coefficients satisfying the given conditions. n = 3; -2 and 2 i are zeros; f(-1) = 15.

Ira Lisetskai [31]

So, n = 3, is a 3rd degree polynomial, roots are -2 and 2i

well 2i is a complex root, or imaginary, and complex root never come all by their lonesome, their sister is always with them, the conjugate, so if 0+2i is there, 0-2i is there too

so, the roots are -2, 2i, -2i

now... $\bf \begin{cases}
x=-2\implies x+2=0\implies &(x+2)=0\\
x=2i\implies x-2i=0\implies &(x-2i)=0\\
x=-2i\implies x+2i=0\implies &(x+2i)=0
\end{cases}
\\\\\\
(x+2)\underline{(x-2i)(x+2i)}=0\\\\
-----------------------------\\\\
\textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)\\\\
-----------------------------\\\\
(x+2)[x^2-(2i)^2]=0\implies (x+2)[x^2-(2^2i^2)]=0
\\\\\\
(x+2)[x^2-(4\cdot -1)]=0\implies (x+2)(x^2+4)=0
\\\\\\
x^3+2x^2+4x+8=0$

now, if we check f(-1), we end up with 5, not 15
hmmm

so, how to turn our 5 to 15? well, 3*5, thus

$\bf 3(x^3+2x^2+4x+8)=f(x)\implies 3(5)=f(-1)\implies 15=f(-1)$

usually, when we get the roots, or zeros, if any common factor that is a constant is about, they get in a division with 0 and get tossed, and aren't part of the roots, thus, we can simply add one, in this case, the common factor of 3, to make the 5 turn to 15

6 0

3 years ago