MATHEMATICS COLLEGE

A box is in the shape of a triangular prism. . The base of the triangular face of the box is 11 centimeters (cm). The height of the triangular face of the box is 7.5 cm. The length of the box is 15 cm. V =1/2hxbxl What is the volume of the box? A 277.5 cm B 577.5 cm C 618.75 cm D1237.5 cm

Answers

Answer 1

Answer:

Answer:

$618.75 cm^(3)$

Step-by-step explanation:

$((11 cm)(7.5 cm)(15 cm))/(2) =618.75 cm^(3)$

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Which function is even? check all that applya. y = csc x
b. y = sin x
c. y = sec x
d. y = cos x

Answers

The function is even will be y = sec x and y = cos x. Then the correct options are C and D.

What is an even function?

Even Function - A true function f(x) is said to be an even function if the output value of f(-x) is the same as the f(x) for all values of x in the domain of f.

The equation should be stored in an even function:

f(-x) = f(x)

Check all that apply.

a. y = csc x, then replace x with -x. Then we have

y = csc -x

y = 1/sin -x

y = - 1/ sin x

y = -csc x

b. y = sin x, then replace x with -x. Then we have

y = sin -x

y = - sin x

c. y = sec x, then replace x with -x. Then we have

y = sec-x

y = 1/cos -x

y = 1/ cos x

y = sec x

d. y = cos x, then replace x with -x. Then we have

y = cos -x

y = cos x

Thus, the function is even will be y = sec x and y = cos x.

Then the correct options are C and D.

More about the even function link is given below.

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Answer: c and d

Step-by-step explanation: they’re both even

Rachel is recording the temperature at her school every morning this week. On Monday, the temperature was 5°C. Part A: On Wednesday, Rachel calculated that the temperature was 13 degrees colder than Monday. What temperature did Rachel record on Wednesday? Part B: On Friday, Rachel calculated that the temperature was 4 degrees warmer than Wednesday. What temperature did Rachel record on Friday?

Answers

Answer:

Part A: -8

Part B: -4

Step-by-step explanation:

$5-13=-8$

$-8+4=-4$

Can someone please help to explain this to me? I have been trying to work at it and have spent time watching multiple videos trying to figure it out, but am unable to figure it out. Any help would be much appreciated! Thank you.

Answers

Answer:

first 1 I think

Step-by-step explanation:

Consider the following functions. f(x) = x − 3, g(x) = x2 Find (f + g)(x). Find the domain of (f + g)(x). (Enter your answer using interval notation.) Find (f − g)(x). Find the domain of (f − g)(x). (Enter your answer using interval notation.) Find (fg)(x). Find the domain of (fg)(x). (Enter your answer using interval notation.) Find f g (x). Find the domain of f g (x). (Enter your answer using interval notation.)

Answers

Answer:

$(f+g)(x)=x-3+x^2$ ; Domain = (-∞, ∞)

$(f-g)(x)=x-3-x^2$ ; Domain = (-∞, ∞)

$(fg)(x)=x^3-3x^2$ ; Domain = (-∞, ∞)

$((f)/(g))(x)=(x-3)/(x^2)$ ; Domain = (-∞,0)∪(0, ∞)

Step-by-step explanation:

The given functions are

$f(x)=x-3$

$g(x)=x^2$

$(f+g)(x)=f(x)+g(x)$

Substitute the values of the given functions.

$(f+g)(x)=(x-3)+x^2$

$(f+g)(x)=x-3+x^2$

The function $(f+g)(x)=x-3+x^2$ is a polynomial which is defined for all real values x.

Domain of (f+g)(x) = (-∞, ∞)

$(f-g)(x)=f(x)-g(x)$

Substitute the values of the given functions.

$(f-g)(x)=(x-3)-x^2$

$(f-g)(x)=x-3-x^2$

The function $(f-g)(x)=x-3-x^2$ is a polynomial which is defined for all real values x.

Domain of (f-g)(x) = (-∞, ∞)

$(fg)(x)=f(x)g(x)$

Substitute the values of the given functions.

$(fg)(x)=(x-3)x^2$

$(fg)(x)=x^3-3x^2$

The function $(fg)(x)=x^3-3x^2$ is a polynomial which is defined for all real values x.

Domain of (fg)(x) = (-∞, ∞)

$((f)/(g))(x)=(f(x))/(g(x))$

Substitute the values of the given functions.

$((f)/(g))(x)=(x-3)/(x^2)$

The function $((f)/(g))(x)=(x-3)/(x^2)$ is a rational function which is defined for all real values x except 0.

Domain of (f/g)(x) = (-∞,0)∪(0, ∞)

$(f + g)(x) = x^2 + x - 3$ , domain: all real numbers.

$(f - g)(x) = -x^2 + x - 3$ , domain: all real numbers.

$(fg)(x) = x^3 - 3x^2$ , domain: all real numbers.

$f(g(x)) = x^2 - 3$ , domain: all real numbers.

To find (f + g)(x), we need to add the functions f(x) and g(x).

The function f(x) = x - 3 and the function $g(x) = x^2.$

So, $(f + g)(x) = f(x) + g(x) = (x - 3) + (x^2).$

Expanding this equation, we get $(f + g)(x) = x^2 + x - 3.$

To find the domain of (f + g)(x), we need to consider the domain of the individual functions f(x) and g(x).

Since both f(x) = x - 3 and $g(x) = x^2$ are defined for all real numbers, the domain of (f + g)(x) is also all real numbers.

To find (f - g)(x), we need to subtract the function g(x) from f(x).

So, $(f - g)(x) = f(x) - g(x) = (x - 3) - (x^2).$

Expanding this equation, we get $(f - g)(x) = -x^2 + x - 3.$

The domain of (f - g)(x) is also all real numbers, since both f(x) and g(x) are defined for all real numbers.

To find (fg)(x), we need to multiply the functions f(x) and g(x).

So, $(fg)(x) = f(x) * g(x) = (x - 3) * (x^2).$

Expanding this equation, we get $(fg)(x) = x^3 - 3x^2.$

The domain of (fg)(x) is all real numbers, since both f(x) and g(x) are defined for all real numbers.

To find f(g(x)), we need to substitute g(x) into the function f(x).

So, $f(g(x)) = f(x^2) = x^2 - 3.$

The domain of f(g(x)) is also all real numbers, as $g(x) = x^2$ is defined for all real numbers, and f(x) = x - 3 is defined for all real numbers.

In summary:

- $(f + g)(x) = x^2 + x - 3$ , domain: all real numbers.

- $(f - g)(x) = -x^2 + x - 3$ , domain: all real numbers.

- $(fg)(x) = x^3 - 3x^2$ , domain: all real numbers.

- $f(g(x)) = x^2 - 3$ , domain: all real numbers.

To Learn more about real numbers here:

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Find the sum 5.5+7.25-5.5+(-7.25) show your work

Answers

Answer:

5.5+7.25−5.5−7.25 = 0

Step-by-step explanation:

To be honest, you shouldn't have to show work for an equation like this. But this is the answer. Hope it helped.

Answer:

5.5+7.25-5.5+(-7.25)

step 1: add 5.5 and 7.25

12.75−5.5−7.25

step 2: subtract 5.5 from 12.75

7.25−7.25

The answer is 0.

Step-by-step explanation:

Brainliest?

If Jim drove 5 miles to work. Kari drove 3 times as many miles as Jim. Sondra drove 4 times as many miles as Kari. How many more miles did Sondra drive than Kari.