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BartSMP [9]
3 years ago
13

How many terms are in 2 + 4x + 7y

Mathematics
2 answers:
TEA [102]3 years ago
7 0
Three terms are there
Nikolay [14]3 years ago
3 0
There are three terms
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Worth 11 points please help me!
Vesnalui [34]

Answer:

124°

I hope it's helps you

7 0
3 years ago
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Consider the following equation, bh+hr=25 when solving for r
Komok [63]
Solve for r.

You want to get r by itself on one side on the equal sign.

bh + hr = 25

Subtract bh from both sides.

hr = 25 - bh

Divide h on both sides.

r = 25 - bh / h

The two h's cancel each other out.

r = 25 - b

Hope this helps!
4 0
3 years ago
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3. A copy machine makes 60 copies per minute. A second
Brums [2.3K]

Answer:

E

Step-by-step explanation:

6 x 480 + 8 x 60 = 960..........

8 0
3 years ago
Choose the domain for which each function is defined
klemol [59]

Answer:

Part 1) f(x)=\frac{x+4}{x} -----> x\neq 0

Part 2) f(x)=\frac{x}{x+4} ----> x\neq -4

Part 3)  f(x)=x(x+4) ----> All real numbers

Part 4) f(x)=\frac{4}{x^2+8x+16} ----> x\neq -4

Step-by-step explanation:

we know that

The domain of a function is the set of all possible values of x

Part 1) we have

f(x)=\frac{x+4}{x}

we know that

In a quotient the denominator cannot be equal to zero

so

For the value of x=0 the function is not defined

therefore

The domain is

x\neq 0

Part 2) we have

f(x)=\frac{x}{x+4}

we know that

In a quotient the denominator cannot be equal to zero

so

For the value of x=-4 the function is not defined

therefore

The domain is

x\neq -4

Part 3) we have

f(x)=x(x+4)

Applying the distributive property

f*(x)=x^2+4x

This is a vertical parabola open upward

The function is defined by all the values of x

therefore

The domain is all real numbers

Part 4) we have

f(x)=\frac{4}{x^2+8x+16}

we know that

In a quotient the denominator cannot be equal to zero

so

Equate the denominator to zero

x^2+8x+16=0

Remember that

x^2+8x+16=(x+4)^2

(x+4)^2=0

The solution is x=-4

so

For the value of x=-4 the function is not defined

therefore

The domain is

x\neq -4

6 0
3 years ago
A city had population 67,255 on january 1, 2000, and its population has been increasing by 2935 people each year since then. A l
irina1246 [14]

We are given: On january 1, 2000 initial population   = 67,255.

Number of people increase each year = 2935 people.

Therefore, 67,255 would be fix value and 2935 is the rate at which population increase.

Let us assume there would be t number of years after year 2000 and population P after t years is taken by function P(t).

So, we can setup an equation as

Total population after t years = Number of t years * rate of increase of population + fix given population.

In terms of function it can be written as

P(t) = t * 2935 + 67255.

Therefore, final function would be

P(t) = 2935t +67255.

So, the correct option is d.P(t) = 67255 + 2935t.

4 0
3 years ago
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