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Zolol [24]
4 years ago
15

What is the range of the function on the graph?

Mathematics
1 answer:
Paladinen [302]4 years ago
6 0
All real numbers less than or equal to 3
You might be interested in
What are the coordinates of the hole in the graph of the function?
MakcuM [25]

Answer:

The hole is at (5,7)

Step-by-step explanation:

x^2 − 3x − 10

-------------------

x−5

Factor the numerator

(x-5)(x+2)

-------------------

x−5

There is a hole at x=5 since it will cancel in the numerator and the denominator

f(x) = x+2  and letting x = 5

f(5) = 5+2

The hole is at (5,7)

4 0
3 years ago
Shoe Boxes A department store is selling its plastic shoe boxes for $1.50 off the
Firdavs [7]
T=4(r-1.5)

4(3.59-1.5)=t= 8.36 dolla

Enjoy.
7 0
3 years ago
Can anyone help me please
guajiro [1.7K]

Answer:

a) 5 people

b) Yes.

Step-by-step explanation:

The best way to solve this problem is to use the amount of work needed each week and divide it by the amount worked by each person (150/32). This will give you 4.6875. However, you cannot have 6/10 of a person, so you must round up to the next whole number, 5. There are 5 people needed to fulfill the amount of hours posted. For Part B, you will be decreasing the hours worked by each person from 32 to 30. Use the same method you did above and divide the total amount of work needed by the amount worked by each person (150/30) to get 5 on the dot. Your answer is Yes.

I hope this helps! Good luck!

6 0
3 years ago
Type an equivalent expression to show the relationship of multiplication and addition.<br><br> 3y
defon

Answer:

y + y + y

Step-by-step explanation:

The term 3y means that the variable y should be added together three times.

Addition is equivalent to multiplication in that given an expression mx, where m is a real number. the expression indicates that the variable x should be added together in m number of times

8 0
3 years ago
Find a particular solution to the differential equation y′′+2y′+y=2t2−t+3e−2t. y′′+2y′+y=2t2−t+3e−2t.
Jlenok [28]
y''+2y'+y=2t^2-t+3e^{-2t}

The characteristic equation is

r^2+2r+1=(r+1)^2=0\implies r=-1

so the characteristic solution is

y_c=C_1e^{-t}+C_2te^{-t}

For the particular solution, we can try looking for a solution of the form

y_p=at^2+bt+c+de^{-2t}
\implies{y_p}'=2at+b-2de^{-2t}
\implies{y_p}''=2a+4de^{-2t}

Substituting into the ODE, we have

(2a+4de^{-2t})+2(2at+b-2de^{-2t})+(at^2+bt+c+de^{-2t})=2t^2-t+3e^{-2t}
at^2+(4a+b)t+(2a+2b+c)+de^{-2t}=2t^2-t+3e^{-2t}
\implies\begin{cases}a=2\\4a+b=-1\\2a+2b+c=0\\d=3\end{cases}\implies a=2,b=-9,c=14,d=3

So the general solution to the ODE is

y=y_c+y_p
y=C_1e^{-t}+C_2te^{-t}+2t^2-9t+14+3e^{-2t}
4 0
3 years ago
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