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3 years ago

The graph of the function f(x) is shown below.

Mathematics

1 answer:

faust18 [17]3 years ago

5 0

Remember, parenthaees are like < and > and brackets ar like ≤ and ≥

domain is how far the x values go
x is left to right

we see they go from -3 to 5, with a filled in dot at -3 and empty dot at 5
means include -3 but not including 5
so like -3≤x<5
or in interval notation
[-3,5) is the domain

range
highest to lowest y value

range is from y=3 to y=-1
we gots full dots so we use brackets
range is [-1,3]

Domain=[-3,5)
Range=[-1,3]

B. read above and understand it

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it took Simon 33 minutes to run 5.5 miles. did he run faster or slower than 1 mile every 5 minutes how can you tell

vitfil [10]

Simon ran slower than 1 mile every 5 minutes.

Because 33 / 5.5 = 6. So he ran 1 mile every 6 minutes, not 5.

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2 years ago

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2 years ago

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Combine and simplify the following radical expression ASAP ASAP ASAP ASAP

Montano1993 [528]

Answer:

$\sqrt[3]{3}$

Step-by-step explanation:

Our expression is: $\frac{1}{3} \sqrt[3]{81}$ .

Let's focus on the cube root of 81 first. What's the prime factorisation of 81? It's simply: 3 * 3 * 3 * 3, or $3^3*3$ . Put this in for 81:

$\sqrt[3]{81} =\sqrt[3]{3^3*3}=\sqrt[3]{3^3} *\sqrt[3]{3}$

We know that the cube root of 3 cubed will cancel out to become 3, but the cube root of 3 cannot be further simplified, so we keep that. Our outcome is then:

$\sqrt[3]{3^3} *\sqrt[3]{3}=3\sqrt[3]{3}$

Now, let's multiply this by 1/3, as shown in the original problem:

$\frac{1}{3}* 3\sqrt[3]{3}=\sqrt[3]{3}$

Thus, the answer is $\sqrt[3]{3}$ .

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3 years ago

In the linear programming graph, where is the feasible region compared with the y = 60 line?

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THE ASWER FOR THIS IS B IM SURE OF IT :)

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2 years ago

An oil tank has to be drained for maintenance. The tank is shaped like a cylinder that is 3ft long with a diameter of 2.2 ft . S

Contact [7]

4 minutes 34 seconds will takes to empty the tank, if the starts completely full and oil drained at a rate of 2.5 $ft^{3}$ per minute.

Step-by-step explanation:

The given is,

Tank is shaped like a cylinder that is 3 ft long with a diameter of 2.2 ft.

Oil drained at a rate of 2.5 $ft^{3}$ per minute.

Step:1

Time taken to dry the oil tank is,

T = $\frac{Volume of oil}{Oil drain rate}$ ....................................(1)

Step:2

Volume of the oil is,

$V = \pi r^{2} h$ .................................................(2)

Where, r - Radius of Cylinder

$r = \frac{d}{2}$

$r = \frac{2.2}{2}$

r = 1.1 ft

From eqn (1),

V = $\pi$ × $1.1^{2}$ × 3

= 11.40398 $ft^{3}$

Step:3

From equation (1)

= $\frac{11.40398}{2.5}$

= 4.56

= 4.56 minutes

T = 4 minutes 34 seconds

Result:

Time taken to dry the oil tank is 4 minutes 34 seconds, if a cylinder is 3 ft long with a diameter of 2.2 ft and oil is drained at a rate of 2.5ft^3 per minute.

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2 years ago