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Naddika [18.5K]
3 years ago
13

What is the range of f(x)=−3x−5?

Mathematics
1 answer:
Murrr4er [49]3 years ago
6 0
The range is (-−
∞
,
∞)
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What is the results when 4x^3+11x^3-9x^2-8x+30 is divided by x+3
sladkih [1.3K]

Answer:

4x^3 -x^2 -6x +10

Step-by-step explanation:

You can perform the division of (4x^4 +11x^3 -9x^2 -8x +30)/(x +3) by polynomial long division or by synthetic division. The latter is shown below.

The result of the division is ...

4x^3 -x^2 -6x +10

4 0
3 years ago
Simplify each expression xy6 / y4x
Rus_ich [418]
Hello there!
Simplify the expression
y^{2}

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7 0
3 years ago
At 2:00 PM a car's speedometer reads 30 mi/h. At 2:20 PM it reads 50 mi/h. Show that at some time between 2:00 and 2:20 the acce
storchak [24]

Answer:

The correct answer will be "60 min/hour".

Step-by-step explanation:

At time 2:00 PM

A car reads = 30 mi/h

At time 2:20 PM

A car reads = 50 mi/h

Now,

At time 2:00 PM

t = 0

At time 2:20 PM

t = \frac{1}{3}

v(0)=30

v(\frac{1}{3})=50

For MVT,

v'(c)=\frac{v(\frac{1}{3} -v(0))}{(\frac{1}{3} )}

       =\frac{20}{\frac{1}{3} }

       =60 \ min/hour

5 0
3 years ago
Please help me with this! Picture is provided and only answer is you know it!
Sophie [7]

Answer:

Option A

Step-by-step explanation:

Number of components assembled by the new employee per day,

N(t) = \frac{50t}{t+4}

Number of components assembled by the experienced employee per day,

E(t) = \frac{70t}{t+3}

Difference in number of components assembled per day by experienced ane new employee

D(t)= E(t) - N(t)

D(t) = \frac{70t}{t+3}-\frac{50t}{t+4}

      = \frac{70t(t+4)-50t(t+3)}{(t+3)(t+4)}

      = \frac{70t^2+280t-50t^2-150t}{(t+3)(t+4)}

      = \frac{20t^2+130t}{(t+3)(t+4)}

      = \frac{10t(2t+13)}{(t+4)(t+3)}

Therefore, Option A will be the answer.

3 0
3 years ago
Which of the following expressions has a value of 2/5 in the picture below?
Anuta_ua [19.1K]

Step-by-step explanation:

<u>Step 1:  List out all of the formulas for the trigonometric functions </u>

<em>sin(x) </em>= opposite/hypotenuse

<em>cos(x) </em>= adjacent/hypotenuse

<em>tan(x) </em>= opposite/adjacent

<u>Step 2:  Find the following expression that has a value of 2/5 </u>

sin(B) = opposite/hypotenuse = ?/5 is FALSE

cos(B) = adjacent/hypotenuse = 2/5 is TRUE

tan(B) = opposite/adjacent = ?/2 is FALSE

Answer:  The correction expression that has a value of 2/5 is option B

5 0
3 years ago
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