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3 years ago

If f(a) = 3a – a2, which of the following are not true statements?

Mathematics

2 answers:

tangare [24]3 years ago

7 0

F(-1)=2 and f(0)=3 are the answers that are not true

svetlana [45]3 years ago

4 0

Answer:

ftjyjytjyujuy

Step-by-step explanation:

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Simplify 3^2/3.3^1/5

san4es73 [151]

ANSWER:

$\frac{1}{5} \\$

EXPLANATION:

$\frac{ \frac{{3}^{2}}{3 \times{3}^{1}}}{5} = \\$

$\frac{ \frac{9}{3 \times 3} }{5} = \\$

$\frac{ \frac{9}{9} }{5} = \\$

$\frac{1}{5} \\$

4 0

3 years ago

Which of the following relationships between two variables could be described using correlation, r?

ladessa [460]

Answer:

C) High temperature of the day and number of zoo visitors that day

Step-by-step explanation:

Correlation describes the relationship between two quantitative variables and both variables are quantitative.

3 0

3 years ago

Last year, a large trucking company delivered 5.5x10^5 tons of goods with an average value of $23,000 per ton. What was the tota

AleksandrR [38]

The total value is:
5. 5 x 10 ^5 * 23 x 10^3 =
= $126.5 * 10^8 =
= $1.265 x 10^10
Answer: B )

4 0

3 years ago

How important is writing the polynomial in standard form when dividing polynomial

Novay_Z [31]

It is verg important because once diving "The length of one cathetus in a right triangle with equal catheti is the squareroot of half of the squared hypotenuse. And yeah

6 0

3 years ago

betty closes the nozzle and fills it completely with a liquid. She then opens the nozzle. If the liquid drips at the rate of 14

natta225 [31]

Answer:

4.71 minutes

Step-by-step explanation:

Incomplete question [See comment for complete question]

Given

Shape: Cone

$r = 3$ -- radius

$h = 7$ --- height

$Rate = 14in^3/min$

Required

Time to pass out all liquid

First, calculate the volume of the cone.

This is calculated as:

$V = \frac{1}{3} \pi r^2h$

This gives:

$V = \frac{1}{3} * 3.14 * 3^2 * 7$

$V = \frac{1}{3} * 197.82$

$V = 65.94in^3$

To calculate the time, we make use of the following rate formula.

$Rate = \frac{Volume}{Time}$

Make Time the subject

$Time= \frac{Volume}{Rate }$

This gives:

$Time= \frac{65.94in^3}{14in^3/min}$

$Time= \frac{65.94in^3}{14in^3}min$

Cancel out the units

$Time= \frac{65.94}{14} min$

$Time= 4.71 min\\$

3 0

3 years ago