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3 years ago

WILL MARK BRAINLIEST!! NEED ANSWER FAST PLEASE!!!!

Mathematics

1 answer:

slega [8]3 years ago

6 0

Answer:

positive,linear association

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The table shows values that represent a quadratic function

Mamont248 [21]

Answer:

- 8

Step-by-step explanation:

The average rate of change of f(x) in the closed interval [ a, b ] is

$\frac{f(b)-f(a)}{b-a}$

here [ a, b ] = [ 3, 5 ]

From the table

f(b) = f(5) = - 26

f(a) = f(3) = - 10

Hence

average rate of change = $\frac{-26-(-10)}{5-3}$ = $\frac{-16}{2}$ = - 8

5 0

3 years ago

The problem has been started for you. 9 StartLongDivisionSymbol 780 EndLongDivisionSymbol minus 72 = 60. 60 minus 54 = a remaind

Fofino [41]

Answer:

86.6

The quotient is the final answer

Step-by-step explanation:

(780/9) = quotient ( 86.66)

3 0

3 years ago

Read 2 more answers

Mrs. Green plants 50 red, pink, and purple tulips. If 35% are red and 27% are pink. how many purple tulips did she plant?

kari74 [83]

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6 0

3 years ago

Read 2 more answers

Help!! i need help asap

umka21 [38]

Y = mx + b
y = -4x -2

trying option 1

-8x + 2y = -4

2y = -4 + 8x

y = -2 + 4x

so its incorrect

Now trying option 2

8x + 2y = -4

2y = -4 - 8x

y = -2 - 4x

y = -4x -2

so Option B is correct

7 0

3 years ago

A bucket that weighs 6 lb and a rope of negligible weight are used to draw water from a well that is 80 ft deep. The bucket is f

zaharov [31]

Answer:

3200 ft-lb

Step-by-step explanation:

To answer this question, we need to find the force applied by the rope on the bucket at time $t$

At $t=0, the weight of the bucket is 6+36=42 \mathrm{lb}$

After $t$ seconds, the weight of the bucket is $42-0.15 t \mathrm{lb}$

Since the acceleration of the bucket is the force on the bucket by the rope is equal to the weight of the bucket.

If the upward direction is positive, the displacement after $t$ seconds is $x=1.5 t$

Since the well is 80 ft deep, the time to pull out the bucket is $\frac{80}{2}=40 \mathrm{~s}$

We are now ready to calculate the work done by the rope on the bucket.

Since the displacement and the force are in the same direction, we can write

$W=\int_{t=0}^{t=36} F d x$

Use $x=1.5 t$ and $F=42-0.15 t$

$W=\int_{0}^{36}(42-0.15 t)(1.5 d t)$

$=\int_{0}^{36} 63-0.225 t d t$

$=63 \cdot 36-0.2 \cdot 36^{2}-0=3200 \mathrm{ft} \cdot \mathrm{lb}$

$=\left[63 t-0.2 t^{2}\right]_{0}^{36}$

$W=3200 \mathrm{ft} \cdot \mathrm{lb}$

4 0

3 years ago