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DochEvi [55]
3 years ago
9

Write an equation of the line in slope intercept form through the points (-3,0) and (0,0)

Mathematics
1 answer:
Helga [31]3 years ago
7 0

Answer:

y=0

Step-by-step explanation:

y=mx+b

do you need more information ?

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Find the absolute minimum and absolute maximum values of f on the given interval. f(x) = (x2 − 1)3, [−1, 2]
Alja [10]

Given :

F(x)=3(x^2-1)

To Find :

the absolute minimum and absolute maximum values of f on the given interval

[-1,2] .

Solution :

Now , getting first order differential equation and equating its equal to zero.

\dfrac{d(3x^2-3)}{dx}=0\\\\6x=0\\x=0

So , x=0 is critical point .

Now , coefficient of x^2 is positive .

Therefore , it is increasing function after  x=0 .

So , min value will be at , x=0.

Min = (0^2-1)\times 3=-3

And maximum value will be the maximum at the x=2 because it is increasing function .

Max=(2^2-1)\times 3=9

Therefore , max and min value is 9 and -3 respectively .

Hence , this is the required solution .

3 0
3 years ago
What is 8h+7m <br> need help please!!!!
Helen [10]

Answer: 14 hm

Step-by-step explanation:

Just add numbers together then put letters in order they show in equation

5 0
3 years ago
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A gardener buys a plant that is 12cm in height. Each week after that the plant grows 10cm. Note: The plant is 12cm high at the b
xxMikexx [17]

formula = height = 12+10t

where t = number of weeks

so for t10

 height = 12+10(10) = 12+100 = 112 cm

for t12

height = 12+10(12) = 12 +120 = 132 cm

4 0
3 years ago
A faucet drips 23 gallon of water in 10 hours.
Oliga [24]

Answer:

  55.2 gallons/day

Step-by-step explanation:

There are 2.4 ten-hour periods in one day, so the total number of gallons per day is ...

  (2.4 periods/day)(23 gallons/period) = 55.2 gallons/day

6 0
3 years ago
Jessie estimated the weight of his cat to be 12 pounds. The actual weight of the cat is 15 pounds.
Studentka2010 [4]

Answer:

20%

Step-by-step explanation:

% of error is given by the following formula:

% Error = \frac{|\textrm {Approx Value  - Exact Value}|}{\textrm {Exact Value}} \times 100

Now, Jessie estimated the weight of his cat to be 12 pounds i.e. Approximate value is 12 pounds.

Again, The actual weight of the cat is 15 pounds i.e. the exact value is 15 pounds.

Therefore, the % error = \frac{|12 - 15|}{15} \times 100 = \frac{3}{15} \times 100 = 20% (Answer)

5 0
3 years ago
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