All categories

2 years ago

The temperature at 1:00 p.m. The temperature at 1:00 p.m. on Tuesday was -13°C. There was an increase of 6º per

Mathematics

1 answer:

kenny6666 [7]2 years ago

3 0

Rate of change=m=6°C
Initial temp=b=-13°C

Equation in slope intercept form

y=mx+b
y=6n+(-13)
y=6n-13

Option A

You might be interested in

Idk what this problem is

MaRussiya [10]

7 multiplied by 1/4.

3 0

2 years ago

Read 2 more answers

Find the measures of the interior angles.

3241004551 [841]

Use angle sum property

x+68+x-24=180
2x+44=180
2x=136
x=68

<A=68

<B=44

<C=68°

7 0

2 years ago

Read 2 more answers

How do you solve this (5/6)(5/6)

Vikentia [17]

Answer:

25/36

Step-by-step explanation:

Use the top top, bottom bottom method.

5*5 = 25

6*6 = 36

6 0

3 years ago

Read 2 more answers

How would I do the steps to solve this?

allsm [11]

Answer:

The maximum revenue is 16000 dollars (at p = 40)

Step-by-step explanation:

One way to find the maximum value is derivatives. The first derivative is used to find where the slope of function will be zero.

Given function is:

$R(p) = -10p^2+800p$

Taking derivative wrt p

$\frac{d}{dp} (R(p) = \frac{d}{dp} (-10p^2+800p)\\R'(p) = -10 \frac{d}{dp} (p^2) +800 \ frac{d}{dp}(p)\\R'(p) = -10 (2p) +800(1)\\R'(p) = -20p+800\\$

Now putting R'(p) = 0

$-20p+800 = 0\\-20p = -800\\\frac{-20p}{-20} = \frac{-800}{-20}\\p = 40$

As p is is positive and the second derivative is -20, the function will have maximum value at p = 40

Putting p=40 in function

$R(40) = -10(40)^2 +800(40)\\= -10(1600) + 32000\\=-16000+32000\\=16000$

The maximum revenue is 16000 dollars (at p = 40)

3 0

2 years ago

Hurry, I need help and please explain

Zepler [3.9K]

Answer:

Question 18

=$3.33

Step-by-step explanation:

26/100 x 4.5

=1.17

(the discount price)

ii)

$4.5-$1.17

=$3.33

5 0

3 years ago

Read 2 more answers