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3 years ago

The sum of three decimal numbers is 6. Exactly one of the numbers is less than 1. What could the numbers be

Mathematics

1 answer:

Temka [501]3 years ago

6 0

Some of the many possible answers can be..
3.3
2.2
0.5

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If 6!+7!+8!=(n)(6!), then the value of n is A) 15 B)16 C)63 D)64 E)15!

gladu [14]

6!+7!+8!=(n)(6!)
calculate the individual values first:
6!=720
7!=5040
8!=40320

plug them into the equation:
720+5040+40320=n720
solve for n
5760+40320=n720
46080=n720
divide both sides by 720 to isolate n
46080/720=n720/720
64=n

6 0

2 years ago

A circle containing the point (6, 1) has its center at (3, 5). Complete the description of the changes in the circumference and

Romashka-Z-Leto [24]

Step-by-step explanation:

eue7e7euydhdhrydyydhdhdhdhhr

8 0

2 years ago

3/6x4/9 plz help meeeeee

Aleonysh [2.5K]

Answer:

12/54

Step-by-step explanation:

4x3=12

6x9=54

3 0

2 years ago

Read 2 more answers

Solve for b. - 32 + 106 = -22

Eddi Din [679]

Answer:

-30+10b=-22

32-22=10b

10b=10

-32+10=-22

hope it helps

6 0

3 years ago

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