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3 years ago

Find the sum. Include all calculations in your answer.

Mathematics

1 answer:

Andreas93 [3]3 years ago

6 0

Answer:

- 2 2/7

Step-by-step explanation:

First, do the conversion:

[-(6*7)+3]/7 = -39/7

[(3*7)+2]/7 = 23/7

Then calculate the result:

23/7 - 39/7 = -16/7 or - 2 2/7

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Problem Page

Orlov [11]

If order matters, then there are 12 ways to do this

If order does not matter, then there are 6 ways to do this

===========================================

We have 4 choices for the first slot and 3 choices for the next (we can't reuse a letter) so that's where 4*3 = 12 comes from

If order doesn't matter, then something like AB is the same as BA. So we are doubly counting each possible combo. To fix this, we divide by 2: 12/2 = 6

To be more formal, you can use nPr and nCr to get 12 and 6 respectively (use n = 4 and r = 2)

7 0

3 years ago

. The diameter of Bubba's bubble is

Oksana_A [137]

Answer:

10.22

Step-by-step explanation:

2.45 * 10 =20.45 - 3.67 * 10 =30.67

30.67 - 20.45

30 - 20 = 10

.67 - .45 = .22

30.67 - 20.45 = 10.22

8 0

2 years ago

How can you use exponents to write 10,000,000?

aleksklad [387]

Let's break it down into something more simple:

What's 10 times 10? 100

What's 10 times 10 times 10? 1000

Do you see a pattern? (Cool trick - The number of 10s match the number of 0s in the answer)

We have seven 0s in our answer, so let's check:

10 x 10 x 10 x 10 x 10 x 10 x 10 = 10,000,000

An exponent tells us that we're multiplying the big number by itself however many times the little number in the corner tells us.

So if we multiply 10 by itself 7 times (what we did above), then our answer will be 10 to the power of 7, or $10^{7}$ . This is equal to 10,000,000.

6 0

3 years ago

The average density of the material in intergalactic space is approximately 2.5 × 10–27 kg/m3. what is the volume of a lead samp

Lesechka [4]

Answer:

e. $1.8\times 10^{-6}m^3$

Step-by-step explanation:

It is given that,

The density of intergalactic space material is $2.5 \times 10^{-27}$ kg per cubic meter.

And the volume of intergalactic space material is $8.0\times 10^{24} m^3$

So the mass of that much intergalactic space material is,

$m= \rho\times v$ , where 'm' is the mass, and 'v' is the volume.

Putting the values we get,

$m=2.5 \times 10^{-27}\times 8.0 \times 10^{24}$

$m=20\times 10^{(-27+24)}=20\times 10^{-3} kg$

It is also given that the mass of lead is the same as the mass of the intergalactic space material. Therefore, the mass of lead is $20 \times 10^{-3}=2 \times 10^{-2}kg$