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hjlf
4 years ago
15

What number has 9 thousands and 4 less hundreds than thens

Mathematics
1 answer:
Reil [10]4 years ago
6 0
..........9500..........
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3 years ago
BRAINIEST TO CORRECT ANSWER<br> (2x^2 +7x)+ (−x^2+10x+3)
Finger [1]

Answer: x2+17x+3

Step-by-step explanation:

2x2+7x−x2+10x+3

=2x2+7x+−x2+10x+3

Combine Like Terms:

=2x2+7x+−x2+10x+3

=(2x2+−x2)+(7x+10x)+(3)

=x2+17x+3

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3 years ago
Drag each example to the correct mathematical property listed.
Murljashka [212]

Answer is in the photo. I can only upload it to a file hosting service. link below!

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3 years ago
Six checkers are placed in three boxes: two red are in a box labeled "RR"; two black are in a box labelled "BB"; a red and a bla
Nezavi [6.7K]

Answe and Step-by-step explanation:

Looking at the question the way it was asked, it is easy because it said they have been labelled, so you don't have to stress yourself. Although, if what is intended is, the labelling got wrong along the way and how do you identify the correct one? Then this is what to do:

Go to the one labelled RB. Since I'm assuming that the labelling got wrong, if you pick a red, it means what we have should be a RR and if we picked a black, it means what we have is a BB and we can't have a RB because it was labelled wrongly.

Let's assume he saw a red,

We know that the BB box was labelled wrongly, and we already determined that the box with RB is a RR. Therefore, the box BB can never be RR(because we've seen it already) and it's certainly not BB (due to the fact that they were labelled wrongly). So we have only one option left which is it being RB and whatever we have left will be BB.

If we assume what was picked from the wrongly labelled RB is black.

We know that the RR box was labelled wrongly, and we already determined that the box with RB is a BB. Therefore, the box RR can never be BB and it's certainly not RR (due to the fact that they were labelled wrongly). So we have only one option left which is it being RB and whatever we have left will be RR.

3 0
3 years ago
Manav is painting a rectangular room and the dimensions of this room are given by x,y and z feet. Manav takes 8 hours to paint a
zhenek [66]

Answer:

B) 36 cubic feet

Step-by-step explanation:

Dimensions of the room are x, y and z feet. It is given that x = 6 feet.

Manav took 8 hours to paint the wall with dimensions x and z.

This means, in 8 hours Manav painted area = xz = 6z

In 1 hour Manav painted area = \frac{6z}{8}

Manav took 4 hours to paint the wall with dimensions y and z.

This means, in 4 hours Manav painted area = yz

In 1 hour Manav painted area = \frac{yz}{4}

Manav took 12 hours to paint the ceiling with dimensions x and y.

This means, in 12 hours Manav painted area = xy = 6y

In 1 hour Manav painted the area = \frac{6y}{12}=\frac{y}{2}

Since, Manav works at constant rate, this means the worked done in 1 hour in all 3 cases would be the same. So we can write:

\frac{6z}{8}=\frac{yz}{4}=\frac{y}{2}

Using the first two equations, we get:

\frac{6z}{8}=\frac{yz}{4}\\\\ \frac{6}{8}=\frac{y}{4}\\\\ y=\frac{6}{8} \times 4\\\\ y = 3

Using the last two equations, we get:

\frac{yz}{4}=\frac{y}{2}\\\\ \frac{z}{4}=\frac{1}{2}\\\\z=2

So, we have the follwoing dimensions of the room:

x = 6 feet

y = 3 feet

z = 2 feet

So the volume of the room is =  6 x 3 x 2 = 36 cubic feet

7 0
3 years ago
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