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Leviafan [203]
3 years ago
15

A woman has two boxes of matches.she uses five matches and has 75 matches left.how many matches were in each box?

Mathematics
1 answer:
Delicious77 [7]3 years ago
8 0

Answer:

40

Step-by-step explanation:

So if you add 5 to 75 you get 80 so that means there were 80 matches all then you divide 80 by 2 and voila 40!

Hoped this helped

P.s. Can I get a happy bday my bday is tmmr

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You dont have to do all but I really just need help but if you do I will give brainiest
sashaice [31]

Answer:

11.V=-2

12.X=6

13.A=9

14.X=-4

15.A=-5

16.P=3

17.K=6

18.M=0

Step-by-step explanation:

Subtract the number on each side to get the variable by itself or add depending on if it has a plus or minus on the side with the variable.

Hope this helps!

Please give me a brainist and add me also make me stars go up. :)

5 0
3 years ago
Read 2 more answers
If p-1 is a factor of p^4+p^2+p-k the value of k is
Ksju [112]
P=1
p^4+p^2+p-k=0

1+1+1-k=0 k=3
7 0
4 years ago
Using compatible number to estimate the quotient 2116÷7
DerKrebs [107]

Answer:

302

Step-by-step explanation:

7 goes into 21 3 times, so the first digit of the quotient is 3.  There's no remainder.  Bringing down the second "1," we find this indivisible by 7, and so bring down the 6.Thus, Remember that 7 goes into 14 2 times.  Thus, a good estimate would be 302.

3 0
3 years ago
Wei is standing in wavy water and notices the depth of
dexar [7]

Answer:

  see the attachment

Step-by-step explanation:

The function can be ...

  D(t) = A +Bcos(C(t-p))

where A is the average depth (55+12)/2 = 33.5 cm,

  B is the peak deviation from average, 55 -33.5 = 21.5 cm,

  C is the horizontal scale factor (2π/T) = (2π/3) for a period (T) of 3 seconds,

and p is the phase offset, given as 1.1 seconds.

The function is ...

  D(t) = 33.5 +21.5cos(2π/3·(t -1.1))

5 0
3 years ago
Read 2 more answers
A polygon is shown: A polygon MNOPQR is shown. The top vertex on the left is labeled M, and rest of the vertices are labeled clo
Mashcka [7]

Answer:

The area of polygon MNOPQR = Area of a rectangle that is 15 square units + Area of a rectangle that is <u>2</u> square units.

Step-by-step explanation:

We are given that a polygon MNOPQR is shown. The top vertex on the left is labeled M, and the rest of the vertices are labeled clockwise starting from the top-left vertex labeled, M. The side MN is parallel to side QR. The side MR is parallel to side PQ.

The side MN is labeled as 5 units. The side QR is labeled as 7 units. The side MR is labeled as 3 units, and the side NO is labeled as 2 units.

Firstly, we will draw a perpendicular line from point O which meets the line RQ at point T. Now, the polygon MNOPQR is divided into two rectangles MNTR and OPQT.

As we know that the area of the rectangle = \text{Length of rectangle} \times \text{Breadth of rectangle}

In the rectangle MNTR, the length (MN) = 5 units and the breadth (MR) = 3 units.

So, the area of the rectangle MNTR = \text{Length (MN)} \times \text{Breadth (MR)}

                                                            = 5 \times 3  = 15 square units

Now, as we know that in rectangle MNTR, the side NT = 3 units and the side NO is labeled as 2 units. This means that the side OT = NT - NO = 3 units - 2 units = 1 unit

Similarly, the side QR is labeled as 7 units and the side RT is labeled as 5 units. This means that the side TQ = QR - RT = 7 - 5 = 2 units

Now, in the rectangle OPQT, the length (TQ) = 2 unit and the breadth (OT) = 2 units.

So, the area of the rectangle OPTQ = \text{Length (TQ)} \times \text{Breadth (OT)}

                                                            = 2 \times 1  = 2 square units

Hence, the area of polygon MNOPQR = Area of a rectangle that is 15 square units + Area of a rectangle that is <u>_2_</u> square units.

3 0
3 years ago
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