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MArishka [77]
3 years ago
15

Find the product. -mnp(3m - 5n 7p)

Mathematics
2 answers:
evablogger [386]3 years ago
5 0
If you would like to find the product, you can calculate this using the following steps:

-mnp * (3m - 5n7p) = -mnp * 3m + mnp * 5n7p = -3m^2np + 35mn^2p^2

The correct result would be -3m^2np + 35mn^2p^2.
Nataliya [291]3 years ago
3 0
-mnp(3m-5n7p)=\\\\-mnp\times3m+mnp\times5n7p=\\\\-3m^2np+35mn^2p^2\\\\\sf\ ANSWER={\boxed{-3m^2np+35mn^2p^2}
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The price of a video game was reduced from $70.00 to $40.00. By what percentage of the price was was the videogame reduced? Just
olya-2409 [2.1K]

Answer:

42.86%.

Step-by-step explanation:

Reduction in price = $70.00 - $40.00 = $30.00

Reduction in percentage = ($30.00 ÷ $70.00) x 100 = 42.86%

The price of the videogame was reduced by 42.86%.

8 0
3 years ago
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two numbered 1 to 6 dice are thrown together and their scores are added. the probability that the sum will be 4 is?​
natulia [17]

Answer:

Pr = \frac{3}{36}

Step-by-step explanation:

Given

2 number die

Required

P(Sum = 4)

The sample space of 2 die is:

S = \{(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6), (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)

(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6), (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)

(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6), (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

n(S)= 36

The pairs that adds up to 4 are:

Sum(4) = \{(1,3), (2,2),(3,1)\}

n(Sum)  =3

So, the probability is:

Pr = \frac{n(Sum)}{n(S)}

Pr = \frac{3}{36}

5 0
2 years ago
QUESTION IN THE ATTACHMENT
eimsori [14]

Answer:

A. The sum of the first 10th term is 100.

B. The sum of the nth term is n²

Step-by-step explanation:

Data obtained from the question include:

Sum of 20th term (S20) = 400

Sum of 40th term (S40) = 1600

Sum of 10th term (S10) =..?

Sum of nth term (Sn) =..?

Recall:

Sn = n/2[2a + (n – 1)d]

Sn is the sum of the nth term.

n is the number of term.

a is the first term.

d is the common difference

We'll begin by calculating the first term and the common difference. This is illustrated below:

Sn = n/2 [2a + (n – 1)d]

S20 = 20/2 [2a + (20 – 1)d]

S20= 10 [2a + 19d]

S20 = 20a + 190d

But:

S20 = 400

400 = 20a + 190d .......(1)

S40 = 40/2 [2a + (40 – 1)d]

S40 = 20 [2a + 39d]

S40 = 40a + 780d

But

S40 = 1600

1600 = 40a + 780d....... (2)

400 = 20a + 190d .......(1)

1600 = 40a + 780d....... (2)

Solve by elimination method

Multiply equation 1 by 40 and multiply equation 2 by 20 as shown below:

40 x equation 1:

40 x (400 = 20a + 190d)

16000 = 800a + 7600. ........ (3)

20 x equation 2:

20 x (1600 = 40a + 780d)

32000 = 800a + 15600d......... (4)

Subtract equation 3 from equation 4

Equation 4 – Equation 3

32000 = 800a + 15600d

– 16000 = 800a + 7600d

16000 = 8000d

Divide both side by 8000

d = 16000/8000

d = 2

Substituting the value of d into equation 1

400 = 20a + 190d

d = 2

400 = 20a + (190 x 2)

400 = 20a + 380

Collect like terms

400 – 380 = 20a

20 = 20a

Divide both side by 20

a = 20/20

a = 1

Therefore,

First term (a) = 1.

Common difference (d) = 2.

A. Determination of the sum of the 10th term.

First term (a) = 1.

Common difference (d) = 2

Number of term (n) = 10

Sum of 10th term (S10) =..?

Sn = n/2 [2a + (n – 1)d]

S10 = 10/2 [2x1 + (10 – 1)2]

S10 = 5 [2 + 9x2]

S10 = 5 [2 + 18]

S10 = 5 x 20

S10 = 100

Therefore, the sum of the first 10th term is 100.

B. Determination of the sum of the nth term.

First term (a) = 1.

Common difference (d) = 2

Sum of nth term (Sn) =..?

Sn = n/2 [2a + (n – 1)d]

Sn = n/2 [2x1 + (n – 1)2]

Sn = n/2 [2 + 2n – 2]

Sn = n/2 [2 – 2 + 2n ]

Sn = n/2 [ 2n ]

Sn = n²

Therefore, the sum of the nth term is n²

6 0
2 years ago
Tell whether (3,5) is a solution of y = 5x – 10.
Ivanshal [37]

Answer:

yes it is

Step-by-step explanation:

plug x and y in

3 * 5 - 10 = 5 <== And this is true

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