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Likurg_2 [28]
3 years ago
10

Sarah lost 12 pounds over the summer by jogging each week. By winter time, she had gained 4 3/8 pounds. Represent this situation

with an expression involving signed numbers. What is the overall change in Sara's weight.
Mathematics
1 answer:
ANTONII [103]3 years ago
7 0
Assume sarah's weight is m.
first, she lost 12 pounds,
this means that her weight became: m - 12 pounds
then, she gained 4 3/8 pounds, therefore, her current weight is: m-12+4 3/8 pounds
the total change in her weight is:
-12+4 3/8 = -7.625 pounds
the negative sign refers to the loss.
this means that the overall change is losing 7.625 pounds
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3x+y=8________(1) and x²+xy=6________(2) solve this​
romanna [79]

Answer:

In first equation if we put the value of x and y (x=2,y=2)than,

3x+y=8

3×x+y=8

3×2+2=8

6+2=8

8=8

in second equation the value of x=2,y=1

x2+xy=6

x2+x×y=6

4+2×1=6

4+2=6

6=6

6 0
3 years ago
Read 2 more answers
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
Pentagon ABCDE ~ Pentagon FGHIJ. If CD=14, DE=19, and IJ=42, what is the length of HI.
xz_007 [3.2K]

It is going to be 41

5 0
3 years ago
Sorry my brain hurts... Most of u love the points anyway
Amanda [17]
I think the answer's C: $7000/$8000
That's a surprisingly easy question for high school
6 0
3 years ago
Read 2 more answers
Identify the augmented matrix for the system of equations.
lord [1]

Answer:

The augmented matrix for the system of equations is \left[\begin{array}{cccc}0&2&-3&1\\7&0&5&8\\4&1&-3&6\end{array}\right].

Step-by-step explanation:

This system consists in three equations with three variables (x, y, z).The augmented matrix of a system of equations is formed by the coefficients and constants of the system of linear equations. In this case, we conclude that the system of equations has the following matrix:

\left[\begin{array}{cccc}0&2&-3&1\\7&0&5&8\\4&1&-3&6\end{array}\right]

The augmented matrix for the system of equations is \left[\begin{array}{cccc}0&2&-3&1\\7&0&5&8\\4&1&-3&6\end{array}\right].

8 0
2 years ago
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