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3 years ago

In a particular week, Mary

Mathematics

2 answers:

Finger [1]3 years ago

7 0

Answer:

B. $ 320

Step-by-step explanation:

[8(3) + 4(2)]10

(24 + 8)10

320

Blababa [14]3 years ago

3 0

THE ANSWER WAS IS: B. $320

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Solve the simultaneous equations 5x+2y=25 5x-y=10

Pavlova-9 [17]

5x + 2y = 25;

5x + 2y = 25;5x - y = 10;

5x + 2y = 25;

5x = 10 + y;

(10 + y) + 2y = 25;

5x = 10 + y;

10 + 3y = 25;

5x = 10 + y;

3y = 15;

5x = 10 + y;

y = 5;

5x = 10 + y;

y = 5;

5x = 10 + 5;

y = 5;

5x = 15;

y = 5;

x = 3.

Answer: (3; 5).

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2 years ago

Linear graphing y= -2x

Vlad1618 [11]

Let x = 0. What do you get for y?
This gives you one point.

Then let x = 1. What do you get for y.
This gives you a second point.

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3 years ago

A quantity with an initial value of 6200 decays continuously at a rate of 5.5% per month. What is the value of the quantity afte

ELEN [110]

Answer:

410.32

Step-by-step explanation:

Given that the initial quantity, Q= 6200

Decay rate, r = 5.5% per month

So, the value of quantity after 1 month, $q_1 = Q- r \times Q$

$q_1 = Q(1-r)\cdots(i)$

The value of quantity after 2 months, $q_2 = q_1- r \times q_1$

$q_2 = q_1(1-r)$

From equation (i)

$q_2=Q(1-r)(1-r) \\\\q_2=Q(1-r)^2\cdots(ii)$

The value of quantity after 3 months, $q_3 = q_2- r \times q_2$

$q_3 = q_2(1-r)$

From equation (ii)

$q_3=Q(1-r)^2(1-r)$

$q_3=Q(1-r)^3$

Similarly, the value of quantity after n months,

$q_n= Q(1- r)^n$

As 4 years = 48 months, so puttion n=48 to get the value of quantity after 4 years, we have,

$q_{48}=Q(1-r)^{48}$

Putting Q=6200 and r=5.5%=0.055, we have

$q_{48}=6200(1-0.055)^{48} \\\\q_{48}=410.32$

Hence, the value of quantity after 4 years is 410.32.

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