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Black_prince [1.1K]
4 years ago
10

**URGENT**

Mathematics
2 answers:
Nostrana [21]4 years ago
7 0
So I set up the equation up as X= computers, x+x+12+2(x+12)=220. 
then you would find the value of x which should be 46. 
so the 1st week he sold 46 computers, for the second week he sold 58 computers, then 116 the third week.

You can plug them into the x spots, for example, the second week he sold 12 more than he did the last week so it would be x+12= number sold so 58.

To check if you got it correct, plug it back into the equation to see if it equals 220 which it does. 
Mama L [17]4 years ago
4 0
1st week they sold 98 computers
2nd week they sold 110 computers
3rd week they sold 220 computers

there you go :)
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A line passes through the point (1,4) and is perpendicular to the line with equations y=-1/5x+2. what's the equation of the line
jekas [21]

To find the slope of the perpendicular line, you can take the negative reciprocal of the slope of the line it is perpendicular to.

Taking the negative reciprocal of -1/5 gives 5.

Now we have y=5x+b, where b is the y-intercept. Since we know that the perpendicular line passes through the point (1,4), we can substitute those values into the equation we have to find b.

y=5x+b

4=5(1)+b

4=5+b

b=-1

Therefore, the equation of the perpendicular line is y=5x-1.

4 0
3 years ago
If allie has five melons and jess take two away how many melons does josh have?​
sashaice [31]
Josh had zero melons because he never got any .
7 0
3 years ago
What is (-3, 4) translated by (x,y) → (x – 6, y – 5).
AlekseyPX

Answer:

(-9,-1)

Step-by-step explanation:

You plug in the x and y values and solve the algebraic expression.

8 0
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salma has been given a list of 5 bands and asked to place a vote. her vote must have the names of her favorite and second favori
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7 0
3 years ago
Initially, there are 40 grams of A and 50 grams of B, and for each gram of B, 2 grams of A is used. It is observed that 15 grams
hram777 [196]

Answer:

X(16)=25.71grams

Step-by-step explanation:

let X(t) denote grams of C formed in  t mins.

For X grams of C we have:

\frac{2}{3}Xg of A and \frac{1}{3}Xg of B

Amounts of A,B remaining at any given time is expressed as:

40-\frac{2}{3}Xg of A and  50-\frac{1}{3}Xg  of B

Rate at which C is formed satisfies:

\frac{dX}{dt} \infty(40-\frac{2}{3}X)(50-\frac{1}{3}X)->\frac{dX}{dt}=k(90-X)\\\therefore \frac{dX}{(90-X)^2}=kdt->\int{\frac{dX}{(90-X)^2}} \, =\int {k} \, dt  \\\therefore \frac{1}{90-X}=kt+c->90-X=\frac{1}{kt+c}\\\\X(t)=90-\frac{1}{kt+c}

Apply the initial condition,X(0)=0 ,to the expression above

0=90-\frac{1}{c} \ \ ->c=\frac{1}{90}\\\therefore\\X(t)=90-\frac{1}{kt+\frac{1}{90}} \ \ ->X(t)=90-\frac{90}{90kt+c}

Now at X(8)=15:

15=90-\frac{90}{90\times 8k+1}  \ ->75=\frac{90}{720k+1}\\k=0.0002778

Substitute  in X(t) to get

X(t)=90-\frac{90}{0.0002778t\times 90+1}\\X(t)=90-\frac{90}{0.25t+1}\\But \ t=16\\\therefore X(t)=90-\frac{90}{0.025\times16+1}\\X(t)=25.71

5 0
3 years ago
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