Lets write as an equation the info provided in the problem:
original area = 200*300 = 60000 squared meters
Adding x meters to the length is: 300 + x
Adding x meters to the width is: 200 + x
If the area doubles in size we have: 2*60000
Now writing as a single equation all info:
(300 + x)(200 + x) = 120000
We have to make the operations and solve:
60000 + 300x + 200x + x^2 = 120000
x^2 + 500x - 60000 = 0
This is a squared trinomial, to solve it we need two numbers that subtracted give us 500 and multiplied -60000:
(x + 600)(x -100) = 0
So there are two solutions, x = -600 and x = 100, we choose the positive one:
x = 100
thererefore the value of x is 100 meters

4 0

3 years ago

Giselle spent 7/10 of her money on a book and 10/100 of her money on food. What fraction of her money did she spend altogether?

aleksklad [387]

Answer:

80/100 or 80%

Step-by-step explanation:

7/10 = 70/100

70/100 + 10/100 = 80/100

8 0

3 years ago

Read 2 more answers

If f(x)=x^2-11x+18 what are the zeros of the function?

UkoKoshka [18]

X= 9 x=2
-2 × -9=18
-2 + -9= -11

8 0

3 years ago

Evaluate the line integral c y3 ds, c.x = t3, y = t, 0 â‰¤ t â‰¤ 3

Temka [501]

$\displaystyle\int_{\mathcal C}y^3\,\mathrm dS=\int_{t=0}^{t=3}t^3\sqrt{1^2+(3t^2)^2}\,\mathrm dt=\int_0^3t^3\sqrt{1+9t^4}\,\mathrm dt$

Take $u=1+9t^4$ so that $\mathrm du=36t^3\,\mathrm dt$ . Then

$\displaystyle\int_{\mathcal C}y^3\,\mathrm dS=\frac1{36}\int_{u=1}^{u=730}\sqrt u\,\mathrm du=\frac{730^{3/2}-1}{54}$

7 0

3 years ago