Write an equation and solve the following problems:?

Question

The length of a rectangle if fixed at 8 feet. For what widths will the area be less than 65.8 ft^2.

Juanita scored 75%, 82% and 78% and her first three tests. What possible scores can she get on the fourth test to have at least an 80% average?

Answer 1

8*x<65.8
x<8.225 width has to be 8.225 feet or less

(75+82+78+x)/4>80
(75+82+78+x)>320
x>320-75-82-78
x>85 she has to score at least 85%

Answer 2

for the rectangle question... remember that the definition of area is a = l * w,

the question states the length is fixed at 8 feet, and that the area has to be less than 65.8 ft^2..

rewrite the equation ...

65.8 = 8*w

now, remember that the equation asks for area 'less than 65.8' feet... so the equation becomes an inequality

65.8 > 8w

65.8/ 8 > w

8.225 > w

so, for all widths less than 8.225 ft will give you a rectangle less than 65.8 ft^2


in the second question, you're looking to find the average of 4 test scores that will give you an overall 80%...

so the equation looks like (75 + 82 + 78 + x) / 4 = 80

simplified... (235 + x) /4 = 80
235 + x = 320 (multiplied both sides by 4)
x = 320 - 235
x = 85

Juanita will need an 85% to get an overall of 80%


Good luck, hope this helps...

Answer 3

For the rectangle draw a figure and label the length L as 8 feet. The short sides as W. The area formula is L * W = A

The condition is set that the area A be less than 65.8 ft^2
so set up that condition as A < 65.8

do you see a connection between these formulas ?

try to combine them with the inequality
so you will get 8 * W < 65.8

from here you are on your own to solve it (and of course check your answer)


Problem 2 If Juanita gets an average of 80 then she must score at least 80*4 points right? or score a total of 320?

so far she has scored how many points? how many more does she need to get to 320 ?

set this up as an equation : 75+82+78 + X > 320
now solve and check it.

Good luck !!

Answer 4

is this exact homework?!

(A)

length = 8 (feet)

if w is the width (the unknown), then the area will be A = 8w

it must be less than 65.8; or, 65.8 > 8w

dividing both sides by 8, we get w < 8.225

so width must be less than 8.23 feet

===============

(B)


let x be the score in the fourth test....(the unknown),

then the average of four tests = (75+82+78+x) / 4
= (235+x)/4

this must be at least 80; so

80 < (235+x) / 4 or 320 < 235+x or 320-235 < x or 85 < x

that is, x must be 85% or more to get average of 80%


=== do more exercises =====

Answer 5

1.) 8L<65.8 L < 8.225

2.) 75+82+78+x / 4 >=80
235 +x >= 80*4
x>=320-235
X>=85

Answer 6

1.let the width be x and so area=8x
the equation8x<65.8
x<65.8/8=>x<8.225
2.let the marks in the 4th test be x%
the equation 75+82+78+x=80*4
x=320-235=85%

Answer 7

1) All widths less than 65.8/8

2) All scores equal to at least 85% = 4*.8 - .75-.82-.78

Answer 8

Area of rect = l * b
l is fixed at 8 and Area < 65.8^2
therefore
rt 65.8^2 < rt8 * rt (b)
65.8 < 2rt2 * rt(b)

80% of 400 (4 tests of 100 each) = 320
Combined numbers of 3 tests = 78+82+78 = 235
320-235=85 therefore she has to get 85% or more to get combined avg of 80%

Answer 9

let w be the width of the rectangle so the equation:
8w = 65.8
65.8/8 = 8.225
so the width has to be less than 8.225 ft^2

let T be the score of the fourth test
(.75+.82+.78+T)/4=.8
(2.35+T)/4=.8
(2.35+T)=3.2

T must be 85% or more

Answer 10

L = 8

Area = 8ft*Wft <65.8 ft^2
W < (65.8/8 )ft

(.75 + .82 + .78 + x )/4 >= 0.80

.75 + .82 + .78 + x >= 3.20
100>= x>= 3.20 - .75 - .82 -.78