All categories

3 years ago

PLEASE HELP BRAINLIES+20 POINTS ASAP!!!!!!!!!!

Mathematics

2 answers:

Cloud [144]3 years ago

4 0

here is the answer + steps! I hope I helped :)

Oksi-84 [34.3K]3 years ago

3 0

Answer:

Step-by-step explanation:

Length = L.

Width = 2L-3.

A = L * W = 170.

L * (2L-3) = 170.

2L^2 - 3L - 170 = 0,

L = (-B +- Sqrt(B^2-4AC))/2A.

L = (3 +- Sqrt(9 + 1360))/4,

L = (3 +- 37)/4 = 0.75 +- 9.25 = 10, and -8.5. In.

Use positive value: L = 10 In.

You might be interested in

72-12+36 divided by 3+14•8

kupik [55]

Answer:

I got 1.25

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Help please I'm lost

yarga [219]

36 1/2 to 37.

45.5 - 36 1/2 = 9

those would be the ones I'd click

3 0

2 years ago

Work out, giving your answer in its simplest form: 5/ 14 x 7/ 30

Elena L [17]

X in (-oo:+oo)

((5/14)*x^7)/30 = 0

1/84*x^7 = 0 // : 1/84

x^7 = 0

x = 0

x = 0

6 0

2 years ago

Given the function f(x) = 4x + 1, EXPLAIN and show how to find the average rate of change between x = 1 and x = 3.

11Alexandr11 [23.1K]

The average rate of change is calculated as the slope of the line between the two points on the graph. Since at x = 1, y = -3, and at x = 3, y = 1, these are the two points which we calculate the slope from. Using the formula

m = (y2 - y1)/(x2 - x1)

m = (1 - (-3))/(3 - 1)

m = 4/2

m = 2

Therefore the slope, or the average rate of change, between these points is 2.

4 0

3 years ago

Read 2 more answers

A huge ice glacier in the Himalayas initially covered an area of 454545 square kilometers. Because of changing weather patterns,

zloy xaker [14]

Answer:

$t=-20ln\left(\dfrac{1}{3}\right)$

Step-by-step explanation:

The relationship between A, the area of the glacier in square kilometers, and t, the number of years the glacier has been melting, is modeled by the equation.:

$A=45e^{-0.05t}$

We want to determine the value of t for which the area, A(t)=15 square kilometers.

$15=45e^{-0.05t}\\$Divide both sides by 45\\\dfrac{15}{45} =\dfrac{45e^{-0.05t}}{45}\\\dfrac{1}{3}=e^{-0.05t}\\$Take the natural logarithm of both sides\\ln\left(\dfrac{1}{3}\right)=ln\left(e^{-0.05t}\right)\\ln\left(\dfrac{1}{3}\right)=-0.05t\\$Divide both sides by -0.05$\\t=-\dfrac{ln\left(\dfrac{1}{3}\right)}{0.05} \\=-\dfrac{ln\left(\dfrac{1}{3}\right)}{0.05}\\t=-20ln\left(\dfrac{1}{3}\right)$

Therefore, the time for which the area will be 15 sqyare kilometers is:

-20 ln(1/3) years.

7 0

3 years ago