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2 years ago

Can you please help me please I need help with this promblem

Mathematics

1 answer:

insens350 [35]2 years ago

8 0

Car B travels 10 more miles per hour than Car A, The line B intersects at 50, While line A intersects at 40 when they hit 1 hour! I Like this Math! Bring me back to this!!!

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3/5 with an exponent of 4

nordsb [41]

Step-by-step explanation:

$\text{If is}\ \dfrac{3}{5}^4,\ \text{then}\ \dfrac{3}{5}^4=\dfrac{3\cdot3\cdot3\cdot3}{5}=\dfrac{81}{5}=16\dfrac{1}{5}\\\\\text{If is}\ \left(\dfrac{3}{5}\right)^4=\left(\dfrac{3}{5}\right)\left(\dfrac{3}{5}\right)\left(\dfrac{3}{5}\right)\left(\dfrac{3}{5}\right)=\dfrac{3\cdot3\cdot3\cdot3}{5\cdot5\cdot5\cdot5}=\dfrac{81}{625}$

7 0

3 years ago

At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel

Alexeev081 [22]

Use Law of Cooling:
$T(t) = (T_0 - T_A)e^{-kt} +T_A$
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
$42 = (80-20)e^{-3k} +20 \\ \\ e^{-3k} = \frac{22}{60}=\frac{11}{30} \\ \\ k = -\frac{1}{3} ln (\frac{11}{30})$

Substituting k back into cooling equation gives:
$T(t) = 60(\frac{11}{30})^{t/3} + 20$

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
$T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\ \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80$
Solve for x:
$x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80$
Sub back into original cooling equation, x = T(t)
$-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20$
Solve for t:
$60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3} = 60 \\ \\ (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\ \\ \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\ \\ \\ t = 4.959$

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM

8 0

3 years ago

Jeremy has 50 ft of fencing to enclose a rectangular-pen for his dog. He plans on using a side of his house for one side of the

kotykmax [81]

The dimensions such as length is 18 and width is 16 or length is 32 and width is 9 will produce an enclosed area of 288ft.

<h3>What is the dimension of length and width?</h3>

We know that,

P = l + 2w

50 = l + 2w

50 - 2w = l

A = l x w

288 = (50 - 2w)w

288 = 50w -2w²

2w² - 50w + 288

By dividing throughout by 2

w² - 25w + 144

This is in quadratic equation form.

By applying quadratic formula,

w = (-(-25) ± $\sqrt{(-25)^{2}-4*1*144$ ) / 2

w = (-(-25) ± 7)/2

w1 = (-(-25) + 7)/2

w1 = (25 + 7) / 2

w1 = 16

w2 = (-(-25) - 7)/2

w2 = (25 - 7) / 2

w2 = 9

By substituting the values in eq1,

A = l x w

if w = 16

288 = l x 16

l = 288/16

l = 18

if w = 9

288 = l x 9

l = 288/9

l = 32

The dimensions will be length is 18 and width is 16 or length is 32 and width is 9.

To learn more about area of rectangle refer to :

brainly.com/question/25292087

#SPJ1

3 0

10 months ago

HELPPPP MEEE PLSSSSSS

wel

Answer: width is 7

Step-by-step explanation:

Formula V=L×W×H

.. 7812= 36×W×31

7812=1116W

Divide by 1116

7=W

5 0

3 years ago

Pliz answer this question

irina1246 [14]

Cent is by the hundred so one percent is 1/100 so .6%=6/1000
which can be reduced to 3/500

7 0

3 years ago