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3 years ago

14

The length of a rectangle is three times its width. The perimeter of the rectangle is 100 inches. What are the dimensions of the

rectangle?

2 answers:

ser-zykov [4K]3 years ago

7 0

Answer:

We have l = 2w and 2l + 2w = 100 inches;

Then 4w + 2w = 100 inches;

6w = 100 inches;

w = 16.6;

l = 3 x 16.6;

l = 49.8 inches;

Step-by-step explanation:

m_a_m_a [10]3 years ago

3 0

Answer:

<h2>The dimensions are 12.5 inches width and 37.5 inches length.</h2>

Step-by-step explanation:

Givens:

Length is three times the width, this can be represented as: $l=3w$
The perimeter is 100 inches. The perimeter is expressed as: $P=2(l+w)$

So, to find the dimensions of the rectangle, that is, the length and width, we have to replace the first equation into the second one:

$100=2(3w+w)\\\frac{100}{2}=4w\\w=\frac{100}{8}= 12.5 \ in$

The length can be found using the first expression:

$l=3(12.5)=37.5$

Therefore, the dimensions are 12.5 inches width and 37.5 inches length.

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Your school is planning a field trip to the zoo. There are two different bus companies that the school can use.

8090 [49]

Answer:

4 students can go to the first one then 2 students can go the second bus

Step-by-step explanation:

$39×4= 156

$79×2=158

5 0

4 years ago

Solve for b<br><br> B=??????

kozerog [31]

Answer:

$\frac{b-7}{b+1}=\frac{b-10}{b+4}$

$\left(b-7\right)\left(b+4\right)=\left(b+1\right)\left(b-10\right)$

$\left(b-7\right)\left(b+4\right)=b^2-3b-28$

$\left(b+1\right)\left(b-10\right)=b^2-9b-10$

$b^2-3b-28=b^2-9b-10$

$b^2-3b-28+28=b^2-9b-10+28$

$b^2-3b=b^2-9b+18$

$b^2-3b-\left(b^2-9b\right)=b^2-9b+18-\left(b^2-9b\right)$

$6b=18$

$\frac{6b}{6}=\frac{18}{6}$

$b=3$

4 0

2 years ago

Find the solution of the given initial value problem. ty' + 2y = sin t, y π 2 = 9, t > 0 y(t) =

Helen [10]

For the ODE

$ty'+2y=\sin t$

multiply both sides by <em>t</em> so that the left side can be condensed into the derivative of a product:

$t^2y'+2ty=t\sin t$

$\implies(t^2y)'=t\sin t$

Integrate both sides with respect to <em>t</em> :

$t^2y=\displaystyle\int t\sin t\,\mathrm dt=\sin t-t\cos t+C$

Divide both sides by $t^2$ to solve for <em>y</em> :

$y(t)=\dfrac{\sin t}{t^2}-\dfrac{\cos t}t+\dfrac C{t^2}$

Now use the initial condition to solve for <em>C</em> :

$y\left(\dfrac\pi2\right)=9\implies9=\dfrac{\sin\frac\pi2}{\frac{\pi^2}4}-\dfrac{\cos\frac\pi2}{\frac\pi2}+\dfrac C{\frac{\pi^2}4}$

$\implies9=\dfrac4{\pi^2}(1+C)$

$\implies C=\dfrac{9\pi^2}4-1$

So the particular solution to the IVP is

$y(t)=\dfrac{\sin t}{t^2}-\dfrac{\cos t}t+\dfrac{\frac{9\pi^2}4-1}{t^2}$

or

$y(t)=\dfrac{4\sin t-4t\cos t+9\pi^2-4}{4t^2}$

6 0

3 years ago

A runner ran 2/3 of a 5 kilometer race in 21 minutes. They ran the entire race at a constant speed.

alekssr [168]

A)
distance covered
5*2/3 = 3.333
time taken = 21min
this means it takes
5*<span>6.306 = 31.53 to covers 5 km
b) </span><span>21min/3.33 = 6.306 time taken to ran 1 km</span>

7 0

3 years ago

Read 2 more answers

math help asap a train travels 100 km in the same the plane covers 350km. If the speed of the plane is 20 km per less than 4 tim

Svet_ta [14]

R=train speed
p=plaane speed
t=time
d=st

100=tr
350=pt
p=4r-20

ok so 100 times 3.5=350
times 1st equaton by 3.5
350=3.5tr
350=pt
sub
3.5tr=pt
divide both sides by t
3.5r=p
sub p=4r-20
3.5r=4r-20
minus 3.4 from both sides
0=0.5r-20
add 20
20=0.5r
times 2
40=r
sub
p=4r-20
p=4(40)-20
p=160-20
p=140

plane=140kmph
train=40kmph

4 0

4 years ago