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3 years ago

A rectangle measures 10ft 10in. tall and 2ft 11 in wide. what is the area of the rectangle?

1 answer:

8 0

Area of a rectangle is just length x height.
It probably looks confusing because of the numbers. So just change them into only inches; 12 inches in a foot.
So the rectangle is 10ft, 10inches tall...or 10*12+10=130inches
The rectangle is 2ft 11inches wide...or 2*12+11=35 inches
So now just multiple 130*35=4550 inches^2

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for v= 4i - 5j, find unit vector u in the direction of v, and write your answer as a linear combination of the standard unit vec

umka21 [38]

Answer:

a. $u=\frac{4\sqrt{41}i }{41}-\frac{5\sqrt{41}j}{41}$

Step-by-step explanation:

The given vector is v= 4i - 5j

The magnitude of this vector is;

$|v|=\sqrt{(-4)^2+(-5)^2}$

$|v|=\sqrt{16+25}$

$|v|=\sqrt{41}$

The unit vector u in the direction of v is;

$u=\frac{v}{|v|}$

$u=\frac{4i - 5j}{\sqrt{41}}$

$u=\frac{4i }{\sqrt{41}}-\frac{5j}{\sqrt{41}}$

We rationalize to get

$u=\frac{4\sqrt{41}i }{41}-\frac{5\sqrt{41}j}{41}$

4 0

3 years ago

Read 2 more answers

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

The trees at a national park have been increasing in numbers. There were 1,000 trees in the first year that the park started tra

ivanzaharov [21]

The sequence forms a Geometric sequence as the rule to obtain the value for the next term is by ratio

Term 1: 1000
Term 2: 200
Term 3: 40

From term 1 to term 2, there's a decrease by $\frac{1}{5}$
From term 2 to term 3, there's a decrease also by $\frac{1}{5}$

The rule to find the $n^{th}$ term in a sequence is
$n^{th}=a r^{n-1}$ , where $a$ is the first term in the sequence and $r$ is the ratio

So, the formula for the sequence in question is
$n_{th}$ term = $1000( \frac{1}{5} ^{n-1} )$

The sequence is a divergent one. We can always find the value of the next term by dividing the previous term by 5 and if we do that, the value of the next term will get closer to 'zero' but never actually equal to zero.

We can find a partial sum of the sequence using the formula
$S_{∞} = \frac{a}{1-r}$ for -1<r<1
Substituting $a=1000$ and $r= \frac{1}{5}$ we have
$S_{∞}$ = $\frac{1000}{1- \frac{1}{5} }$ = 1250

Hence, the correct option is option number 1

6 0

3 years ago

Let make sum friendss..

sukhopar [10]

Answer:

hi!

Step-by-step explanation:

u look beautiful

Thanks for the pts

✌️❤️✌️

3 0

2 years ago

Read 2 more answers

6 yd 4 yd 7 yd 4 yd Perimeter: Area: Help me find the area and the permiter

morpeh [17]

Step-by-step explanation:

Assuming figure to be trapezoid,

Perimeter=6+4+7+4yd

=21 yd

Area of trapezoid=((6+7)/2) x ((4)^2-(0.5)^2)

=(13/2)x(16-0.25)

=6.5 x 15.75

=102.375 Sq yards

8 0

3 years ago