For this, we need to find the area of the larger rectangle (The room) and subtract from that the area of the smaller, shaded rectangle (The rug).
A=LW (L=Length, W=Width, A=Area)

The larger rectangle first:
A=(15)(10)
A=150 feet squared

Now the Rug:
A=(12)(7)
A=84 feet squared

Now to find the area that is NOT covered by the rug, we subtract:
150-84
= 66 feet squared will NOT be covered by the rug.

Hope this Helps!
-Sinnamin

4 0

3 years ago

Please help logarithms!

nlexa [21]

Given:

$\log_34\approx 1.262$

$\log_37\approx 1.771$

To find:

The value of $\log_3\left(\dfrac{4}{49}\right)$ .

Solution:

We have,

$\log_34\approx 1.262$

$\log_37\approx 1.771$

Using properties of log, we get

$\log_3\left(\dfrac{4}{49}\right)=\log_34-\log_349$ $\left[\because \log_a\dfrac{m}{n}=\log_am-\log_an\right]$

$\log_3\left(\dfrac{4}{49}\right)=\log_34-\log_37^2$

$\log_3\left(\dfrac{4}{49}\right)=\log_34-2\log_37$ $[\log x^n=n\log x]$

Substitute $\log_34\approx 1.262$ and $\log_37\approx 1.771$ .

$\log_3\left(\dfrac{4}{49}\right)=1.262-2(1.771)$

$\log_3\left(\dfrac{4}{49}\right)=1.262-3.542$

$\log_3\left(\dfrac{4}{49}\right)=-2.28$

Therefore, the value of $\log_3\left(\dfrac{4}{49}\right)$ is $-2.28$ .

5 0

2 years ago

Find the remaining trigonometric ratios of θ if csc(θ) = -6 and cos(θ) is positive

VikaD [51]

Now, the cosecant of θ is -6, or namely -6/1.

however, the cosecant is really the hypotenuse/opposite, but the hypotenuse is never negative, since is just a distance unit from the center of the circle, so in the fraction -6/1, the negative must be the 1, or 6/-1 then.

we know the cosine is positive, and we know the opposite side is -1, or negative, the only happens in the IV quadrant, so θ is in the IV quadrant, now

$\bf csc(\theta)=-6\implies csc(\theta)=\cfrac{\stackrel{hypotenuse}{6}}{\stackrel{opposite}{-1}}\impliedby \textit{let's find the \underline{adjacent side}}
\\\\\\
\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
\pm\sqrt{6^2-(-1)^2}=a\implies \pm\sqrt{35}=a\implies \stackrel{IV~quadrant}{+\sqrt{35}=a}$

recall that

$\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad\qquad 
cos(\theta)=\cfrac{adjacent}{hypotenuse}
\\\\\\
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}
\qquad \qquad 
% cotangent
cot(\theta)=\cfrac{adjacent}{opposite}
\\\\\\
% cosecant
csc(\theta)=\cfrac{hypotenuse}{opposite}
\qquad \qquad 
% secant
sec(\theta)=\cfrac{hypotenuse}{adjacent}$

therefore, let's just plug that on the remaining ones,

$\bf sin(\theta)=\cfrac{-1}{6}
\qquad\qquad 
cos(\theta)=\cfrac{\sqrt{35}}{6}
\\\\\\
% tangent
tan(\theta)=\cfrac{-1}{\sqrt{35}}
\qquad \qquad 
% cotangent
cot(\theta)=\cfrac{\sqrt{35}}{1}
\\\\\\
sec(\theta)=\cfrac{6}{\sqrt{35}}$

now, let's rationalize the denominator on tangent and secant,

$\bf tan(\theta)=\cfrac{-1}{\sqrt{35}}\implies \cfrac{-1}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{-\sqrt{35}}{(\sqrt{35})^2}\implies -\cfrac{\sqrt{35}}{35}
\\\\\\
sec(\theta)=\cfrac{6}{\sqrt{35}}\implies \cfrac{6}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{6\sqrt{35}}{(\sqrt{35})^2}\implies \cfrac{6\sqrt{35}}{35}$

3 0

3 years ago

A small combination lock on a suitcase has 4 wheels, each labeled with the 10 digits 0 to 9. how many 4 digit combinations are

tamaranim1 [39]

This is the easiest way to solve this problem:

Imagine this represents how many combinations you can have for each of the 4 wheels (each blank spot for one wheel): __ __ __ __

For the first situation it says how many combos can we make if no digits are repeated.
We have 10 digits to use for the first wheel so put a 10 in the first slot
10 __ __ __
Since no digit can be repeated we only have 9 options for the second slot
10 9_ __ __
Same for the third slot, so only 8 options
10 9 8 __
4th can't be repeated so only 7 options left
10 9 8 7

Multiply the four numbers together: 10*9*8*7 = 5040 combinations

For the next two do the same process as the one above.

If digits can be repeated? You have ten options for every wheel so it would look like this: 10 10 10 10

10*10*10*10 = 10,000 combinations

If successive digits bust be different?
We have 10 for the first wheel, but second wheel only has 9 options because 2nd number can't be same as first. The third and fourth wheels also has 9 options for the same reason.

10 9 9 9 

10*9*9*9 = 7290 combinations

8 0

2 years ago