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alukav5142 [94]
3 years ago
5

Million dollar bussyy jk help

Mathematics
1 answer:
nata0808 [166]3 years ago
3 0

Answer:

2. 2

3. 4

4. 3/4

5. 5/12

6. 6

7. 1/2

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A rectangular park is 40 m long and 24 m wide. A path 3m wide is constructed outside the park. Find the area of the path?
satela [25.4K]

Answer:

area of path is  area of park with path - area of rectangular park                      

area of park =l x b= 45 x 30=1350 cm2

area of park with path is =45+2.5+2.5 x 30+2.5+2.5 = 1750cm2

area of path = 1750-1350=400 cm2

Step-by-step explanation:

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4 0
3 years ago
I need help can you help me please
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Just multiple them together and u get 693
6 0
3 years ago
The cafeteria workers are preparing snack bags of strawberries for the students. They have a package of 82 strawberries and a pa
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Step-by-step explanation:

8 0
3 years ago
Is sin theta=5/6, what are the values of cos theta and tan theta?
mina [271]

let's bear in mind that sin(θ) in this case is positive, that happens only in the I and II Quadrants, where the cosine/adjacent are positive and negative respectively.

\bf sin(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{hypotenuse}{6}}\qquad \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{6^2-5^2}=a\implies \pm\sqrt{36-25}\implies \pm \sqrt{11}=a \\\\[-0.35em] ~\dotfill

\bf cos(\theta )=\cfrac{\stackrel{adjacent}{\pm\sqrt{11}}}{\stackrel{hypotenuse}{6}} \\\\\\ tan(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{adjacent}{\pm\sqrt{11}}}\implies \stackrel{\textit{and rationalizing the denominator}~\hfill }{tan(\theta )=\pm\cfrac{5}{\sqrt{11}}\cdot \cfrac{\sqrt{11}}{\sqrt{11}}\implies tan(\theta )=\pm\cfrac{5\sqrt{11}}{11}}

5 0
3 years ago
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