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2 years ago

I need help to find the Area

Mathematics

2 answers:

Alenkinab [10]2 years ago

8 0

The answer is 58.4 or at least in the 50s

hram777 [196]2 years ago

6 0

To find the area you have to (13.2x18)+(4x4)

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Ice cream consumption was measured over 30 four-week periods from March 18, 1951 to July 11, 1953. The purpose of the study was

natka813 [3]

Answer:Temperature

Step-by-step explanation:

7 0

2 years ago

At the beginning of a walk, roberto and juanita are 7.77.7 miles apart. if they leave at the same time and walk in the same di

dem82 [27]

Roberto overtakes Juanita at the rate of (7.7 mi)/(11 h) = 0.7 mi/h. This is the difference in their speeds. The sum of their speeds is (7.7 mi)/1 h) = 7.7 mi/h.

Roberto walks at the rate (7.7 + 0.7)/2 = 4.2 mi/h.
Juanita walks at the rate 4.2 - 0.7 = 3.5 mi/h.

_____
In a "sum and diference" problem, one solution is half the total of the sum and difference. If we let R and J be the respective speeds of Roberto and Juanita, we have
R + J = total speed
R - J = difference speed
Adding these two equations, we have
2R = (total speed + difference speed)
R = (total speed + difference speed)2 . . . . . as computed above

4 0

3 years ago

Solve triangle ABC

Goshia [24]

That figure obviously doesn't go with this problem. It doesn't matter; this is triangle ABC labeled the usual way.

c = 10, B = 35°, C = 65°

We have two angles and a side. The third angle is obviously

A = 180° - 35°- 65° = 80°

The remaining sides are given by the Law of Sines,

$\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$

$a = \dfrac{c \sin A}{\sin C} = \dfrac{10 \sin 80}{\sin 65} = 10.866$

$b = \dfrac{c \sin B}{\sin C} = \dfrac{10 \sin 35}{\sin 65} = 6.328$

Answer: A=80°, a=10.9, b=6.3, third choice

4 0

2 years ago

It takes Wayne 1/10 hours to walk 2/6 mile park loop. What is Wayne's unit rate, in miles per hour?

-BARSIC- [3]

If it takes .1 hours to walk 2/6 miles, then it takes one hour to walk 20/6 miles.
20/6 miles = 3.33 miles
So that is 3.33 miles per hour.

5 0

2 years ago

Intergal of 8sin^4(x)dx

Romashka-Z-Leto [24]

$\int\limits8sin^4(x)dx =$

Take the constant out $\int\limits a*f(x)dx=a* \int\limitsf(x)dx$

$= 8 \int\limits sin^4(x)dx$

$\int\limits sin^4(x)dx = - \frac{cos(x)sin^3(x)}{4} = \frac{3}{4} \int\limits sin^2(x)dx$

$= 8( -\frac{cos(x)sin^3(x)}{4}+ \frac{3}{4} \int\limits sin^2(x)dx)
$

Use the following identify : $sin^2(x) = \frac{1-cos(2x)}{2}$

$= 8(- \frac{cos(x)sin^3(x)}{4} + \frac{3}{4} \frac{1}{2}( \int\limits 1 - cos(2x)dx)$

$\int\limits 1dx = x$

$\int\limits cos(2x)dx = \frac{1}{2} sin(2x)$

Apply the sum rule :

$=8(- \frac{cos(x)sin^3(x)}{4} + \frac{3}{4} \frac{1}{2} (x- \frac{1}{2} sin(2x))$

Simplify :

$8( \frac{3}{8} (x- \frac{1}{2} sin(2x)) - \frac{1}{4} sin^3(x)cos(x))$

Therefore add a constant to the solution:

$= 8 ( \frac{3}{8} (x- \frac{1}{2} sin(2x)) - \frac{1}{4} sin^3(x)cos(x))+C$

hope this helps!

3 0

2 years ago