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DIA [1.3K]
3 years ago
6

If Ann travels 50 meters North, 200 meters East and 150 meters South, what is her total distance travelled ? A. 150 meters B.400

meters C. 250 meters D. 0 meter
Mathematics
2 answers:
Sophie [7]3 years ago
4 0
400 total distance by adding.
olga2289 [7]3 years ago
3 0

Answer:

400 meters

Step-by-step explanation:

50 + 200 + 150 = 400

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Answer:

46 \times 2 = 92 \: points \\ 100 - 92 = 8 \: points \\ 4 \times 2 = 8 \\ two \: of \: the \: 4 \: points \: questions \:  \\ were \: answered \: correctly

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2 years ago
Which function is an odd function?
hjlf
A function is odd if:
f ( - x ) = - f ( x )
A ) f ( - x )= 0.8 ( - x ) ^3 = - 0.8 x^3 = - f ( x )
Answer: A ) f ( x ) = 0.8 x^3 is an odd function. 
Functions B ) and C ) are even and D) is neither even nor odd.
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3 years ago
Which to temperatures have a sum of 0°C?
Murljashka [212]

Answer:

2°C and -2°C

Step-by-step explanation:

A number and its opposite sum to zero. (The "opposite" is the number with a minus sign instead of a plus sign, or vice versa.)

4 0
3 years ago
The flagpole casts a 8 foot shadow as shown in the table.At the same time, the oak tree casts a 12 foot shadow.How tall is the o
just olya [345]
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8 0
3 years ago
Evaluate the integral following ​
alina1380 [7]

Answer:

\displaystyle{4\tan x + \sin 2x - 6x + C}

Step-by-step explanation:

We are given the integral of:

\displaystyle{\int 4(\sec x - \cos x)^2 \, dx}

First, we can use a property to separate a constant out of integrand:

\displaystyle{4 \int (\sec x - \cos x)^2 \, dx}

Next, expand the expression (integrand):

\displaystyle{4 \int \sec^2 x - 2\sec x \cos x + \cos^2 x \, dx}

Since \displaystyle{\sec x = \dfrac{1}{\cos x}} then it can be simplified to:

\displaystyle{4 \int \dfrac{1}{\cos^2 x} - 2\dfrac{1}{\cos x} \cos x + \cos^2 x \, dx}\\\\\displaystyle{4 \int \dfrac{1}{\cos^2 x} - 2 + \cos^2 x \, dx}

Recall the formula:

\displaystyle{\int \dfrac{1}{\cos ^2 x} \, dx = \int \sec ^2 x \, dx = \tan x + C}\\\\\displaystyle{\int A \, dx = Ax + C \ \ \tt{(A \ and \ C \ are \ constant.)}

For \displaystyle{\cos ^2 x}, we need to convert to another identity since the integrand does not have a default or specific integration formula. We know that:

\displaystyle{2\cos^2 x -1 = \cos2x}

We can solve for \displaystyle{\cos ^2x} which is:

\displaystyle{2\cos^2 x = \cos2x+1}\\\\\displaystyle{\cos^2x = \dfrac{\cos 2x +1}{2}}

Therefore, we can write new integral as:

\displaystyle{4 \int \dfrac{1}{\cos^2 x} - 2 + \dfrac{\cos2x +1}{2} \, dx}

Evaluate each integral, applying the integration formula:

\displaystyle{\int \dfrac{1}{\cos^2x} \, dx = \boxed{\tan x + C}}\\\\\displaystyle{\int -2 \, dx = \boxed{-2x + C}}\\\\\displaystyle{\int \dfrac{\cos 2x +1}{2} \, dx = \dfrac{1}{2}\int \cos 2x +1 \, dx}\\\\\displaystyle{= \dfrac{1}{2}\left(\dfrac{1}{2}\sin 2x + x\right) + C}\\\\\displaystyle{= \boxed{\dfrac{1}{4}\sin 2x + \dfrac{1}{2}x + C}}

Then add all these boxed integrated together then we'll get:

\displaystyle{4\left(\tan x - 2x + \dfrac{1}{4}\sin 2x + \dfrac{1}{2} x\right) + C}

Expand 4 in the expression:

\displaystyle{4\tan x - 8x +\sin 2x + 2 x + C}\\\\\displaystyle{4\tan x + \sin 2x - 6x + C}

Therefore, the answer is:

\displaystyle{4\tan x + \sin 2x - 6x + C}

4 0
11 months ago
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