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AveGali [126]
3 years ago
12

Based on this scatterplot, if Carley runs in a 4-kilometer race and she maintains the same pace, how many minutes will it take h

er to run 4 kilometers?
Mathematics
2 answers:
ololo11 [35]3 years ago
8 0

The question is incomplete, the complete question is;

Based on the graph, if Carley runs in a 4-  kilometer race and she maintains the  same pace, how many minutes will it take  her to run 4 kilometers?

A 50 minutes

B 45 minutes

C 35 minutes

D 20 minutes

Answer:

C 35 minutes

Step-by-step explanation:

If we look at the scatter plot closely, we can trace the distance of 4 Km on the vertical axis down to its matching point on the horizontal axis.

If this is done correctly, you will discover that the point on the time axis corresponding to the 4 Km distance on the vertical (distance) axis is 35 minutes (the point lies mid-way between 30 minutes and 40 minutes points on the horizontal axis).

Therefore, it will take Carey 35 minutes to run 4 Kilometres according to the scatter plot.

Anarel [89]3 years ago
4 0

Answer:35 mins

Step-by-step explanation:

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Free_Kalibri [48]

Answer:

65°

Step-by-step explanation:

To obtain Angle A, we use the cosine rule ;

Cos A = (b² + c² - a²) / 2bc

Cos A = (12² + 14² - 14²) / 2(12)(14)

Cos A = 144 / 336

A = Cos^-1(144/336)

A = 64.62°

A = 65°

4 0
2 years ago
Lalasa and yasmin are designing a triangular banner to hang in the school gymnasiums. They first draw the design on paper. The t
jasenka [17]
<span>The triangle on paper is 5 inches wide by 7 inches high.</span><span>
</span><span>
If 1 inch on paper is equal to 1.5 feet on paper, then you need to multiply all dimensions by 18 because 1.5 feet is equal to 18 inches and you are getting 18 inches on the real banner for every inch on the paper banner. </span><span>

5 * 18 = 90 inches from the base of the rear triangle.</span><span>
7 * 18 = 125 inches for the height of the rear triangle. </span><span>
</span><span>The area of the paper triangle will be 5 * 7 / 2 = 35/2 square inches.</span><span>
The area of the rear triangle will be 90 * 125 / 2 = 5625 square inches. </span><span>

regarding feet: 

The base of the real banner will be 125/12 = 10.41666666....... feet.</span><span>
the height of the real banner will be 90 / 12 = 7.5 feet.</span><span>
the area of the real banner will be 5625 / 12^1 = 5625 / 144 = 39.0625 square feet.</span><span>
</span>

3 0
2 years ago
Test the series for convergence or divergence (using ratio test)​
Triss [41]

Answer:

    \lim_{n \to \infty} U_n =0

Given series is convergence by using Leibnitz's rule

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given series is an alternating series

∑(-1)^{n} \frac{n^{2} }{n^{3}+3 }

Let   U_{n} = (-1)^{n} \frac{n^{2} }{n^{3}+3 }

By using Leibnitz's rule

   U_{n} - U_{n-1} = \frac{n^{2} }{n^{3} +3} - \frac{(n-1)^{2} }{(n-1)^{3}+3 }

 U_{n} - U_{n-1} = \frac{n^{2}(n-1)^{3} +3)-(n-1)^{2} (n^{3} +3) }{(n^{3} +3)(n-1)^{3} +3)}

Uₙ-Uₙ₋₁ <0

<u><em>Step(ii):-</em></u>

    \lim_{n \to \infty} U_n =  \lim_{n \to \infty}\frac{n^{2} }{n^{3}+3 }

                       =  \lim_{n \to \infty}\frac{n^{2} }{n^{3}(1+\frac{3}{n^{3} } ) }

                    = =  \lim_{n \to \infty}\frac{1 }{n(1+\frac{3}{n^{3} } ) }

                       =\frac{1}{infinite }

                     =0

    \lim_{n \to \infty} U_n =0

∴ Given series is converges

                       

                     

 

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the first term is -2

\bf n^{th}\textit{ term of a geometric sequence}\\\\&#10;a_n=a_1\cdot r^{n-1}\qquad &#10;\begin{cases}&#10;n=n^{th}\ term\\&#10;a_1=\textit{first term's value}\\&#10;r=\textit{common ratio}\\&#10;----------\\&#10;a_1=-2\\&#10;r=-4&#10;\end{cases}\implies a_n=-2(-4)^{n-1}
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3 years ago
I need the answer just put the work or answer
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Answer:

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Step-by-step explanation:


4 0
2 years ago
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