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Verdich [7]
4 years ago
12

13/25 divided by 13/35

Mathematics
1 answer:
DiKsa [7]4 years ago
3 0

Answer:

\frac{7}{5}

Step-by-step explanation:

To divide fractions, flip, then multiply:

\frac{13}{25}  \div  \frac{13}{35}  =  \frac{13}{25}  \times \frac{35}{13}

To multiply fractions, divide the product of the numerators by the product of the denominators.

\frac{13}{25} \times  \frac{35}{13}  =  \frac{13 \times 35}{25 \times 13}  =  \frac{455}{325}  =  \frac{7}{5}

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The answer to your question is No

Step-by-step explanation:

To answer this question use the Vertical line test. This test states that if we draw a vertical line and intercepts the function only in one point it is a function.

It the vertical line intersects the graph in two or more points, it is not a function.

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Step-by-step explanation:

I just did a quiz with this question and got it right.

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4 years ago
Which of the following is true regarding the angle shown?
Marat540 [252]

Answer:

C

Step-by-step explanation:

We know that A is not true because the angle is formed by rays, not line segments. We know this because the ends of the lines are arrowheads, which indicates that they are rays. You might be saying that A is true because BA and BC form the angle but actually, even if you removed points A and C, you would still have the angle. B is not correct because according to the diagram, the vertex is point B, not point A. The vertex of an angle is the common endpoint of the two rays that form the angle, therefore it would be point B. D is not correct because angles can have multiple names, the letters do not have to be in alphabetical order. Therefore, C is correct.

5 0
3 years ago
Find all relative extrema of the function. Use the Second Derivative Test where applicable. (If an answer does not exist, enter
Lapatulllka [165]

Answer:

Relative minimum: \left(-\frac{5}{2}, -\frac{33}{4}\right), Relative maximum: DNE

Step-by-step explanation:

First, we obtain the First and Second Derivatives of the polynomic function:

First Derivative

f'(x) = 2\cdot x + 5 (1)

Second Derivative

f''(x) = 2 (2)

Now, we proceed with the First Derivative Test on (1):

2\cdot x + 5 = 0

x = -\frac{5}{2}

The critical point is -\frac{5}{2}.

As the second derivative is a constant function, we know that critical point leads to a minimum by Second Derivative Test, since f\left(-\frac{5}{2}\right) > 0.

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f\left(-\frac{5}{2} \right) = -\frac{33}{4}

There are relative maxima.

6 0
3 years ago
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