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3 years ago

I need the answer fast pls

Mathematics

2 answers:

puteri [66]3 years ago

7 0

Answer:

Step-by-step explanation:

The median in a box plot is the line in the middle of the box

The median is 58

Montano1993 [528]3 years ago

6 0

Answer:

Step-by-step explanation:

The median in a box plot is the line in the middle of the box

The median is 58

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Is (2, 2) a solution of y < 4x - 6?<br> Choose 1 answer<br> yes or no

zalisa [80]

No, (2,2) is not solution of y < 4x - 6

Step-by-step explanation:

To solve above linear problem, we use standard method of Back- Substitution.

First of all, A point (2,2) only becomes solution of equation, if it satisfies equation $y < 4x - 6$ .

Here, the $y < 4x - 6$ seems to be like a linear equation.

In this Equation, using Back- Substitution method,

Point (2,2) : first digit 2 corresponds to x- coordinate and second digit 2 corresponds to y-coordinate.

$y < 4x - 6.....(1) \\\ x= 2 .....(2) \\\ y= 2 .....(3)$

By putting value of x and y in equation (1),

Equation becomes,

$2 < 4 \times 2 - 6$

$2 < 8 - 6$

$2 < 2$ .... not true mathematically.

Finally, the result came is not true. Therefore, (2,2) is not solution for equation

$y < 4x - 6$

4 0

3 years ago

What is the area of the trapezoid below?

TEA [102]

The answer is A. 36 $\sqrt{3} cm^{2}$

3 0

3 years ago

Read 2 more answers

The fraction: 340 | 3/4

Effectus [21]

Answer:

It's A I hope that helps

Step-by-step explanation:

8 0

3 years ago

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babunello [35]

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3 years ago

For what value of constant c is the function k(x) continuous at x = 0 if k =

nlexa [21]

The value of constant c for which the function k(x) is continuous is zero.

<h3>What is the limit of a function?</h3>

The limit of a function at a point k in its field is the value that the function approaches as its parameter approaches k.

To determine the value of constant c for which the function of k(x) is continuous, we take the limit of the parameter as follows:

$\mathbf{ \lim_{x \to 0^-} k(x) = \lim_{x \to 0^+} k(x) = 0 }$

$\mathbf{\implies \lim_{x \to 0 } \ \ \dfrac{sec \ x - 1}{x}= c }$

Provided that:

$\mathbf{\implies \lim_{x \to 0 } \ \ \dfrac{sec \ x - 1}{x}= \dfrac{0}{0} \ (form) }$

Using l'Hospital's rule:

$\mathbf{\implies \lim_{x \to 0} \ \ \dfrac{\dfrac{d}{dx}(sec \ x - 1)}{\dfrac{d}{dx}(x)}= \lim_{x \to 0} sec \ x \ tan \ x = 0}$

Therefore:

$\mathbf{\implies \lim_{x \to 0 } \ \ \dfrac{sec \ x - 1}{x}=0 }$

Hence; c = 0

Learn more about the limit of a function x here:

brainly.com/question/8131777

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2 years ago