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3 years ago

How many 13s are in 223?

Mathematics

1 answer:

Masja [62]3 years ago

4 0

Answer: There are 17 times 13 in 223. The answer you divide 223 by 13 which would get you 17.153846153846.

Step-by-step explanation:

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For the functions f(x) = x − 3 and g\left(x\right)=\frac{x+7}{2} g ( x ) = x + 7 2 , find [g∘f](x). Group of answer choices \fra

pychu [463]

Answer:

$\textbf{$g(f(x)) = \frac{2x + 1}{2}$ }\\$

Step-by-step explanation:

$$ f(x) = x - 3 $$$$ g(x) = x + \frac{7}{2} $$\textup{To compute g(f(x))}$ \implies g(f(x)) = g(x - 3) $\\\textup{i.e., to substitute $x - 3$ instead of x in g(x)}\\$ \therefore g(x - 3) = x - 3 + \frac{7}{2} $ \\\textup{Simplifying this we get:}\\\textbf{$$g(f(x)) = \frac{2x + 1}{2}$$}$

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Tanner walks 42 kilometers each morning. How many meters does he walk each morning?

hichkok12 [17]

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3 years ago

Find an equation for the perpendicular bisector of the line segment whose endpoints

TEA [102]

Answer:

y= -2x -8

Step-by-step explanation:

I will be writing the equation of the perpendicular bisector in the slope-intercept form which is y=mx +c, where m is the gradient and c is the y-intercept.

A perpendicular bisector is a line that cuts through the other line perpendicularly (at 90°) and into 2 equal parts (and thus passes through the midpoint of the line).

Let's find the gradient of the given line.

$\boxed{gradient = \frac{y1 -y 2}{x1 - x2} }$

Gradient of given line

$= \frac{1 - ( - 5)}{3 - ( - 9)}$

$= \frac{1 + 5}{3 + 9}$

$= \frac{6}{12}$

$= \frac{1}{2}$

The product of the gradients of 2 perpendicular lines is -1.

(½)(gradient of perpendicular bisector)= -1

Gradient of perpendicular bisector

= -1 ÷(½)

= -1(2)

= -2

Substitute m= -2 into the equation:

y= -2x +c

To find the value of c, we need to substitute a pair of coordinates that the line passes through into the equation. Since the perpendicular bisector passes through the midpoint of the given line, let's find the coordinates of the midpoint.

$\boxed{midpoint = ( \frac{x1 + x2}{2} , \frac{y1 + y2}{2}) }$