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2 years ago

Given A(-2, 5) and B(13, -7), find the midpoint of AB.

Mathematics

1 answer:

Dmitrij [34]2 years ago

4 0

Answer:

The mid-point is (9,-9/2)

Step-by-step explanation:

You would use the mid-point formula for this

$(\frac{x_{2} +x_{1} }{2} , \frac{y_{2}+y_{1} }{2} )$

if you plug that in it is ( $\frac{13-2}{2}, \frac{-7+5}{2}$ )

resulting in (11/2,-2/2) = (11/2,-1)

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stepladder [879]

Answer:

h(x) ≤ 5

Step-by-step explanation:

The function and boundaries given are;

h(x) = x + 2, x < 3

And ;

h(x) = -x + 8, x ≥ 3

Now,in the first condition, the maximum value of x will be 2. This means the maximum value of h(x) there is;

h(2) = 2 + 2 = 4

For the second condition, x ≥ 3. Thus, the maximum value of x is 3.

This means that the maximum value of h(x) here is;

h(3) = -3 + 8 = 5

Foe the 2 conditions combined it is clear that the maximum value of h(x) is 5.

Thus,we can say; h(x) ≤ 5

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3 years ago

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Answer:

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Step-by-step explanation:

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4 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

(-2, 8) (9, 10) (1,7) (3, 9) (10, 12)FunctionNot a Function

lesya692 [45]

First of all, we need to know what a function is.

A function is a binary relation between two sets that

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1 year ago

Find the derivative of <img src="https://tex.z-dn.net/?f=tan%5E%7B-1%7D%20x" id="TexFormula1" title="tan^{-1} x" alt="tan^{-1} x

sladkih [1.3K]

$\huge{\color{magenta}{\fbox{\textsf{\textbf{Answer}}}}}$

$\frak {\huge{ \frac{1}{1 + {x}^{2} } }}$

Step-by-step explanation:

$\sf let \: f(x) = { \tan }^{ - 1} x \\ \\ \sf f(x + h) = { \tan}^{ - 1} (x + h)$

$\sf f'(x) = \frac{f(x+h) - f(x) }{h}$

$\sf \implies \lim_{ h \to 0 } \frac{ { \tan }^{ - 1}(x + h) - { \tan}^{ - 1}x }{h} \\ \\ \\ \sf \implies \lim_ {h \to 0} \frac{ { \tan}^{ - 1} \frac{x + h - x}{1 + (x + h)x} }{h}$

By using

$\sf { \tan}^{ - 1} x - { \tan}^{ - 1} y = \\ \sf { \tan}^{ - 1} \frac{x - y}{1 + xy} formula$

$\sf \implies \large \lim_{h \to0 } \frac{ { \tan}^{ - 1} \frac{h}{1 + hx + {x}^{2} } }{h} \\ \\ \\ \sf \implies \large{\lim_{h \to0} } \frac{ { \tan}^{ - 1} \frac{h}{1 + hx + {x}^{2} } }{ \frac{h}{1 + hx + {x}^{2} } \times (1 + hx + {x}^{2} )} \\ \\ \\ \sf \implies \large \lim_{h \to0} \frac{ { \tan}^{ - 1} \frac{h}{1 + hx + {x}^{2} } }{ \frac{h}{1 + hx + {x}^{2} } } + \lim_{h \to0} \frac{1}{1 + hx + {x}^{2} }$

Now putting the value of h = 0

$\sf \large \implies 0 + \frac{1}{1 + 0 + {x}^{2} } \\ \\ \\ \purple{ \boxed { \implies \frac{1}{1 + {x}^{2} } }}$

6 0

2 years ago

Given A(-2, 5) and B(13, -7), find the midpoint of AB.​

Given A(-2, 5) and B(13, -7), find the midpoint of AB.