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2 years ago

Hey guys can I get a little help on this problem please, thank you.

Mathematics

1 answer:

DedPeter [7]2 years ago

5 0

Point g because it’s in the middle.

midpoint= the middle.

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What values of x makes the expression (x-6)(x+3) positive?

Georgia [21]

Answer:

This means that for (x-6)(x+3) to be positive, the value of x has to be greater than 6; x>6

Step-by-step explanation:

Step 1

Given the equation;

(x-6)(x+3)

Step 2

To determine the values of x that make the equation positive, the equation can be expressed as;

(x-6)(x+3)>0

x-6>0/(x+3)

x-6>0

x>0+6

x>6

This means that for (x-6)(x+3) to be positive, the value of x has to be greater than 6; x>6

4 0

3 years ago

Limit definition for slope of the graph, equation of tangent line point for<br> f(x)=2x^2 at x=(-1)

Tems11 [23]

The slope of the tangent line to $f$ at $x=-1$ is given by the derivative of $f$ at that point:

$f'(-1)=\displaystyle\lim_{x\to-1}\frac{f(x)-f(-1)}{x-(-1)}=\lim_{x\to-1}\frac{2x^2-2}{x+1}$

Factorize the numerator:

$2x^2-2=2(x^2-1)=2(x-1)(x+1)$

We have $x$ approaching -1; in particular, this means $x\neq-1$ , so that

$\dfrac{2x^2-2}{x+1}=\dfrac{2(x-1)(x+1)}{x+1}=2(x-1)$

Then

$f'(-1)=\displaystyle\lim_{x\to-1}\frac{2x^2-2}{x+1}=\lim_{x\to-1}2(x-1)=2(-1-1)=-4$

and the tangent line's equation is

$y-f(-1)=f'(-1)(x-(-1))\implies y-4x-2$

6 0

3 years ago

An equivalent expression for n x a using only addition

Rina8888 [55]

<u /><u /><u /> $n\times a\quad =\quad (a+a+a+a+a+.....+a)$

6 0

3 years ago

Read 2 more answers

Instructions: Find the missing side. Round your answer to the nearest

ch4aika [34]

Answer:

Step-by-step explanation:

Tan(51) = opposite / adjacent

The adjacent side makes up the reference angle (51o)

The opposite side is not part of the reference angle

adjacent side = x

opposite side = 16

Tan(51) = 16 / x Multiply both sides by x

x * Tan(51) = 16 Divide by tan 51

x = 16 / tan(51)

tan(51) = 1.2349

x = 16 / 1.2349

x = 12.96

7 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B%20%5Cfrac%7B3%7D%7B16%7D%20%7D%20" id="TexFormula1" title=" \sqrt{ \frac{3}{16}

sweet-ann [11.9K]

Hello!
When you have a non-exact root (irrational number)

$\sqrt{ \frac{3}{16} } \to \frac{ \sqrt{3} }{ \sqrt{16} } \to \boxed{\boxed{\frac{ \sqrt{3} }{4} }}\Longleftarrow(irrational\:number)\end{array}}\qquad\quad\checkmark$

7 0

4 years ago

Hey guys can I get a little help on this problem please, thank you.​

Hey guys can I get a little help on this problem please, thank you.