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3 years ago

What is the point slope of an equation with slope of 3/5 that passes through the point (10,-2)

Mathematics

2 answers:

soldier1979 [14.2K]3 years ago

6 0

Y - y_1 = m(x - x_1)
y - (-2) = 3/5(x - 10)
y + 2 = 3/5(x - 10)

The answer is B.

Verdich [7]3 years ago

4 0

The answer would be B. It gives you the slope and a point on the graph so use that to your advantage. All of the answers show the slope and the slope in the right place, so for now it can be any of them. Then we get to the point (10,-2). For point slope form you would switch the signs for both x and y. So the Y would now be +2 and x would be -10. Y goes together with y, and X goes together with x. In this case y+-y= slope(X+-x) Substitute in for them. Y+2=3/5(X-10)

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Question number 8. Prove that 3 + √5 is an irrational number.

Marina86 [1]

Answer :

Let's assume the opposite of the statement i.e., 3 + √5 is a rational number.

$\\ \sf \: 3 + \sqrt{5} = \frac{a}{b} \\ \\ \qquad \: \tiny \sf{(where \: \: a \: \: and \: \: b \: \: are \: \: integers \: \: and \: \: b \: \neq \: 0)} \\$

$\\ \sf \: \sqrt{5} = \frac{a}{b} - 3 = \frac{a - 3b}{b} \\$

Since, a, b and 3 are integers. So,

$\\ \sf \: \frac{p - 3b}{b} \\ \\ \qquad \tiny \sf{ \: (is \: \: a \: \: rational \: \: number \:) } \\$

Here, it contradicts that √5 is an irrational number.

because of the wrong assumption that 3 + √5 is a rational number.

$\\$

Hence, 3 + √5 is an irrational number.

7 0

3 years ago

Can you guy help for this questing <br> 1234

postnew [5]

Answer:

1. Median

2. Altitude

3. Altitude

4. Neither

Step-by-step explanation:

Median- splits the side into two congruent parts

Altitude- Makes a right angle

1. Splits the side. Median

2. Makes a right angle. Altitude

3. Makes a right angle. Altitude

4. Doesn't make a right angle or split the sides. Neither

4 0

2 years ago

A recipe calls for 2 3/4 cups of sugar and 4 1/3 cups of flour. How many cups of sugar are used for every 1 cup of flour?

worty [1.4K]

4 1/3 cup of flour = 2 3/4 cup of sugar

1 cup of flour = 2 3/4 ÷ 4 1/3 = 33/52 cup of sugar <=== answer

6 0

3 years ago

our entire family me , my mom , grandma , sister , and two brothers went to dinner . we each had a coupon for 3$ off the cost of

hichkok12 [17]

Answer:

$79.50

Step-by-step explanation:

5 0

3 years ago

For <img src="https://tex.z-dn.net/?f=e%5E%7B-x%5E2%2F2%7D" id="TexFormula1" title="e^{-x^2/2}" alt="e^{-x^2/2}" align="absmiddl

nevsk [136]

I'm assuming you're talking about the indefinite integral

$\displaystyle\int e^{-x^2/2}\,\mathrm dx$

and that your question is whether the substitution $u=\dfrac x{\sqrt2}$ would work. Well, let's check it out:

$u=\dfrac x{\sqrt2}\implies\mathrm du=\dfrac{\mathrm dx}{\sqrt2}$
$\implies\displaystyle\int e^{-x^2/2}\,\mathrm dx=\sqrt2\int e^{-(\sqrt2\,u)^2/2}\,\mathrm du$
$=\displaystyle\sqrt2\int e^{-u^2}\,\mathrm du$

which essentially brings us to back to where we started. (The substitution only served to remove the scale factor in the exponent.)

What if we tried $u=\sqrt t$ next? Then $\mathrm du=\dfrac{\mathrm dt}{2\sqrt t}$ , giving

$=\displaystyle\frac1{\sqrt2}\int \frac{e^{-(\sqrt t)^2}}{\sqrt t}\,\mathrm dt=\frac1{\sqrt2}\int\frac{e^{-t}}{\sqrt t}\,\mathrm dt$

Next you may be tempted to try to integrate this by parts, but that will get you nowhere.

So how to deal with this integral? The answer lies in what's called the "error function" defined as

$\mathrm{erf}(x)=\displaystyle\frac2{\sqrt\pi}\int_0^xe^{-t^2}\,\mathrm dt$

By the fundamental theorem of calculus, taking the derivative of both sides yields

$\dfrac{\mathrm d}{\mathrm dx}\mathrm{erf}(x)=\dfrac2{\sqrt\pi}e^{-x^2}$

and so the antiderivative would be

$\displaystyle\int e^{-x^2/2}\,\mathrm dx=\sqrt{\frac\pi2}\mathrm{erf}\left(\frac x{\sqrt2}\right)$

The takeaway here is that a new function (i.e. not some combination of simpler functions like regular exponential, logarithmic, periodic, or polynomial functions) is needed to capture the antiderivative.

3 0

3 years ago