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3 years ago

Greatest common factor of 20 and 10.

Mathematics

1 answer:

Anna71 [15]3 years ago

7 0

Answer:

1,2,5,10

Step-by-step explanation:

Greatest common factor is 10

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If f(x) = 3x - 2 and g(x) = 2x + 1, find (f - g)(x)

morpeh [17]

Answer:

(f-g)(x) = x-3

Step-by-step explanation:

Given

f(x) = 3x-2

and

g(x) = 2x+1

We have to find (f-g)(x)

So,

(f-g)(x) = f(x)-g(x)

= 3x-2 - (2x+1)

= 3x-2-2x-1

=x-3

Hence,

(f-g)(x) = x-3

6 0

3 years ago

Read 2 more answers

A survey conducted in July 2015 asked a random sample of American adults whether they had ever used online dating (either an onl

Nadusha1986 [10]

Answer: $(0.081,\ 0.163)$

Step-by-step explanation:

Let $\hat{p}$ be the sample proportion.

As per given , we have

n= 411

$\hat{p}=\dfrac{50}{411}=0.121654501217\approx0.122$

Standard error : se=0.016

Critical value for 99% confidence : $z_{\alpha/2}=2.576$

Confidence interval for population proportion :-

$\hat{p}\pm z_{\alpha/2} (se)\\\\=0.122\pm (2.576)(0.016)\\\\=0.122\pm0.041216\\\\=(0.122-0.041216,\ 0.122+0.041216)\\\\=(0.080784,\ 0.163216)\approx(0.081,\ 0.163)$

Hence, the 99% confidence interval for the proportion of all US adults ages 55 to 64 to use online dating : $(0.081,\ 0.163)$

8 0

3 years ago

Four integers are consecutive terms of an arithmetic sequence. The sum of these four numbers is 24 and the product is 945.Find t

Marta_Voda [28]

When common difference, $d=\sqrt{39}$ then the largest term of these four terms will be $6+3\sqrt{39}$

Let four consecutive terms of AP is,

$a-3d, a-d, a+d, a+3d$

Since, sum of all four terms is 24

So, $a-3d+a-d+a+d+a+3d=24\\\\a=6$

And product is 945

So, $(a-3d)(a+3d)(a-d)(a+d)=945\\\\9d^4-360d^2+351=0\\\\d^4-40d^2+39=0\\\\(d^2-1)(d^2-39)=0\\\\d=1,-1\\d=\sqrt{39}, -\sqrt{39}$

When d= 1, -1 then largest term is 9

When $d=\sqrt{39}, -\sqrt{39}$ then largest term is $6+3\sqrt{39}$

Learn more:

https://brainly.in/question/3719232

7 0

2 years ago

Consider the curve given by the equation y^2-2x^2y=3.

Vsevolod [243]

Answer:

Part A)

$\displaystyle \frac{dy}{dx}=\frac{2xy}{y-x^2}$

Part B)

$y=x-2$

Part C)

$(0, \sqrt{3})\text{ and } (0, -\sqrt3)$

Part D)

$\displaystyle \frac{d^2y}{dx^2}_{(1, -1)}=-\frac{1}{2}$

Step-by-step explanation:

We have the equation:

$y^2-2x^2y=3$

Part A)

We want to find the derivative of the equation. So, dy/dx.

Let’s take the derivative of both sides with respect to x. Therefore:

$\displaystyle \frac{d}{dx}\Big[y^2-2x^2y\Big]=\frac{d}{dx}[3]$

Differentiate. We will need to implicitly different on the left. The second term will also require the product rule. Therefore:

$\displaystyle 2y\frac{dy}{dx}-4xy-2x^2\frac{dy}{dx}=0$

Rearranging gives:

$\displaystyle \frac{dy}{dx}\Big(2y-2x^2\Big)=4xy$

Therefore:

$\displaystyle \frac{dy}{dx}=\frac{4xy}{2y-2x^2}$

And, finally, simplifying:

$\displaystyle \frac{dy}{dx}=\frac{2xy}{y-x^2}$

Part B)

We want to write the equation for the line tangent to the curve at the point (1, -1).

So, we will require the slope of the tangent line at (1, -1). Substitute these values into our derivative. So:

$\displaystyle \frac{dy}{dx}_{(1, -1)}=\frac{2(1)(-1)}{(-1)-(1)^2}=1$

Now, we can use the point-slope form:

$y-y_1=m(x-x_1)$

Substitute:

$y-(-1)=1(x-1)$

Simplify:

$y+1=x-1$

So:

$y=x-2$

Part C)

If the line tangent to the curve is horizontal, this means that dy/dx=0. Hence:

$\displaystyle 0=\frac{2xy}{y-x^2}$

Multiplying both sides by the denominator gives:

$0=2xy$

Assuming y is not 0, we can divide both sides by y. Hence:

$2x=0$

Then it follows that:

$x=0$

Going back to our original equation, we have:

$y^2-2x^2y=3$

Substituting 0 for x yields:

$y^2=3$

So:

$y=\pm\sqrt{3}$

Therefore, two points where the derivative equals 0 is:

$(0, \sqrt{3})\text{ and } (0, -\sqrt3)$

However, we still have to test for y. Let’s go back. We have:

$0=2xy$

Assuming x is not 0, we can divide both sides by x. So:

$0=2y$

Therefore:

$y=0$

And going back to our original equation and substituting 0 for y yields:

$(0)^2-2x^2(0)=0\neq3$

Since this is not true, we can disregard this case.

So, our only points where the derivative equals 0 is at:

$(0, \sqrt{3})\text{ and } (0, -\sqrt3)$

Part D)

Our first derivative is:

$\displaystyle \frac{dy}{dx}=\frac{2xy}{y-x^2}$

Let’s take the derivative of both sides again. Hence:

$\displaystyle \frac{d^2y}{dx^2}=\frac{d}{dx}\Big[\frac{2xy}{y-x^2}\Big]$

Utilize the quotient and product rules and differentiate:

$\displaystyle \frac{d^2y}{dx^2}=\frac{(2y+2x\frac{dy}{dx})(y-x^2)-2xy(\frac{dy}{dx}-2x)}{(y-x^2)^2}$

Let dy/dx=y’. Therefore:

$\displaystyle \frac{d^2y}{dx^2}=\frac{(2y+2xy^\prime)(y-x^2)-2xy(y^\prime-2x)}{(y-x^2)^2}$

For (1, -1), we already know that y’ is 1 at (1, -1). Therefore:

$\displaystyle \frac{d^2y}{dx^2}_{(1, -1)}=\frac{(2(-1)+2(1)(1))((-1)-(1)^2)-2(1)(-1)((1)-2(1))}{((-1)-(1)^2)^2}$

Evaluate:

$\displaystyle \frac{d^2y}{dx^2}_{(1, -1)}=-\frac{1}{2}$

7 0

3 years ago

Ron is a car salesman. He earns a monthly wage of $1,000 plus a commission equal to 3% of the total price of the cars he sells.

Leya [2.2K]

1000 + .03p = r

You write "which equation represents" and yet you did not write any of the choices:
Could be .03p + 1000=r
or r= .03p +1000
or r= 1000 +.03p
They are all equivalent equations.

8 0

3 years ago