(5x + k)2 = mx2 + 70x + 49<br> Find m<br> Find k

I don’t understand Please help!

AlekseyPX

Answer:

The answer is water.........

5 0

3 years ago

What is the mode of the data set?

strojnjashka [21]

The data set is

20

32, 34, 36

40, 42, 44, 48

55

65

We can see that each value only shows up one time. Therefore there is no mode. To have a mode, we need to have a value show up more than once, and it must be the most frequent value. For example, the set {1,2,3,3,4} has a mode of 3 since it shows up twice, the most of any value in that set. However we don't have that occur for the data set your teacher gave you.

<h3>Final Answer: There is no mode for this data set</h3>

5 0

2 years ago

Nadia is a stockbroker. She earns 13% commission each week. Last week, she sold $4,500 worth of stocks. How much did she make

m_a_m_a [10]

Multiply amount sold by commission rate:

4500 x 0.13 = 585

She made $585 last week.

Multiply amount made in 1 week by 52 weeks ( 1 year = 52 weeks)

585 x 52 = 30,420

She made $30,420 in 2011

6 0

2 years ago

Find the distance of the point (3,4) from the midpoint of the line joining (8,10) and (4,6)

Elis [28]

Answer:

Distance between point $(3,4)$ and midpoint of line joining $(8,10)$ and $(4,6)$ = $5$ units.

Step-by-step explanation:

Given:

Points:

$A(3,4)\\B(8,10)\\C(4,6)$

To find distance from point A to midpoint of BC.

Midpoint M of BC:

$M=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})\\$

$M=(\frac{8+4}{2},\frac{10+6}{2})\\$ [Plugging in points $B(8,10)\ and\ C(4,6)$ ]

$M=(\frac{12}{2},\frac{16}{2})\\$

$M=(6,8)\\$

Distance between A and M:

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\$

$D=\sqrt{(6-3)^2+(8-4)^2} \\$ [Plugging in points $A(3,4)\ and\ M(6,8)$ ]

$D=\sqrt{(3)^2+(4)^2} \\$

$D=\sqrt{9+16} \\$

$D=\sqrt{25} \\$

$D=\pm5$

Since distance is always positive ∴ $D= 5$ units

4 0

3 years ago

Use implicit differentiation to find the points where the parabola defined by x2−2xy+y2+4x−8y+20=0 has horizontal and vertical t

Komok [63]

Answer:

The parabola has a horizontal tangent line at the point (2,4)

The parabola has a vertical tangent line at the point (1,5)

Step-by-step explanation:

Ir order to perform the implicit differentiation, you have to differentiate with respect to x. Then, you have to use the conditions for horizontal and vertical tangent lines.

-To obtain horizontal tangent lines, the condition is:

$\frac{dy}{dx}=0$ (The slope is zero)

--To obtain vertical tangent lines, the condition is:

$\frac{dy}{dx}=\frac{1}{0}$ (The slope is undefined, therefore the denominator is set to zero)

Derivating respect to x:

$\frac{d(x^{2}-2xy+y^{2}+4x-8y+20)}{dx} = \frac{d(x^{2})}{dx}-2\frac{d(xy)}{dx}+\frac{d(y^{2})}{dx}+4\frac{dx}{dx}-8\frac{dy}{dx}+\frac{d(20)}{dx}$ = $2x -2(y+x\frac{dy}{dx})+2y\frac{dy}{dx}+4-8\frac{dy}{dx}= 0$