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3 years ago

What is the length of DD'?

Mathematics

2 answers:

Blababa [14]3 years ago

8 0

Answer:

The length of DD' is $\sqrt{29}$ or 5.39 units.

Step-by-step explanation:

The distance formula is

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

From the given figure it is clear that the coordinates of D are (2,0) and coordinates of D' are (7,2).

Using distance formula we get

$DD'=\sqrt{(7-2)^2+(2-0)^2}$

$DD'=\sqrt{25+4}$

$DD'=\sqrt{29}$

$DD'\approx 5.39$

Therefore the length of DD' is $\sqrt{29}$ or 5.39 units.

zavuch27 [327]3 years ago

6 0

The length of DD' is 5.39 units.

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Step-by-step explanation:

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3 years ago

One of the roots of the quadratic equation dx^2+cx+p=0 is twice the other, find the relationship between d, c and p

scZoUnD [109]

Answer:

$c^2 = 9dp$

Step-by-step explanation:

Given

$dx^2 + cx + p = 0$

Let the roots be $\alpha$ and $\beta$

So:

$\alpha = 2\beta$

Required

Determine the relationship between d, c and p

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Divide through by d

$\frac{dx^2}{d} + \frac{cx}{d} + \frac{p}{d} = 0$

$x^2 + \frac{c}{d}x + \frac{p}{d} = 0$

A quadratic equation has the form:

$x^2 - (\alpha + \beta)x + \alpha \beta = 0$

So:

$x^2 - (2\beta+ \beta)x + \beta*\beta = 0$

$x^2 - (3\beta)x + \beta^2 = 0$

So, we have:

$\frac{c}{d} = -3\beta$ -- (1)

and

$\frac{p}{d} = \beta^2$ -- (2)

Make $\beta$ the subject in (1)

$\frac{c}{d} = -3\beta$

$\beta = -\frac{c}{3d}$

Substitute $\beta = -\frac{c}{3d}$ in (2)

$\frac{p}{d} = (-\frac{c}{3d})^2$

$\frac{p}{d} = \frac{c^2}{9d^2}$

Multiply both sides by d

$d * \frac{p}{d} = \frac{c^2}{9d^2}*d$

$p = \frac{c^2}{9d}$

Cross Multiply

$9dp = c^2$

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Hence, the relationship between d, c and p is: $c^2 = 9dp$

8 0

3 years ago

Can anyone explain this to me??

finlep [7]

It is the first option. DF = DE. It is referring to the lengths of each line, DE and DF look like they are the same. It is not a scalene triangle it is a right triangle. EF is way too long to be equal to DE or DF

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