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g100num [7]
3 years ago
8

What is the divisor of 50=25 x 2 + 0

Mathematics
1 answer:
aleksley [76]3 years ago
5 0

Step-by-step explanation:

First 25 x 2 = 50

second 50+0=50

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Step-by-step explanation:

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Choose the constant term that completes the perfect square trinomial.<br><br> y2 + y
Oduvanchick [21]

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First off, please write y2 + y as y^2 + y. The " ^ " symbol denotes exponentiation, whereas y2 is meaningless.

To find the constant term in question, take half of the coefficient of y (that is, take 1/2) and square it. Then we have y2 + y + 1/4.

The constant term in question is 1/4.


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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

dx^2 + cx + p = 0

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

or

c^2 = 9dp

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

8 0
3 years ago
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