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aniked [119]
3 years ago
10

PLEASE ASAP ILL GIVE BRAINLIEST.

Mathematics
1 answer:
iren [92.7K]3 years ago
6 0

Answer:

first one

Step-by-step explanation:

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Please help and thanks ​
pav-90 [236]

Answer: Bottom left corner

Angle DEC and Angle DEH

=============================================================

Explanation:

I recommend opening your favorite paint program, and using different colors to mark on the drawing as I have done below (see attached image). Note how angle DEC is in red and angle DEH is in blue. The two angles are adjacent, meaning they share the same ray (in this case, ray ED) and they form a straight angle.

Because they form a straight angle, or straight line, this means the two angles are supplementary. Supplementary angles always add to 180.

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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

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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Mice21 [21]
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Read 2 more answers
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