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Delvig [45]
3 years ago
6

In ADEF, the measure of ZF=90°, DF = 40, ED = 41, and FE = 9. What ratio

Mathematics
1 answer:
frez [133]3 years ago
6 0

Answer:

sec(∠F) = \frac{41}{9}

Step-by-step explanation:

In triangle EFD,

m∠F = 90°

Adjacent side of ∠E = EF = 9 units

Opposite side of ∠E = DF = 40 units

Hypotenuse = DE = 41 units

For secant of ∠E ,

sec(∠E) = \frac{\text{Hypotenuse}}{\text{Adjacent side}}

            = \frac{41}{9}

Therefore, \frac{41}{9} is the ratio of secant of ∠F.

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katovenus [111]

Since \alpha and \beta are roots of ax^2+bx+c, we can factorize the quadratic in terms of the roots as

ax^2+bx+c = a(x-\alpha)(x-\beta)

Expanding the right side gives

ax^2+bx+c = ax^2-a(\alpha+\beta)x+a\alpha\beta

so that

\alpha + \beta = -\dfrac ba \\\\ \alpha\beta = \dfrac ca

A polynomial with roots \alpha^2 and \beta^2 would be

(x-\alpha^2)(x-\beta^2)

and expanding this gives

x^2 - (\alpha^2+\beta^2)x+\alpha^2\beta^2

Now,

\alpha\beta = \dfrac ca \implies \alpha^2\beta^2 = (\alpha\beta)^2 = \dfrac{c^2}{a^2}

and

(\alpha+\beta)^2 = \alpha^2 + 2\alpha\beta + \beta^2 \\\\ \implies \alpha^2+\beta^2 = \left(-\dfrac ba\right)^2 -2\left(\dfrac ca\right) = \dfrac{b^2-2ac}{a^2}

So we can write the second quadratic in terms of a,b,c as

(x-\alpha^2)(x-\beta^2) = x^2 - \dfrac{b^2-2ac}{a^2}x + \dfrac{c^2}{a^2}

and to make things look cleaner, scale the whole expression by a^2 to get

\boxed{a^2x^2 + (2ac-b^2)x + c^2}

4 0
3 years ago
During the grand opening of an electronics store, any customer has a 17% chance of winning a free laptop. What is the chance tha
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6 0
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5 0
3 years ago
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a_sh-v [17]
So we are given the mean and the s.d.. The mean is 100 and the sd is 15 and we are trying the select a random person who has an I.Q. of over 126. So our first step is to use our z-score equation:

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1 - 0.9582 = 0.0418

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Hopes this helps!
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3 years ago
A model aeroplane is made to a scale of 1:50. If the span of the wings of the model is 30 cm , what is the actual length of the
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