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2 years ago

In this diagram, ABAC – AEDF. If the area of ABAC = 6 in?, what is the area of AEDF?

Mathematics

1 answer:

shepuryov [24]2 years ago

3 0

Answer:

2.7 in²

Step-by-step explanation:

similar triangles have the same angles, and all side lengths (or other distances) of one triangle have the same scaling factor to the side lengths of the other triangle.

so, we know the relation between the 2 baselines is 2/3, as this is the factor to turn the baseline of the large triangle into the baseline of the smaller triangle.

in other words

EF = BC × 2/3

2 = 3 × 2/3

correct

how do we calculate the area of a triangle ?

Area = baseline × height / 2

from BAC we know

Area = 6

baseline = 3

height = ?

6 = 3 × height / 2

12 = 3 × height

height = 4

aha !

now, EDF has a height too that we need to calculate is Area. and this height has the same scaling factor compared to the larger triangle as the side lengths : 2/3

so, for EDF we know

Area = ?

baseline = 2

height = 4 × 2/3 = 8/3

therefore, the area is

Area = (2 × 8/3) / 2 = (16/3) / 2 = 8/3 = 2.66666... ≈ 2.7

the shirt answer would be :

we know from the 2 baselines that the scaling factor for each distance is 2/3.

for the area we need to multiply 2 distances, so that means we have to multiply both by 2/3. and so on the formula for the area we have to use 2/3 × 2/3.

2/3 × 2/3 = 4/9

Area small = Area large × 4/9 = 6 × 4/9 = 24/9 = 8/3 ≈ 2.7

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Step-by-step explanation:

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3 years ago

Do you agree with the statement that " cubic functions must have at least 1 x-intercept but not more than 3, whereas quadratic f

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Answer:

Yes, I agree

Step-by-step explanation:

For the cubic function

A cubic function is represented as:

$f(x)=ax^3+bx^2+cx+d$

A cubic function may have 1, 2 or 3 x intercepts. This is shown below

For 3 x intercepts

$y = x^3 - x$

Equate y to 0

$x^3 - x = 0$

Expand

$x(x^2 - 1) = 0$

Express $x^2 - 1$ as difference of two squares

$x(x - 1)(x +1 ) = 0$

x = 0 or 1 or -1

For 2 x intercepts

$y = x^3 - x$

$y =(x-5)^2)(x+7)$

Equate y to 0

$(x-5)^2(x+7) = 0$

Expand

$(x-5)(x-5)(x+7) = 0$

x= 5 or x = -7

For 1 x intercept

$y = x^3$

Equate y to 0

$x^3 = 0$

Take cube roots of both sides

$x = 0$

It has been shown above that a cubic function may have 1, 2 or 3.

So, I agree to the statement

For the quadratic function

A quadratic function will not have any x intercept when the function can not be factorized;

E.g.

$y = x^2 + x + 17$

The above function has no x intercept.

A quadratic function will have at least 1 x intercept when the function can be factorized;

E.g.

$y = x^2- 6x + 9$

Equate y to 0

$x^2- 6x + 9 = 0$

Expand

$x^2 - 3x - 3x + 9 = 0$

$(x - 3)(x-3) = 0$

$x = 3$

We've shown that a quadratic may have no x intercept, and it may also have x intercept(s)

Hence, I agree to both statement

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2 years ago

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3 years ago

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Answer:

The correct option is;

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Step-by-step explanation:

The given parameters are;

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3 years ago