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3 years ago

Help plz !!!!! Isosceles and Equilateral.

Mathematics

1 answer:

Ilya [14]3 years ago

8 0

Answer: 40

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

Explanation:

The angle we want is QPR (bottom left), which is one of the base angles. The other base angle is QRP (bottom right). These two angles are equal because PQR is an isosceles triangle (PQ = RQ)

So if we can find angle QRP, then we have found angle QPR

Note how angle QRP and the 140 degree angle combine to form a straight 180 degree angle. Therefore these two angles add to 180 degrees

(angle QRP) + (140) = 180
(angle QRP) + 140 - 140 = 180-140 ... subtract 140 from both sides
angle QRP = 40

Since,
angle QPR = angle QRP
this means
angle QPR = 40
and
b = 40

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Answer:

a) $P (x) = (x + 3) (x-1) (x-4)$

b) $P (x) = (2x + 5) (5x - 4) (x-6)$

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Step-by-step explanation:

For the question a * you need to find a polynomial of degree 3 with zeros in -3, 1 and 4.

This means that the polynomial P(x) must be zero when x = -3, x = 1 and x = 4.

Then write the polynomial in factored form.

$P (x) = (x + 3) (x-1) (x-4)$

Note that this polynomial has degree 3 and is zero at x = -3, x = 1 and x = 4.

For question b, do the same procedure.

Degree: 3

Zeros: -5/2, 4/5, 6.

The factors are

$x = -\frac{5}{2}\\\\x +\frac{5}{2} = 0\\\\(2x +5) = 0$

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$x =\frac{4}{5}\\\\x-\frac{4}{5} = 0\\\\(5x-4) = 0$

--------------------------------------

$x = 6\\\\(x-6) = 0$

--------------------------------------

$P (x) = (2x + 5) (5x - 4) (x-6)$

Finally for the question c we have

Degree: 5

Zeros: -3, 1, 4, -1

Multiplicity 2 in -1

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

--------------------------------------

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

--------------------------------------

$x = 4\\\\(x-4) = 0$

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

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$P (x) = (x-3) (x-1) (x-4) (x + 1) ^ 2$

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