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2 years ago

Find the parity of the following if a and b have different parities. (5a - 3b)(a² + 5b) + 2

Mathematics

1 answer:

sp2606 [1]2 years ago

4 0

9514 1404 393

Answer:

odd

Step-by-step explanation:

If 'a' and 'b' have different parity, the result has odd parity.

Assume a=odd, b=even

(5·odd -3·even) = odd

(odd^2 +5·even) = odd

(odd)(odd) +2 = odd

Assume a=even, b=odd

(5·even -3·odd) = odd

(even^2 +5(odd)) - odd

(odd)(odd) +2 = odd

_____

The attached verifies this with numbers.

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The temperature in the core of the sun is 27,000,000 •F

AlekseyPX

Answer:

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Step-by-step explanation:

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3 years ago

Please help with geometry. 20 points, and another question worth 98 on my profile!

vovikov84 [41]

Answer:

Part 1) m∠1 =(1/2)[arc SP+arc QR]

Part 2) $PR^{2} =PS*PT$

Part 3) PQ=PR

Part 4) m∠QPT=(1/2)[arc QT-arc QS]

Step-by-step explanation:

Part 1)

we know that

The measure of the inner angle is the semi-sum of the arcs comprising it and its opposite.

we have

m∠1 -----> is the inner angle

The arcs that comprise it and its opposite are arc SP and arc QR

m∠1 =(1/2)[arc SP+arc QR]

Part 2)

we know that

The Intersecting Secant-Tangent Theorem, states that the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.

In this problem we have that

$PR^{2} =PS*PT$

Part 3)

we know that

The Tangent-Tangent Theorem states that if from one external point, two tangents are drawn to a circle then they have equal tangent segments

In this problem

PQ=PR

Part 4)

we know that

The measurement of the outer angle is the semi-difference of the arcs it encompasses.

In this problem

m∠QPT -----> is the outer angle

The arcs that it encompasses are arc QT and arc QS

therefore

m∠QPT=(1/2)[arc QT-arc QS]

4 0

2 years ago

HELP! Identify the equation of the circle that has its center at (-3, -4) and passes through the origin.

Lunna [17]

Answer:

Option B. (x+3) ^2+(y+4)^2=25

Step-by-step explanation:

we know that

The equation of the circle in standard form is equal to

$(x-h)^{2}+(y-k)^{2}=r^{2}$

where

(h,k) is the center and r is the radius

step 1

Find the radius of the circle

Remember that the distance of the center and any point on the circle is equal to the radius of the circle

Find the distance between the points (-3,-4) and (0,0)

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

substitute

$r=\sqrt{(0+4)^{2}+(0+3)^{2}}$

$r=\sqrt{(4)^{2}+(3)^{2}}$

$r=\sqrt{25}$

$r=5\ units$

step 2

Find the equation of the circle

$(x-h)^{2}+(y-k)^{2}=r^{2}$

the center is the point (-3,-4) and the radius is r=5 units

substitute

$(x+3)^{2}+(y+4)^{2}=5^{2}$

$(x+3)^{2}+(y+4)^{2}=25$

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2 years ago

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3 years ago