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2 years ago

A circle with the are 16pi has a sector with the central angel of 8/5 radians

Mathematics

1 answer:

dolphi86 [110]2 years ago

7 0

Answer:

3/2 units^2

Step-by-step explanation:

The area of the entire circle is 16π. What is the area of a sector of this circle if the central angle (not angel) is 8/5 radians?

Formula: (fraction of circle represented by the sector)(area of the circle

Area of the sector described above is

8/5 radians 8*16π square units

----------------------- * 16π square units = ------------------------------ = 3/2 units^2

2π radians 10π

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If Y varies directly as X and Y= 79 when X= 49, Find Y if X = 9

ryzh [129]

Answer:

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Step-by-step explanation:

given that y varies directly as x

mathematically,

y ∝ x

y = kx .............................................. where k is the constant of proportionality.

k = y /x

given that

y = 79

x = 49

k = 79/49

k = 1.61

to find y when x = 9

from

y = kx

y = 1.61 × 9

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therefore the value of y when x = 9 in ths variation is evaluated to be equals to 14.5

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Drag the tiles to the correct boxes to complete the pairs not all tiles will be used match each quadratic graph to its respectiv

ch4aika [34]

Answer:

Part 1) The function of the First graph is $f(x)=(x-3)(x+1)$

Part 2) The function of the Second graph is $f(x)=-2(x-1)(x+3)$

Part 3) The function of the Third graph is $f(x)=0.5(x-6)(x+2)$

See the attached figure

Step-by-step explanation:

we know that

The quadratic equation in factored form is equal to

$f(x)=a(x-c)(x-d)$

where

a is the leading coefficient

c and d are the roots or zeros of the function

Part 1) First graph

we know that

The solutions or zeros of the first graph are

x=-1 and x=3

The parabola open up, so the leading coefficient a is positive

The function is equal to

$f(x)=a(x-3)(x+1)$

Find the value of the coefficient a

The vertex is equal to the point (1,-4)

substitute and solve for a

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

$-4=a(-2)(2)$

$a=1$

therefore

The function is equal to

$f(x)=(x-3)(x+1)$

Part 2) Second graph

we know that

The solutions or zeros of the first graph are

x=-3 and x=1

The parabola open down, so the leading coefficient a is negative

The function is equal to

$f(x)=a(x-1)(x+3)$

Find the value of the coefficient a

The vertex is equal to the point (-1,8)

substitute and solve for a

$8=a(-1-1)(-1+3)$

$8=a(-2)(2)$

$a=-2$

therefore

The function is equal to

$f(x)=-2(x-1)(x+3)$

Part 3) Third graph

we know that

The solutions or zeros of the first graph are

x=-2 and x=6

The parabola open up, so the leading coefficient a is positive

The function is equal to

$f(x)=a(x-6)(x+2)$

Find the value of the coefficient a

The vertex is equal to the point (2,-8)

substitute and solve for a

$-8=a(2-6)(2+2)$

$-8=a(-4)(4)$

$a=0.5$

therefore

The function is equal to

$f(x)=0.5(x-6)(x+2)$

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3 years ago