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3 years ago

Use the equation below to identify the value fo each variable for the circle.

Mathematics

1 answer:

natita [175]3 years ago

7 0

Answer:

The equation of the circle in standard form is: (x - 2)² + (y - 4)² = 9

Step-by-step explanation:

* Lets revise the standard form of the equation of the circle

- If the center of the circle is point (h , v) and the radius of the

circle is r, then the standard form of the equation of the circle

is (x - h)² + (y - v)² = r²

- (x , y) a general point on the circle

* Lets look to the picture

- The center of the circle is point (2 , 4)

- The highest point on the circle is (2 , 7) and the lowest point

on the circle is (2 , 1)

∴ The diameter of the circle = 7 - 1 = 6

∵ The radius = 1/2 the diameter

∴ The radius of the circle = 1/2 × 6 = 3

* Now we can write the equation of the circle

∵ h = 2 and v = 4

∵ r = 3

∴ (x - 2)² + (y - 4)² = 3²

∴ The equation of the circle in standard form is:

(x - 2)² + (y - 4)² = 9

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A rectangular area adjacent to a river is fenced in; no fence is needed on the river side. The enclosed area is 1500 square fee

ZanzabumX [31]

Answer:

a) C(x) = 15000/x + 6x +80

b) Domain of C(x) { R x>0 }

Step-by-step explanation:

We have:

Enclosed area = 1500 ft² = x*y from which y = 1500 / x (a) where x is perpendicular to the river

Cost = cost of sides of fenced area perpendicular to the river + cost of side parallel to river + cost of 4 post then

Cost = 10*y + 2*3*x + 4*20 and accoding to (a) y = 1500/x

Then

C(x) = 10* ( 1500/x ) + 6*x + 80

C(x) = 15000/x + 6x +80

Domain of C(x) { R x>0 }

5 0

2 years ago

For the cost function c equals 0.1 q squared plus 2.1 q plus 8, how fast does c change with respect to q when q equals 11? Det

Soloha48 [4]

Answer:

Rate of change of c with respect to q is 4.3

Percentage rate of change c with respect to q is 9.95%

Step-by-step explanation:

Cost function is given as, $c=0.1\:q^{2}+2.1\:q+8$

Given that c changes with respect to q that is, $\dfrac{dc}{dq}$ . So differentiating given function,

$\dfrac{dc}{dq}=\dfrac{d}{dq}\left (0.1\:q^{2}+2.1\:q+8 \right)$

Applying sum rule of derivative,

$\dfrac{dc}{dq}=\dfrac{d}{dq}\left(0.1\:q^{2}\right)+\dfrac{d}{dq}\left(2.1\:q\right)+\dfrac{d}{dq}\left(8\right)$

Applying power rule and constant rule of derivative,

$\dfrac{dc}{dt}=0.1\left(2\:q^{2-1}\right)+2.1\left(1\right)+0$

$\dfrac{dc}{dt}=0.1\left(2\:q\right)+2.1$

$\dfrac{dc}{dt}=0.2\left(q\right)+2.1$

Substituting the value of $q=11$ ,

$\dfrac{dc}{dt}=0.2\left(11\right)+21.$

$\dfrac{dc}{dt}=2.2+2.1$

$\dfrac{dc}{dt}=4.3$

Rate of change of c with respect to q is 4.3

Formula for percentage rate of change is given as,

$Percentage\:rate\:of\:change=\dfrac{Q'\left(x\right)}{Q\left(x\right)}\times 100$

Rewriting in terms of cost C,

$Percentage\:rate\:of\:change=\dfrac{C'\left(q\right)}{C\left(q\right)}\times 100$

Calculating value of $C\left(q \right)$

$C\left(q\right)=0.1\:q^{2}+2.1\:q+8$

Substituting the value of $q=11$ ,

$C\left(q\right)=0.1\left(11\right)^{2}+2.1\left(11\right)+8$

$C\left(q\right)=0.1\left(121\right)+23.1+8$

$C\left(q\right)=12.1+23.1+8$

$C\left(q\right)=43.2$

Now using the formula for percentage,

$Percentage\:rate\:of\:change=\dfrac{4.3}{43.2}\times 100$

$Percentage\:rate\:of\:change=0.0995\times 100$

$Percentage\:rate\:of\:change=9.95%$

Percentage rate of change of c with respect to q is 9.95%

7 0

3 years ago

How we can differentiate between direct proportion and inverse proportion?

Yakvenalex [24]

Answer:

Please check explanations for answer

Step-by-step explanation:

Here, we want to differentiate between inverse and direct proportion

Let x and y be two quantities

If they are directly proportional;

x/y = k

x = ky

where k is proportionality constant

If inversely;

xy = k

In verbal terms;

For direct proportionality, when one value increase, the other too does so and vice versa

For inverse, when one value inverse, the other decrease

4 0

3 years ago

PLEASE HELP ASAP!! 50 POINTS!!!

Andre45 [30]

Quantity Equation: t + s = 20 gallons
Lemon content Eq: 20t+70s = 40*20 gallons
Modify for elimination:
2t + 2s = 2*20
2t + 7s = 4*20
Subtract and solve for "s":
5s = 2*20
s = 8 gallons (amt. of 70% mix needed)
t = 20-8 = 12 gallons (amt of 20% mix needed)

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3 years ago

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500,000=_____x10 I'm confused on this

aalyn [17]

Answer:

5 x $10^{5}$

Step-by-step explanation:

if its scientific notation

8 0

2 years ago