All of the following are ways to prove triangles congruent except

Find the area of the shaded region in the figure below, if the radius of the outer circle is 4 and the radius of the inner circl

denis-greek [22]

TheThe area of the shaded region if the radius of the outer circle is 4 and the radius of the inner circle is 2 is 12π.

<h3>Area of the shaded region</h3>

Area of a circle = πr²

r =radius of the circle

Area of the outer circle:

Area of the outer circle = π(4)²

Area of the outer circle = 16π

Area of the inner circle:

Area of the inner circle = π (2)²

Area of the inner circle = 4 π

Area of the shaded Region :

Area of the shaded Region = 16π - 4 π

Area of the shaded Region = 12π

Therefore the area of the shaded region if the radius of the outer circle is 4 and the radius of the inner circle is 2 is 12π.

Learn more about area of the shaded region here: brainly.com/question/19577281

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5 0

2 years ago

Find the absolute minimum and absolute maximum values of f on the given interval. f(t) = t 25 − t2 , [−1, 5]

Zina [86]

Answer: Absolute minimum: f(-1) = -2 $\sqrt{6}$

Absolute maximum: f( $\sqrt{12.5}$ ) = 12.5

Step-by-step explanation: To determine minimum and maximum values in a function, take the first derivative of it and then calculate the points this new function equals 0:

f(t) = $t\sqrt{25-t^{2}}$

f'(t) = $1.\sqrt{25-t^{2}}+\frac{t}{2}.(25-t^{2})^{-1/2}(-2t)$

f'(t) = $\sqrt{25-t^{2}} -\frac{t^{2}}{\sqrt{25-t^{2}} }$

f'(t) = $\frac{25-2t^{2}}{\sqrt{25-t^{2}} }$ = 0

For this function to be zero, only denominator must be zero:

$25-2t^{2} = 0$

t = ± $\sqrt{2.5}$

$\sqrt{25-t^{2}}$ ≠ 0

t = ± 5

Now, evaluate critical points in the given interval.

t = $-\sqrt{2.5}$ and t = - 5 don't exist in the given interval, so their f(x) don't count.

f(t) = $t\sqrt{25-t^{2}}$

f(-1) = $-1\sqrt{25-(-1)^{2}}$

f(-1) = $-\sqrt{24}$

f(-1) = $-2\sqrt{6}$

f( $\sqrt{12.5}$ ) = $\sqrt{12.5} \sqrt{25-(\sqrt{12.5} )^{2}}$

f( $\sqrt{12.5}$ ) = 12.5

f(5) = $5\sqrt{25-5^{2}}$

f(5) = 0

Therefore, absolute maximum is f( $\sqrt{12.5}$ ) = 12.5 and absolute minimum is

f(-1) = $-2\sqrt{6}$ .

8 0

3 years ago

An object is moving at a speed of 4 meters per month. Express this speed in inches

BartSMP [9]

Answer:

<h2><em>0.22inches/hour</em></h2>

Step-by-step explanation:

Given the speed of an object to be 4metres/month, we are to express the speed in inches/hour.

First we will need to convert 4meters to inches

From the conversion factors given,

1 foot (ft) = 0.305 meters (m)

x ft = 4metres

cross multiply

x * 0.305 = 4*1

x = 4/0.305

x = 13.115ft

4 metres = 13.115ft

Then we will convert 13.115ft to inches

From the conversion factors,

1 foot (ft) = 12 inches (in)

13.115ft = y

cross multiply

y = 13.115*12

y = 157.377 inches

Hence<em> 4 metres = 157.77inches.</em>

For 1 month to hour

First we need to convert month to days and from the conversion factor

1 month = 30 days

Next is to convert 30days to hours

Since 1 day = 24hours

30days = z

z = 30* 24

z = 720hours

Hence <em>1 month = 720hours</em>

On converting 4metres/month to inches/hour

4metres/month = 157.77inches/720hours

4metres/month = 0.219inches/hour

<em>Hence the speed of the object in inches per hour to the nearest hundredth is 0.22inches/hour</em>

7 0

3 years ago

Please help me with this thanks

Natalija [7]

.no lo sé, pero como necesito puntos, ¿sí?

6 0

2 years ago

Assume the resting heart rates for a sample of individuals are normally distributed with a mean of 75 and a standard deviation o

Shalnov [3]

Answer:

a) 0.975

b) 0.16

c) 0.475

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 75

Standard deviation = 5

We also have that since the normal distribution is symmetric, 50% of the measures are below the mean and 50% are above.

a. The relative frequency of rates less than 85 using the 68-95-99.7 rule is

85 is two standard deviations above the mean.

So all the 50% below the mean is below 85

And of the 50% above the mean, 95% is less than 85. So

0.5 + 0.95*0.5 = 0.975

b. The relative frequency of rates greater than 80 using the 68-95-99.7 rule is

80 is one standard deviations above the mean.

Of the 50% above the mean, 68% is less than 80. So 100-68 = 32% is more than 80.

0.32*0.5 = 0.16.

c. The relative frequency of rates between 65 and 75 using the 68-95-99.7 rule is

Within the mean and two standard deviations below the mean.

So 95% of 50%

0.95*0.5 = 0.475

6 0

4 years ago