6 times 1/5 less than 1 , equal to 1, or greater than 1?

Write an equation for a circle with a diameter that has endpoints at (–10, 1) and (–8, 5). Round to the nearest tenth if necessa

satela [25.4K]

The standard form for the equation of a circle is :

(x−h)^2+(y−k)^2=r2 ----------- EQ(1)
 where handk are the x and y coordinates of the center of the circle and r is the radius.
 The center of the circle is the midpoint of the diameter.

So the midpoint of the diameter with endpoints at (−10,1)and(−8,5) is :

((−10+(−8))/2,(1+5)/2)=(−9,3)

So the point (−9,3) is the center of the circle.

Now, use the distance formula to find the radius of the circle:

r^2=(−10−(−9))^2+(1−3)^2=1+4=5

⇒r=√5

Subtituting h=−9, k=3 and r=√5 into EQ(1) gives :

(x+9)^2+(y−3)^2=5

7 0

3 years ago

Use two different methods to find an explain the formula for the area of a trapezoid that has parallel sides of length a and B a

evablogger [386]

Answer:

Formula of Trapezoid:

A = (a + b) × h / 2

The formula can be derived in different ways. for now, we have discussed two ways:

1. By using the formula of a triangle

2. By dividing into different sections

Step-by-step explanation:

1. By using the formula of a triangle

One of the ways to explain a formula for an area of a trapezoid using a formula for a triangle can be as follows.

Assume a trapezoid PQRS with lower base SR and upper base PQ (they are parallel) and sides PS and QR.

The image is attached below.

Connect vertices P and R with a diagonal.

Consider triangle ΔPQR as having a base PQ and an altitude from vertex R down to point M on base PQ (RM⊥PQ).

Its area is

S1= $\frac{1}{2} *PQ*RM$

Consider triangle ΔPRS as having a base SR and an altitude from vertex P up to point N on-base SR (PN⊥SR).

Its area is

S2= $\frac{1}{2} *SR*PN$

Altitudes RM and PN are equal and constitute the distance between two parallel bases PQ and SR.

They both are equal to the altitude of the trapezoid h.

Therefore, we can represent areas of our two triangles as

S1= $\frac{1}{2}*PQ*h$

S2= $\frac{1}{2}*SR*h$

Adding them together, we get the area of the whole trapezoid:

S=S1+S2= $\frac{1}{2} (PQ+SR)h,$

which is usually represented in words as "half-sum of the bases times the altitude".

2. By dividing into different sections

Trapezoid PQRS is shown below, with PQ parallel to RS.

Figure 1 - Trapezoid PQRS with PQ parallel to RS(image is attached below.)

We are going to derive the area of a trapezoid by dividing it into different sections.

If we drop another line from Q, then we will have two altitudes namely PT and QU.

Figure 2 - Trapezoid PQRS divided into two triangles and a rectangle. (image is attached below.)

From Figure 2, it is clear that Area of PQRS = Area of PST + Area of PQUT + Area of QRU. We have learned that the area of a triangle is the product of its base and altitude divided by 2, and the area of a rectangle is the product of its length and width. Hence, we can easily compute the area of PQRS. It is clear that

=> $A_{PQRS} = (\frac{ah}{2}) + b_{1}h + \frac{ch}{2}$

Simplifying, we have

=> $A= \frac{ah+2b_{1+C} }{2}$

Factoring we have,

=> $A_{PQRS} = (a+ 2b_{1} + c)\frac{h}{2} \\= > {(a+ b_{1} + c) + b_{1} }\frac{h}{2}$

But, $a+ b_{1} + c$ is equal to $b_{2}$ , the longer base of our trapezoid.

Hence, $A_{PQRS}= (b_{1} + b_{2} )\frac{h}{2}$

We have discussed two ways by which we can derive area of a trapezoid.

Read to know more about Trapezoid

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5 0

2 years ago

The outside temperature can be estimated based on how fast crickets chirp. At 104 chirps per minute, the temperature is 63 degre

Rasek [7]

Answer:

For 124 chirps per minute the temperature is 68 ºF.

For 68 chirps per minute the temperature is 54 ºF.

Step-by-step explanation:

Linear functions are those whose graph is a straight line. A linear function has the following form

$f(x)=b+mx$

b is the constant term or the y intercept. It is the value of the dependent variable when x = 0.

m is the coefficient of the independent variable. It is also known as the slope and gives the rate of change of the dependent variable.

We know that

At 104 chirps per minute, the temperature is 63 ºF.

At 176 chirps per minute, the temperature is 81 ºF.

This information can be converted to Cartesian coordinates (x, y). Where x = the number of chirps per minute and y = the temperature in ºF.

To find a linear function that let us find the outside temperature from how fast crickets chirp we must:

Find the slope:

$m=\frac{y_2-y_1}{x_2-x_1}=\frac{81-63}{176-104}=\frac{1}{4}$

Find the equation:

$81=\frac{1}{4}\cdot 104+b$

Solving for b

$b=81-\frac{1}{4} (176)=37$

Therefore, the linear function is

$y=\frac{1}{4} \cdot x+37$

Now, using this linear function we can know the temperature when we know the chirps per minute:

For 124 chirps per minute the temperature is:

$y=\frac{1}{4} \cdot (124)+37=68$

For 68 chirps per minute the temperature is:

$y=\frac{1}{4} \cdot (68)+37=54$

6 0

3 years ago

Mr. Vectadore is buying two new cowboy hats. All

Bess [88]

Answer:

$50.08

Step-by-step explanation:

Let's use x to represent the lesser expensive hat and 2x to represent the more expensive hat since the more expensive hat is twice the price.

So, our equation would be 2x + x = 75.12

We can combine them together to make 3x.

Then, we make the 3x equal to 75.12

3x = 75.12

Now, we solve it.

Divide both sides by 3 to solve for x.

75.12/3 = 25.04

x = 25.04

Lastly, plug it in for 2x to get the price of the more expensive hat.

2(25.04) = ?

2(25.04) = 50.08

Therefore, the price of the more expensive hat is $50.08

8 0

3 years ago

The diameter of a circle is 20 centimeters. What is the circumference?

kati45 [8]

C ≈ 62.83cm is the answer!

Hope this helps!

3 0

3 years ago

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