Given: overline LM cong overline ON and overline LO cong overline MN Prove : LMNO is a parallelogram

To solve this question, we have to find the equation of the circle with given center and where it passes. Doing this, we get that the equation of the circle is:

[tex](x - 16)^2 + (y - 30)^2 = 1156[/tex]

Equation of a circle:

The equation of a circle with center [tex](x_0, y_0)[/tex] and radius r is given by:

[tex](x - x_0)^2 + (y - y_0)^2 = r^2[/tex]

Center at (16, 30)

This means that [tex]x_0 = 16, y_0 = 30[/tex]

Thus

[tex](x - 16)^2 + (y - 30)^2 = r^2[/tex]

Passes through the origin:

We use this to find the radius squared, as this means that [tex]x = 0, y = 0[/tex] is part of the circle. Thus

[tex](x - 16)^2 + (y - 30)^2 = r^2[/tex]

[tex](0 - 16)^2 + (0 - 30)^2 = r^2[/tex]

[tex]r^2 = 16^2 + 30^2 = 1156[/tex]

Thus, the equation of the circle is:

[tex](x - 16)^2 + (y - 30)^2 = 1156[/tex]

For another example to find the equation of a circle, you can look at https://brainly.com/question/23719612

Answer:

√(p²-1)

Step-by-step explanation:

sin (27°) = √(p²-1) / 1

= √(p²-1)

Answer:

3.1

Step-by-step explanation:

5 x5 =25

25+3=28

the other children are 4 years old

Given:

The vertices of the triangle ABC are A(2, 1), B(0,-1), C(2, -1).

To find:

The vertices of the image of triangle ABC if ABC is translated two units down.

Solution:

It is given that the triangle ABC is translated two units down. So, the rule of translation is:

[tex](x,y)\to (x,y-2)[/tex]

Using this rule, we get

[tex]A(2,1)\to A'(2,1-2)[/tex]

[tex]A(2,1)\to A'(2,-1)[/tex]

Similarly,

[tex]B(0,-1)\to B'(0,-1-2)[/tex]

[tex]B(0,-1)\to B'(0,-3)[/tex]

And,

[tex]C(2.-1)\to C'(2,-1-2)[/tex]

[tex]C(2.-1)\to C'(2,-3)[/tex]

The vertices of the image are A'(2,-1), B'(0,-3), C'(2,-3).

Therefore, the correct option is C.

Answer:

C.

Step-by-step explanation:

Answer:

True

Step-by-step explanation:

A cylinder has two circles as its bases, and the diameter formula is:

2r = d

So if you plug in:

2r = 18

---    ---

2     2

r = 9

So it will be true.

Hope this helped.

Answer:

B

Step-by-step explanation:

We are given the equation:

[tex]\displaystyle x(b-c) = y+x[/tex]

And that:

[tex]2b=3c=7[/tex]

And we want to find the value of y / x.

To start, subtract x from both sides in the first equation:

[tex]x(b-c) -x = y[/tex]

Divide both sides by x:

[tex]\displaystyle \frac{x(b-c)-x}{x}=\frac{y}{x}[/tex]

Simplify:

[tex]\displaystyle (b-c)-1 = \frac{y}{x}[/tex]

Next, in the second equation, divide everything by two:

[tex]\displaystyle b = \frac{3}{2} c = \frac{7}{2}[/tex]

Substitute:

[tex]\displaystyle \left(\frac{3}{2} c - c \right) - 1= \frac{y}{x}[/tex]

Simplify:

[tex]\displaystyle \frac{1}{2} c - 1 = \frac{y}{x}[/tex]

From the modified second equation, we can multipy both sides by 1/3:

[tex]\displaystyle \frac{1}{2} c = \frac{7}{6}[/tex]

Substitute:

[tex]\displaystyle \left(\frac{7}{6}\right) -1 = \frac{y}{x}[/tex]

Subtract:

[tex]\displaystyle \frac{y}{x} = \frac{7}{6} - \frac{6}{6} = \frac{1}{6}[/tex]

Therefore, our answer is B.

Answer:

C: 119 square inches

Step-by-step explanation:

We are given;

Slant height; L = 5 in

Base area; B = 49 in²

Since it's a square pyramid, the base portion has a square shape.

Thus, area of base = x²

Where x is a side of the square.

Thus;

x² = 49

x = √49

x = 7

Perimeter of base = 7 × 4 = 28 in

Area of pyramid = ½PL + B

Plugging in the relevant values;

Area of pyramid = (½ × 28 × 5) + 49

Area of pyramid = 119 in²

Answer:

We can write a ratio between two quantities, x and y, as:

x to y.

To find if two ratios:

"a to b" and "c to d" are equal, we need to see if the quotientes:

a/b and c/d are equal.

Here we know that the ratios are:

4 apples to 5 strawberries, this gives the quotient 4/5 = 0.8

7 apples to 9 strawberries, this gives the quotient 7/9 = 0.78

So the quotients are different, which means that the ratios are not equal.

Now we want to see who used a greater ratio of apples to strawberries.

notice that in the numerator we used the number of apples, so as larger is the quotient, larger is the ratio of apples to strawberries.

We can see that the quotient of Angie is larger, then Angie used the greater ratio of apples to strawberries.

Answer:

y= 3x+8

Step-by-step explanation:

not a 100% sure...

sry if its wrong

(try using Math-way, its rly helpful)

Answer:

Step-by-step explanation:

y=mx+b

To find slope: -4+1/-4+3

Slope=3

y=3x+b

Plug in either points ,as an example, i'll plug in (-3,-1)

-1=3(-3)+b

-1=-9+b

8=b

Finished formula: y=3x+8

Answer:

x = 14.63

Step-by-step explanation:

Answer:

37.5 cm^2

Step-by-step explanation:

The area of a parallelogram is

A = bh  where b is the base and h is the height

A = 7.5 * 5

A = 37.5 cm^2

Answer:

A = 37.5 cm²

Step-by-step explanation:

The area of a parallelogram is calculated as

A = bh ( b is the base and h the perpendicular height )

Here b = 7.5 and h = 5 , then

A = 7.5 × 5 = 37.5 cm²

Answer:

y = -1/2x

Step-by-step explanation:

If two lines are perpendicular to each other, they have opposite slopes.

The first line is y = 2x + 5. Its slope is 2. A line perpendicular to this one will have a slope of -1/2.

Plug this value (-1/2) into your standard point-slope equation of y = mx + b.

y = -1/2x + b

To find b, we want to plug in a value that we know is on this line: in this case, it is (-4, 2). Plug in the x and y values into the x and y of the standard equation.

2 = -1/2(-4) + b

To find b, multiply the slope and the input of x (-4)

2 = 2 + b

Now, subtract 2 from both sides to isolate b.

0 = b

Plug this into your standard equation.

y = -1/2x + 0 or y = -1/2x

This equation is perpendicular to your given equation (y = 2x + 5) and contains point (-4, 2)

Hope this helps!

Answer: D. Angles that form a linear pair are sometimes supplementary.

That is not true, linear pairs are always supplementary.

Answer:

D. Angles that form a linear pair are sometimes supplementary.

~OAmalOHopeO

Answer:

No Major League ballparks are exactly alike, but certain aspects of the field of play must be uniform across baseball.

The infield must be a square that is 90 feet on each side, and the outfield is the area between the two foul lines formed by extending two sides of said square (though the dirt portion of the field that runs well past the 90-foot basepaths in all Major League parks is also commonly referred to as the infield). The field must be constructed so that the bases are the same level as home plate.

The rulebook states that parks constructed by professional teams after June 1, 1958, must have a minimum distance of 325 feet between home plate and the nearest fence, stand or other obstruction on the right- and left-field foul lines, and 400 feet between home plate and the nearest fence, stand or other obstruction in center field. However, some clubs have been permitted to construct parks after that date with dimensions shorter than those specified.

The pitcher's plate must be a 24-inch by 6-inch slab of whitened rubber that is 10 inches above the level of home plate and 60 feet, 6 inches away from the back point of home plate. It is placed 18 inches behind the center of the mound -- which is erected within an 18-foot diameter circle -- and surrounded by a level area that is 5 feet by 34 inches. The slope of the pitcher's mound begins 6 inches in front of the pitcher's plate and must gradually decrease by 1 inch every foot for 6 feet in the direction of home plate.

Home plate is a 17-inch square of whitened rubber with two of the corners removed so that one edge is 17 inches long, two adjacent sides are 8 1/2 inches each and the remaining two sides are 12 inches each and set at an angle to make a point. The 17-inch side faces the pitcher's plate, and the two 12-inch edges coincide with the first- and third-base lines. The back tip of home plate must be 127 feet, 3 and 3/8 inches away from second base.

The other bases must be 15-inch squares that are between 3 and 5 inches thick, covered by white canvas or rubber and filled with soft material.

Step-by-step explanation:

Answer:

Step-by-step explanation:

Scientific notation is always written as a number between 1 and 9.9999999 repeating. There follows a power of 10. I'm not sure which answer would help you see it clearer but

4.8 * 10^4 the decimal is shifted 4 places the right which is 48000 where you have no numbers, you use zeros.

2.01 * 10^6 means that you shift the decimal 6 places to the right.

2010000

So 2 million is larger than 48 thousand.

The other way to do it is just to look at the power. 6 is more than 4. 6 describes the number of places you move. So does 4. 6>4 so that's the larger number.

Answer:

Annual payment wil be 2,001 and total amount after compounded annually will be 27,001

Answer:

49/1

Step-by-step explanation:

The point is on (1,49)

Answer:

8th term is -15309

Step-by-step explanation:

[tex]{ \boxed{ \bf{u_{n} = a( {r}^{n - 1} ) }}} \\ { \tt{u_{8} = 7( {( - 3)}^{8 - 1}) }} \\ { \tt{u_{8} = 7( - 2187)}} \\ { \tt{u _{8} = - 15309}}[/tex]

r is the common difference, r = -21/7 = -3

Answer:

a₈ = - 15309

Step-by-step explanation:

The nth term of a geometric sequence is

[tex]a_{n}[/tex] = a₁ [tex](r)^{n-1}[/tex]

where a₁ is the first term and r the common ratio

Here a₁ = 7 and r = [tex]\frac{a_{2} }{a_{1} }[/tex] = [tex]\frac{-21}{7}[/tex] = - 3 , then

a₈ = 7 × [tex](-3)^{7}[/tex] = 7 × - 2187 = - 15309

Answer:

the answer = um ok

Step-by-step explanation:

9514 1404 393

Answer:

  9.23%

Step-by-step explanation:

The account balance for simple interest is given by ...

  A = P(1 +rt) . . . . . principal P invested for t years at rate r

  554 = 379(1 +r·5)

  554 = 379 + 1895r . . . . eliminate parentheses

  175 = 1895r . . . . . . . . . subtract 379

  r = 175/1895 ≈ 0.092348 ≈ 9.23%

Jennifer's account had an interest rate of about 9.23%.

Answer:

4/25

Step-by-step explanation:

All the amount of recipes is 11+10+4=25 . There are 4 recipes for cakes

so probability is 4/25

Answer:

[tex] \rm \displaystyle x \approx \bigg \{ {59.3}^{ \circ} + \frac{n\pi}{2} , - {14.3}^{ \circ} + \frac{n\pi}{2} \bigg \}[/tex]

Step-by-step explanation:

we would like to solve the following trigonometric equation:

[tex] \rm \displaystyle \sin(x) \cos(3x) + \cos(x) \sin(3x) = \tan( {140}^{ \circ} ) [/tex]

the left hand side can be rewritten using angle sum indentity of sin which is given by:

[tex] \rm \displaystyle \sin( \alpha + \beta ) = \sin( \alpha ) \cos( \beta ) + \cos( \alpha ) \sin( \beta ) [/tex]

therefore Let

Thus substitute:

[tex] \rm \displaystyle \sin(x + 3x) = \tan( {140}^{ \circ} ) [/tex]

simplify addition:

[tex] \rm \displaystyle \sin(4x) = \tan( {140}^{ \circ} ) [/tex]

keep in mind that sin(t)=sin(π-t) saying that there're two equation to solve:

[tex] \begin{cases} \rm \displaystyle \sin(4x) = \tan( {140}^{ \circ} ) \\ \\ \displaystyle \sin(\pi - 4x) = \tan( {140}^{ \circ} ) \end{cases}[/tex]

take inverse trig and that yields:

[tex] \begin{cases} \rm \displaystyle 4x= { \sin}^{ - 1} ( \tan( {140}^{ \circ} ) ) \\ \\ \displaystyle \pi - 4x = { \sin}^{ - 1}( \tan( {140}^{ \circ} ) ) \end{cases}[/tex]

add π to both sides of the second equation and that yields:

[tex] \begin{cases} \rm \displaystyle 4x= { \sin}^{ - 1} ( \tan( {140}^{ \circ} ) ) \\ \\ \displaystyle - 4x = { \sin}^{ - 1}( \tan( {140}^{ \circ} ) ) + \pi\end{cases}[/tex]

sin function has a period of 2nπ thus add the period:

[tex] \begin{cases} \rm \displaystyle 4x= { \sin}^{ - 1} ( \tan( {140}^{ \circ} ) ) + 2n\pi\\ \\ \displaystyle - 4x = { \sin}^{ - 1}( \tan( {140}^{ \circ} ) ) + \pi + 2n\pi\end{cases}[/tex]

divide I equation by 4 and II by -4 which yields:

[tex] \begin{cases} \rm \displaystyle x= \frac{ { \sin}^{ - 1} ( \tan( {140}^{ \circ} ) ) }{4} + \frac{n\pi}{2} \\ \\ \displaystyle x = - \frac{{ \sin}^{ - 1}( \tan( {140}^{ \circ} ) ) + \pi}{4} - \frac{n\pi}{2} \end{cases}[/tex]

recall that,-½(nπ)=½(nπ) therefore,

[tex] \begin{cases} \rm \displaystyle x= \frac{ { \sin}^{ - 1} ( \tan( {140}^{ \circ} ) ) }{4} + \frac{n\pi}{2} \\ \\ \displaystyle x = - \frac{{ \sin}^{ - 1}( \tan( {140}^{ \circ} ) ) + \pi}{4} + \frac{n\pi}{2} \end{cases}[/tex]

by using a calculator we acquire:

[tex] \begin{cases} \rm \displaystyle x \approx - {14.3}^{ \circ} + \frac{n\pi}{2} \\ \\ \displaystyle x \approx {59.3}^{ \circ} + \frac{n\pi}{2} \end{cases}[/tex]

hence,

the general solution for: for the trig equation are

[tex] \rm \displaystyle x \approx \bigg \{ {59.3}^{ \circ} + \frac{n\pi}{2} , - {14.3}^{ \circ} + \frac{n\pi}{2} \bigg \}[/tex]

Answer:

y = - x + 5

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Here gradient (slope) = - 1 and (0, 5) ⇒ c = 5

y = - x + 5 ← equation of line

Answer:

Step-by-step explanation:

Answer:

4, yes through the middle    5, yes through the middle 6, yes through the middle  all of them reflect from the center

Step-by-step explanation:

Answer:

0.04

Step-by-step explanation:

2 / 50 is the same as 1 / 25.

1 / 25 = 0.04

Answer:

0.04

Step-by-step explanation:

Answer:

383.54 m

Step-by-step explanation:

The length of the training track running around the field = circumference of the circle formed by the two semicircles + 2(length of the rectangle)

The two semicircles forms a fill circle with diameter (d) = width of rectangle = 61 m

Length of rectangle (L) = 96 m

π = 3.14

The length of the training track running around the field = πd + 2(L)

Substitute the values

The length of the training track running around the field = 3.14*61 + 2(96)

= 191.54 + 192

= 383.54 m

Answer: Day 4 and 15

Step-by-step explanation:

Set up system of equations:

[tex]25x+136=-x^2+44x+76[/tex]

Simplify:

[tex]x^2-19x+60=0[/tex]

Factor: Both factors must be positive as the days can only be natural numbers.

[tex](x-15)(x-4)=0[/tex]

x=4,15