What is the radius of a circle that has a diameter of 199
meters?
Do not include a unit of measure with your response;
provide just a numeric answer.

[tex]\qquad \qquad\huge \underline{\boxed{\sf Answer}}[/tex]

For the first figure ~

The diagonals of a kite intersect each other at 90°

So, we can apply Pythagoras theorem here :

[tex]\qquad \sf  \dashrightarrow \: CD² = OC² + OD²[/tex]

[tex]\qquad \sf  \dashrightarrow \: CD² = {7}^{2} + {9}^{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \: CD² = 49 + 81[/tex]

[tex]\qquad \sf  \dashrightarrow \: CD² = 130[/tex]

[tex]\qquad \sf  \dashrightarrow \: CD=x = \sqrt{ 130}[/tex]

For the second figure ;

we have same concept of kite, and use of Pythagoras theorem !

Also, the diagonal QS bisects diagonal PR

Hence,

[tex]\qquad \sf  \dashrightarrow \: PR = 2 \times OR [/tex]

[tex]\qquad \sf  \dashrightarrow \: 10 = 2 \times OR [/tex]

[tex]\qquad \sf  \dashrightarrow \: OR = 10 \div 2[/tex]

[tex]\qquad \sf  \dashrightarrow \: OR = 5 \: mm[/tex]

now, apply pythagoras theorem ~

[tex]\qquad \sf  \dashrightarrow \: QR² = OR² + OQ²[/tex]

[tex]\qquad \sf  \dashrightarrow \: QR² = {5}^{2} + {6}^{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \: QR² = 25 + 36[/tex]

[tex]\qquad \sf  \dashrightarrow \: QR² = 61[/tex]

[tex]\qquad \sf  \dashrightarrow \: QR=x = \sqrt{61} \: mm[/tex]

here, 2 OR = 2 OP = PR

so, similarly OP = 5 mm

Applying pythagoras theorem again ;

[tex]\qquad \sf  \dashrightarrow \: SP² = OS² + OP²[/tex]

[tex]\qquad \sf  \dashrightarrow \: SP² = {10}^{2} + {5}^{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \: SP² = {100}^{} + 25[/tex]

[tex]\qquad \sf  \dashrightarrow \: SP² = 125[/tex]

[tex]\qquad \sf  \dashrightarrow \: SP = \sqrt{125}[/tex]

[tex]\qquad \sf  \dashrightarrow \: SP = y = 5\sqrt{5} \: mm[/tex]

Answer:

4.5 L

Step-by-step explanation:

4500/1000 =4.5

Have an amazing day!

Please mark brainliest!

Answer:

b

Step-by-step explanation:

Answer:

10.5

Step-by-step explanation:

8.4^2 = 70.56

6.3^2 = 39.69

110.25 = c^2

c = 10.5

Answer:

V =2860 in ^3

Step-by-step explanation:

The volume of a rectangular prism is given by

V = l*w*h  where l is the length, w is the width and h is the height.

V = 20 * 11 * 13

V =2860 in ^3

Answer:

Therefore the equation is y=-x+4 (correct optlon: C)

Step-by-step explanation:

In order to find the equation that passes through both points, we can use the slope-intercept form of the linear equation:

y=mx+b

Where m is the slope and b is the y-intercept.

Using the given points on this equation, we have:

(-4,8):

(-4,8):8=m*(-4)+b

(-4,8):8=m*(-4)+bb=8+4m

(-4,8):8=m*(-4)+bb=8+4m(1,3):

(-4,8):8=m*(-4)+bb=8+4m(1,3):3=m+b

(-4,8):8=m*(-4)+bb=8+4m(1,3):3=m+b3=m+8+4m

(-4,8):8=m*(-4)+bb=8+4m(1,3):3=m+b3=m+8+4m5m=3-8

(-4,8):8=m*(-4)+bb=8+4m(1,3):3=m+b3=m+8+4m5m=3-85m=-5

(-4,8):8=m*(-4)+bb=8+4m(1,3):3=m+b3=m+8+4m5m=3-85m=-5m=-1

(-4,8):8=m*(-4)+bb=8+4m(1,3):3=m+b3=m+8+4m5m=3-85m=-5m=-1b=8+4*(-1)=8-4=4

Therefore the equation is y=-x+4 (correct optlon: C)

Answer:

[tex] a=\frac{9c}{2}-b [/tex]

Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variable.

Answer:

[tex]a=\frac{9c}{2}-b[/tex]

Step-by-step explanation:

Isoate the variable to solve

Answer:

from the figure find the length unknown side

Answer: 2/4 simplified to 1/2

Step-by-step explanation:

Answer:

-1/2

Step-by-step explanation:

A(-3, 4)

B(1, 2)

slope = (y2 - y1)/(x2 - x1)

slope = (4 - 2)/(-3 - 1)

slope = 2/(-4)

slope = -1/2

Answer:

circumference=2πr

2×3.14×5

10×3.14

=31.4cm

Answer: Area=78.54

circumference=31.42

Step-by-step explanation:

[tex]\textit{internal division of a line segment using ratios} \\\\\\ A(-7,2)\qquad B(1,-6)\qquad \qquad \stackrel{\textit{ratio from A to B}}{5:3} \\\\\\ \cfrac{A\underline{P}}{\underline{P} B} = \cfrac{5}{3}\implies \cfrac{A}{B} = \cfrac{5}{3}\implies 3A=5B\implies 3(-7,2)=5(1,-6)[/tex]

[tex](\stackrel{x}{-21}~~,~~ \stackrel{y}{6})=(\stackrel{x}{5}~~,~~ \stackrel{y}{-30})\implies P=\underset{\textit{sum of the ratios}}{\left( \cfrac{\stackrel{\textit{sum of x's}}{-21+5}}{5+3}~~,~~\cfrac{\stackrel{\textit{sum of y's}}{6-30}}{5+3} \right)} \\\\\\ P=\left( \cfrac{-16}{8}~~,~~\cfrac{-24}{8} \right)\implies P=(-2~~,~~-3)[/tex]

Answer:

C+3

Step-by-step explanation:

[tex]\qquad \qquad\huge \underline{\boxed{\sf Answer}}[/tex]

In the given question, the two angles are complementary Angle pair, so the sum of their values will add up to 90°

that is ~

[tex]\qquad \sf  \dashrightarrow \: (3x + 30) \degree + x \degree = 90 \degree[/tex]

[tex]\qquad \sf  \dashrightarrow \: 3x + x + 30 \degree = 90 \degree[/tex]

[tex]\qquad \sf  \dashrightarrow \: 4x = 90 \degree - 30 \degree[/tex]

[tex]\qquad \sf  \dashrightarrow \: 4x = 60 \degree[/tex]

[tex]\qquad \sf  \dashrightarrow \: x = 60 \div 4[/tex]

[tex]\qquad \sf  \dashrightarrow \: x = 15 \degree[/tex]

Answer:

x = 15 (third option)

Step-by-step explanation:

Complementary angles sum up to 90°.

∠1 + ∠2 = 90°

x° + (3x + 30)° = 90°

x + 3x + 30 = 90

4x = 90 - 30

4x = 60

x = 60/4

x = 15

Hope it helps ⚜

For function 1

p=r+7

On comparing to y=mx+b

For function 2

y intercept is 8

Function 2 has greater intercept

Answer:

0.78

Step-by-step explanation:

To know which one is the greatest, convert all of them into decimals.

3/4 is the same as 0.75

50% is the same as 0.50

0.78 doesn't needs to be changed

We now have 0.75, 0.50 and 0.78.

Out of all these numbers, 0.78 is the greatest number.

Hope this helps.

Answer:

Andy basically has two stickers

Answer:

√12

Step-by-step explanation:

Use the Phythagorean Theorem.

2² + b² = 4²

= 4 + b² = 16

 -4            -4

= b² = 12

√12

Answer:

[tex]\sqrt{12}[/tex] or [tex]2\sqrt{3}[/tex] (simplified)

Step-by-step explanation:

Pythagorean theorem is

α²+b²=c²

a is the leg, which is 2 in this triangle

b is the base, which x in this triangle

c is the hypotenuse, which is 4 in this triangle

So, we can plug in the values:

2²+x²=4²

4+x²=16

-4       -4

x²=12

x=[tex]\sqrt{12}[/tex], or if simplified: [tex]2\sqrt{3}[/tex]

Hope this helps!

The equation a/b x b/a = 1 is a product equation

It is true that a/b x b/a = 1

The product expression is given as:

a/b x b/a = 1

Rewrite the product expression properly as follows:

[tex]\frac ab * \frac ba = 1[/tex]

Multiply the numerator of the fractions

[tex]\frac {ab}b * \frac 1a = 1[/tex]

Multiply the denominator of the fractions

[tex]\frac {ab}{ab} = 1[/tex]

Evaluate the quotient

[tex]1= 1[/tex]

Hence, it is true that a/b x b/a = 1

Read more about mathematical proofs at:

https://brainly.com/question/1788884

SOLUTION:

=) 49/64 is the answer I think

Answer:

Dilate, Reflect

Not congruent, dilations are used

Step-by-step explanation:

Dilate the figure by a scale factor of 1/3

Reflecting the resulting shape will map the figure

Step 1: Dilate by 1/3

Step 2: Reflect over the y-axis

Since dilations are used, the figures are not the same, as it changes area and side lengths.

-Chetan K

We need x and y

JKLM is a rectangle not triangle

Opposite sides of rectangle are equal

Now

[tex]\\ \rm\Rrightarrow 4y+5=2y+35[/tex]

[tex]\\ \rm\Rrightarrow -2y=-30[/tex]

[tex]\\ \rm\Rrightarrow y=-30/-2[/tex]

[tex]\\ \rm\Rrightarrow y=15[/tex]

And

[tex]\\ \rm\Rrightarrow x+31=5x-9[/tex]

[tex]\\ \rm\Rrightarrow -4x=-40[/tex]

[tex]\\ \rm\Rrightarrow x=10[/tex]

Answer:

  (c)  $63.75

Step-by-step explanation:

The discounted price will be 100% -25% = 75% of the marked price. Mike's final price will be ...

  ($80 +5)×0.75 = $63.75

Answer:

The cost of one ticket will be £8.20

It looks like the integral might be

[tex]\displaystyle \int (9 - x^2)^{3/2} \, dx[/tex]

or perhaps

[tex]\displaystyle \int \frac1{(9 - x^2)^{3/2}} \, dx[/tex]

Take note of the fact that both integrands are defined only over the interval -3 < x < 3.

For either integral, we substitute x = 3 sin(θ) and dx = 3 cos(θ) dθ.

Note that we want this substitution to be reversible, so we must restrict -π/2 ≤ θ ≤ π/2, an interval over which sine has an inverse. Then θ = arcsin(x/3).

The first case then reduces to

[tex]\displaystyle \int (9 - (3\sin(\theta))^2)^{3/2} (3 \cos(\theta) \, d\theta) = 3 \times 9^{3/2} \int (1 - \sin^2(\theta))^{3/2} \cos(\theta) \, d\theta \\\\ = 81 \int (\cos^2(\theta))^{3/2} \cos(\theta) \, d\theta \\\\ = 81 \int |\cos^3(\theta)| \cos(\theta) \, d\theta[/tex]

By definition of absolute value,

[tex]\displaystyle 81 \int |\cos^3(\theta)| \cos(\theta) \, d\theta = \begin{cases}\displaystyle 81 \int \cos^4(\theta) \, d\theta & \text{if }\cos(\theta) \ge 0 \\ \displaystyle -81 \int \cos^4(\theta) \, d\theta & \text{if }\cos(\theta) < 0\end{cases}[/tex]

and these cases correspond to 0 ≤ θ < π/2 and π/2 < θ ≤ π, respectively. But we are assuming -π/2 ≤ θ ≤ π/2, so the negative case doesn't matter to us.

You can compute the remaining antiderivative by exploiting the half-angle identity for cosine,

[tex]\cos^2(\theta) = \dfrac{1 + \cos(2\theta)}2[/tex]

Then

[tex]\cos^4(\theta) = \left(\cos^2(\theta)\right)^2 = \dfrac{1 + 2\cos(2\theta) + \cos^2(2\theta)}4 = \dfrac{3 + 4\cos(2\theta) + \cos(4\theta)}8[/tex]

and so

[tex]\displaystyle \int \cos^4(\theta) \, d\theta = \dfrac{12\theta + 8\sin(2\theta) + \sin(4\theta)}{32} + C[/tex]

We can simplify this using the double angle identity for (co)sine,

sin(2θ) = 2 sin(θ) cos(θ)

cos(2θ) = 1 - 2 sin²(θ)

as well as the relations,

sin(arcsin(x/3)) = x/3

cos(arcsin(x/3)) = √(9 - x²)/3

which gives us

[tex]\displaystyle \int \cos^4(\theta) \, d\theta = \dfrac{12\theta + 16 \sin(\theta) \cos(\theta) + 4 \sin(\theta) \cos(\theta) (1 - 2\sin^2(\theta))}{32} + C[/tex]

Putting this in terms of x, we get

[tex]\displaystyle \int (9 - x^2)^{3/2} \, dx \\ = 81 \times \dfrac{12\arcsin\left(\frac x3\right) + 16 \times \frac x3 \times \frac{\sqrt{9-x^2}}3 + 4\times\frac x3\times\frac{\sqrt{9-x^2}}3 \left(1 - 2\left(\frac x3\right)^2\right)}{32} + C[/tex]

[tex]\displaystyle \int (9 - x^2)^{3/2} \, dx = 81 \times \dfrac{12\arcsin\left(\frac x3\right) + \frac{16x\sqrt{9-x^2}}9 + \frac{4x(9-2x^2)\sqrt{9-x^2}}{81}}{32} + C[/tex]

[tex]\boxed{\displaystyle \int (9 - x^2)^{3/2} \, dx = \dfrac{12\arcsin\left(\frac x3\right) + (180x-8x^3)\sqrt{9-x^2}}{32} + C}[/tex]

If you were asking about the other integral, the first few steps are similar and you end up with the far more trivial integral and antiderivative

[tex]\displaystyle \frac19 \int \frac{d\theta}{\cos^2(\theta)} = \frac19 \int \sec^2(\theta) \, d\theta = \frac19 \tan(\theta) + C[/tex]

Putting it back in terms of x, we get

[tex]\displaystyle \int \frac1{(9 - x^2)^{3/2}} \, dx = \frac19 \tan\left(\arcsin\left(\frac x3\right)\right) + C[/tex]

Recall that tan(θ) = sin(θ)/cos(θ), so

[tex]\displaystyle \int \frac1{(9 - x^2)^{3/2}} \, dx = \frac19 \times \frac{\frac x3}{\frac{\sqrt{9-x^2}}3} + C[/tex]

[tex]\boxed{\displaystyle \int \frac1{(9 - x^2)^{3/2}} \, dx = \frac{x}{9\sqrt{9-x^2}} + C}[/tex]

Answer:

2

Step-by-step explanation:

Degree of the polynomial is the highest power of the term.

As degree of the polynomial is 3, m value should be 2

Answer: its b.) 2

Step-by-step explanation:

I got it right om test

Answer:

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

Step-by-step explanation: