Given that 32 = 9, write this in logarithm form.

Question 8 options:

log2 3 = 9


log3 9 = 2


log3 2 = 9


log2 9 = 3

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

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

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

[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]

[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]

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:

Andy basically has two stickers

Answer:

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

Step-by-step explanation:

Answer:

from the figure find the length unknown side

[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

SOLUTION:

=) 49/64 is the answer I think

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

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:

[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:

4.5 L

Step-by-step explanation:

4500/1000 =4.5

Have an amazing day!

Please mark brainliest!

answer:

$47.92

explanation:

$3.88 × 16=$62.08

$110-$62.08=$47.92

the 16 is the days

110is the budget

3.88 is the cost

hope that helps

The total money Helen have left after 16 days will be $47.92

The unitary method is a method for solving a problem by the first value of a single unit and then finding the value by multiplying the single value. If an event can occur in m different ways and if following it, a second event can occur in n different ways, then the two events in succession can occur in m × n different ways.

Given that Helen budgets for coffee from the coffee bar located in the lobby of her apartment building was  $110. And she stops by the bar every morning and her average order costs $3.88.

The budget = 110 dollars

The average cost = 3.88 dollars

Then the average order costs for 16 days;

$3.88 × 16 = $62.08

Therefore, the money she have left after 16 days;

$110-$62.08=$47.92

Hence, The total money she have left after 16 days will be $47.92

Learn more about the unitary method, please visit the link given below;

https://brainly.com/question/23423168

#SPJ2

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:

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:

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:

√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!

Answer:

The cost of one ticket will be £8.20

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:

C+3

Step-by-step explanation:

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:

b

Step-by-step explanation: