What is the solution to the equation startfraction 3 over m 3 endfraction minus startfraction m over 3 minus m endfraction = startfraction m squared 9 over m squared minus 9 endfraction?

ED and AC are congruent

Write Statement as:

AB*BC=EB*BD

8(4x+2)=9(4x)

32x+16=36x

16=4x

4=x

Hope it helps!

Answer:

[tex]\textsf{B.}\quad -2(\sqrt{2}-1)[/tex]

Step-by-step explanation:

Given integral:

[tex]\displaystyle \int^{\pi}_{\frac{\pi}{2}}\dfrac{\cos \theta}{\sqrt{1+ \sin \theta}}\:\:d\theta[/tex]

Solve by using Integration by Substitution

Substitute u for one of the functions of [tex]\theta[/tex] to give a function that's easier to integrate.

[tex]\textsf{Let }u=1+\sin \theta[/tex]

Find the derivative of u and rewrite it so that [tex]d \theta[/tex] is on its own:

[tex]\implies \dfrac{du}{d \theta}=\cos \theta[/tex]

[tex]\implies d \theta=\dfrac{1}{\cos \theta}\:du[/tex]

Use the substitution to change the limits of the integral from [tex]\theta[/tex]-values to u-values:

[tex]\textsf{When }\theta=\pi \implies u=1[/tex]

[tex]\textsf{When }\theta=\dfrac{\pi}{2} \implies u=2[/tex]

Substitute everything into the original integral and solve:

[tex]\begin{aligned}\displaystyle \int^{\pi}_{\frac{\pi}{2}}\dfrac{\cos \theta}{\sqrt{1+ \sin \theta}}\:\:d\theta & =\int^{1}_2}\dfrac{\cos \theta}{\sqrt{u}}\:\cdot \dfrac{1}{\cos \theta}\:\:du\\\\& =\int^{1}_{2}\dfrac{1}{\sqrt{u}} \:\:du \\\\& =\int^{1}_{2} u^{-\frac{1}{2}}\:\:du \\\\& = \left[ 2u^{\frac{1}{2}} \right]^{1}_{2}\\\\& = \left(2(1)^{\frac{1}{2}}\right)-\left(2(2)^{\frac{1}{2}}\right)\\\\& = 2-2\sqrt{2}\\\\& = -2(\sqrt{2}-1)\end{aligned}[/tex]

Answer:

A

Step-by-step explanation:

Comment (and answer)

Both start with the plus x axis as one of the arms. The other arm of 110o goes anti clockwise until it is 30 degrees into the second quadrant.

The answer can be found by using 110 - 360 = negative angle.

So 110 - 360 = - 250 which goes clockwise from the plus x axis. The answer must be A.

Make sure you understand the concept of clockwise and anti clockwise. If you don't know, ask your teacher, write it down, and try not to forget it. The distinction comes up a lot.

Answer:

6 square units

Step-by-step explanation:

base = 4 units

height = 3 units

 [tex]\sf \boxed{\bf Area \ of \ triangle = \dfrac{1}{2}*base*height}[/tex]

                                  [tex]\sf =\dfrac{1}{2}*4*3\\\\ = 2*3\\\\ = 6 \ square units[/tex]

[tex]\Large❏ \: \large\begin{gathered} {\underline{\boxed{ \rm {\blue{Area \: of \: triangle \: = \: \frac{1}{2} \: \times \: Base \: × \: Height }}}}}\end{gathered}[/tex]

[tex]\rm \large \red{\: Area \: of \: triangle }\large\purple\implies \tt \large \: \frac{1}{2} \: \times \: Base \: × \: Height [/tex]

[tex]\rm \large \red{\: Area \: of \: triangle }\large\purple\implies \tt \large \: \frac{1}{2} \: \times \: 4 \: × \: 3 [/tex]

[tex]\rm \large \red{\: Area \: of \: triangle }\large\purple\implies \tt \large \: \frac{1}{ \cancel2} \: \times \: \cancel{4} \: ^{\green2} \: × \: 3 [/tex]

[tex]\rm \large \red{\: Area \: of \: triangle }\large\purple\implies \tt \large \: 2 \: \times \: 3[/tex]

[tex]\rm \large \red{\: Area \: of \: triangle }\large\purple\implies \tt \large \: 6[/tex]

Hence , the area of triangle is 6 units.

From the given information, we have

[tex]13\sin(\theta) = 12\cos(\theta) \implies \dfrac{\sin(\theta)}{\cos(\theta)} = \dfrac{12}{13}[/tex]

In the expression of interest, divide through everything by [tex]\cos^2(\theta)[/tex] to get

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

Then plugging in the given info, we get

[tex]\dfrac{2\sin(\theta)\cos(\theta)}{\cos^2(\theta) - \sin^2(\theta)} = \dfrac{2\times \frac{12}{13}}{1 - \left(\frac{12}{13}\right)^2} = \boxed{\dfrac{312}{25}}[/tex]

Answer:

The answer is 1: 50000... If you look down here↓:

Step-by-step explanation:

With a scale of 1:20000, a unit length on the map represents a length of 20000. So 1 cm^2 = 1 cm* 1 cm which represents 20000*20000 cm^2 will be = 4*10^8 cm^2.

50 cm^2 represents that 4*50*10^8 cm^2 = 200* 10^8 cm^2.

Now, on the other hand, the map has the same. 200* 10^8 is represented by 8 cm^2. So the scale is sqrt [ 200* 10^8 / 8 ] = sqrt [ 25* 10^8 ] = 5* 10^4 = 50000.

Therefore the scale on the second map is 1: 50000.

Based on the amount of cash that Elias Rubio brings, we can calculate that his total cash deposit is $1,323

This can be found as:

= (75 x 1) + (78 x 5) + (83 x 10) + (80 x 0.25) + (32 x 0.10 or 10 cents) + (95 x 0.05 or 5 cents) + (5 x 0.01 or 1 cent)

Solving gives:

= 75 + 390 + 830 + 20 + 3.20 + 4.75 + 0.05

= $1,323

Remaining part of the question:

How much did he deposit in cash?

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The resulting polynomial when 5x -6 is subtracted from 3x² -6x +2 is; 3x² -11x +8.

It follows from the task content that the polynomial 5x -6 is to be subtracted from 3x² -6x +2.

Hence, the subtraction goes thus; 3x² -6x +2 -(5x-6).

Hence, 3x² -6x +2 -5x+6

= 3x² -11x +8.

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The equation of the parabola will be y = x² – 6x + 9. Then the correct option is C.

The complete question is attached below.

It's the locus of a moving point that keeps the same distance between a stationary point and a specified line. The focus is a non-movable point, while the directrix is a non-movable line.

The equation of a quadratic function, of vertex (h, k), is given by:

y = a(x – h)² + k

where a is the leading coefficient.

The vertex of the parabola is at (3, 0). Then the equation of the parabola will be

y = (x – 3)² + 0

Then open the bracket, we have

y = x² – 6x + 9

Then the correct option is C.

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Using limits, it is found that the end behavior of the graph is given as follows:

It rises to the left, and stays constant at y = -4 to the right.

It is given by the limits of f(x) as x goes to infinity.

In this problem, the function is given by:

[tex]f(x) = 4\left(\frac{2}{5}\right)^{x + 3} - 4[/tex]

Hence:

[tex]\lim_{x \rightarrow -\infty} f(x) = \lim_{x \rightarrow -\infty} 4\left(\frac{2}{5}\right)^{x + 3} - 4 = 4\left(\frac{5}{2}\right)^{\infty + 3} - 4 = \infty - 4 = \infty[/tex]

[tex]\lim_{x \rightarrow \infty} f(x) = \lim_{x \rightarrow -\infty} 4\left(\frac{2}{5}\right)^{x + 3} - 4 = 4\left(\frac{2}{5}\right)^{\infty + 3} - 4 = 0 - 4 = -4[/tex]

Hence:

It rises to the left, and stays constant at y = -4 to the right.

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Answer:

38.8 litres

Step-by-step explanation:

22.6 + 18.7 = 41.3

41.3 - 2.5 = 38.8

38.8 litres

Answer:

38.8 liters.

Step-by-step explanation:

Word problems with addition and subtraction:

To find the quantity of water inside the bath tub, add the quantity of hot water and cold water. Then from this subtract the water that is spilled.

Quantity of water in the bathtub = 22.6 + 18.7 - 2.5

                                                      = 41.3 - 2.5

                                                      = 38.8 liters

Answer:

20ft²

Step-by-step explanation:

Answer:

the answer is 5

Step-by-step explanation:

if you plug in everything, then the equation is the square root of 2 to the 3rd power minus 7 plus the absolute value of 4.

So first 2 to the third power is 8

Then the absolute value of negative 4 is 4

So 8-7 is 1

The square root of 1 is 1

1 plus 4 is 5, the answer is 5

Answer:

1st option

Step-by-step explanation:

the equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k ) are the coordinates of the centre and r is the radius

here (h, k ) = (- 2, 1 ) and r = 3 , then

(x - (- 2) )² + (y - 1)² = 3² , that is

(x + 2)² + (y - 1)² = 9 ← expand factors using FOIL

x² + 4x + 4 + y² - 2y + 1 = 9

x² + 4x + y² - 2y + 5 = 9 ( subtract 9 from both sides )

x² + 4x + y² - 2y - 4 = 0 , that is

x² + y² + 4x - 2y - 4 = 0 ← in general form

Answer:

[tex]\textsf{1)} \quad x^2+y^2+4x-2y-4=0[/tex]

Step-by-step explanation:

Equation of a circle

[tex](x-a)^2+(y-b)^2=r^2[/tex]

where:

From inspection of the graph:

Substitute the found values into the formula:

[tex]\implies (x-(-2)^2+(y-1)^2=3^2[/tex]

[tex]\implies (x+2)^2+(y-1)^2=9[/tex]

Expand and simplify:

[tex]\implies (x+2)^2+(y-1)^2=9[/tex]

[tex]\implies x^2+4x+4+y^2-2y+1=9[/tex]

[tex]\implies x^2+y^2+4x-2y+5=9[/tex]

[tex]\implies x^2+y^2+4x-2y-4=0[/tex]

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Answer:

Given - Length is nine feet more

Area is 252

To find - Value of length and width

Solution-

let the width of the carpet be X

therefore length = 9 + X

Area of carpet = Length * width

252 = (9+x) * X

252 = 9x + x²

Solving the equation by splitting middle term

x² + 9x - 252 = 0

x²- 21x + 12x - 252= 0

x ( x - 21) + 12( x - 21) = 0

(x -21)( x + 12) = 0

X = 21 or X = (-12)

Length = 9 + X

= 30 or (-3)

negative is not possible so

length = 30

width = 21

Answer:

2x(x^2 - 3)(x^2 + 9)

Step-by-step explanation:

2x^5 + 12x^3 − 54x

2x(x^4 + 6x - 27)

Since -3 + 9 = 6 and -3 x 9 = -27:

2x(x^2 - 3)(x^2 + 9)

Answer:

[tex]2x(x^2-3)(x^2+9)[/tex]

Step-by-step explanation:

Given polynomial:

[tex]2x^5+12x^3-54x[/tex]

Factor out the common term [tex]2x[/tex]:

[tex]\implies 2x(x^4+6x^2-27)[/tex]

To factor the trinomial [tex]x^4+6x^2-27[/tex]:

[tex]\textsf{Let }u=x^2 \implies u^2+6u-27[/tex]

Factor the quadratic by finding two numbers that multiply to -27 and sum to 6:  9 and -3

Rewrite the middle term as the sum of these two numbers:

[tex]\implies u^2+9u-3u-27[/tex]

Factorize the first two terms and the last two terms separately:

[tex]\implies u(u+9)-3(u+9)[/tex]

Factor out the common term (u + 9):

[tex]\implies (u-3)(u+9)[/tex]

Substitute back [tex]u=x^2[/tex]:

[tex]\implies (x^2-3)(x^2+9)[/tex]

Therefore, the factored form of the given polynomial is:

[tex]\implies 2x(x^2-3)(x^2+9)[/tex]

The x-coordinate of the point (50, 55) is 50

A graph can be defined as a pictorial representation or a diagram that represents data or values.

The distance of point (50, 55) perpendicular to y-axis

Thus, perpendicular distance = 50 units

The x-coordinate of the point (50, 55) is 50

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Answer:

0.5 hours

Explanation:

Given:

Formula:

Solve:

The direct proportion is shown by table (B)

Direct proportion or direct variation is the relation between two quantities where the ratio of the two is equal to a constant value.

Here the second table shows the direct proportion relation.

This is because the ratio of x/y remain same.

4/2= 7/3.5= 10/5= 11/5.5= 15/7.5 = 2/1

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Answer:

4x82x929wjdjdejdjekke

[tex] \frac{1}{x + 2} + \frac{1}{x + 3} = \frac{2}{x + 9 } \\ \frac{(x + 3 ) + (x + 2)}{ {x}^{2} + 5x + 6 } = \frac{2}{x + 9} \\ \frac{2x + 5}{ {x}^{2} + 5x + 6 } = \frac{2}{x + 9} \\ 2( {x}^{2} + 5x + 6) = (x + 9)(2x + 5) \\ 2 {x}^{2} + 10x + 12 = 2 {x}^{2} + 5x + 18x + 45 \\ 2 {x}^{2} + 10x + 12 = 2 {x}^{2} + 23x + 45 \\ 2 {x}^{2} - 2 {x}^{2} + 10x - 23x + 12 - 45 = 0 \\ - 13x - 33 = 0 \\ - 13x = 33 \\ x = \frac{33}{ - 13} \\ x = - 2.54[/tex]

Hope it helps

Please give brainliest

Answer:

See below

Step-by-step explanation:

Period = 1 day      

Periodic interest = .045 / 365       (since there are 365 days in a year)

V(t) = 6000 *  ( 1 + .045/365)^t      where t is in days

V(t) = 6000 * (1+.045/365)^(t * 365)       where t is in years

Step-by-step explanation:

We can solve this kind of trigonometric problem

easily by the followings method :

Given :

cos^4a + cos^2a = 1

or, cos4^a = 1 - cos^2a

Therefore; cos^4a = sin^2a

Again,

To prove: cot^4a - cot^2a = 1

L.H.S = Cot^4a - cot^2a

= Cos^4a÷ sin^4a - Cos^2a ÷ sin^2a

= Sin^2a ÷ Sin^4a - cos^2a ÷ Sin^2a

= 1 ÷ Sin^2a - cos^2a ÷ Sin^2a

= 1 - cos^2a ÷ Sin^2a

= Sin^2a ÷ Sin^2a

= 1 = R.H.S proved.

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Answer:

(-4,-2)

x=-4

y=-2

Step-by-step explanation:

-3x-4y=20

3(x-10y=16)=3x-30y=48

The reason I multiplied 3 to the second equation is for when we add the equations together the x will cancel out.

  -3x-4y=20

+  3x-30y=48

  -34y=68

Divide -34 from both sides.

y=-2

To find x you need to plug in -2 for y into one of the equations.

x-10y=16

x-10(-2)=16

Remember a negative times a negative equals a positive.

x+20=16

Subtract 20 from both sides.

x=-4

Hope this helps!

If not, I am sorry.

Answer:

it takes 6 days for 1 machine...

The correct answer is option b which is the exponential grows at the same rate as the quadratic.

The complete question is given below with the graph attached:-

An exponential function and a quadratic function are graphed below. Which of the following is true of the growth rate of the functions over the interval

a. The exponential grows at half the rate of the quadratic.

b.The exponential grows at the same rate as the quadratic.

c.The exponential grows at twice the rate of the quadratic.

d.The exponential grows at four times the rate of the quadratic.

Inequality is the relationship between two expressions that are not equal, employing a sign such as ≠ “not equal to,” > “greater than,” or < “less than.”.

Given an exponential function, say f(x), such that f(0) = 1 and f(1) = 2 and a quadratic finction, say g(x), such that g(0) = 0 and g(1) = 1.

The rate of change of a function f(x) over an interval

a ≤ x ≤ b

is given by

[tex]\dfrac{f(b)-f(a)}{b-a}[/tex]

Thus, the rate of change (growth rate) of the exponential function, f(x) over the interval

0 ≤ x ≤ 1

is given by

[tex]\dfrac{f(1)-f(0)}{1-0}= \dfrac{2-1}{1}=1[/tex]

Similarly, the rate of change (growth rate) of the quadratic function, g(x) over the interval

0 ≤ x ≤ 1

is given by

[tex]\dfrac{g(1)-g(0)}{1-0}=\dfrac{1-0}{1}=1[/tex]

Therefore, the exponential grows at the same rate as the quadratic in the interval .

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Answer:it is o to 10/5 which is 1/

Step-by-step explanation:

Answer:

Step-by-step explanation:

Is there answer choices?

Answer:

Step-by-step explanation:

The last one: 79, 79, 79, 79, 79, 79, 79, 79, 79, 79 because there isn't any difference in the numbers.

S.D.: 0

Answer: 45

Step-by-step explanation: I used a graphing calculator and input all the equations and all the restraints, and I found that you can’t use the first equation since x has to be less than or equal to -3, and the question calls for x to be -2. So, in the second equation, there’s a point (-2, 16) and in the third equation there’s a point (1, -3). You know you have to use these two points since in the second equation, the restraint only lets you use numbers greater than -3 or less than 0, which cannot be one, and in the third equation, the restraint only lets you use numbers greater than 0, which can’t be -2. I hope that made some sense. So then, with substitution the equation would be 3(16)+(-3) which equals 45. 3(16)=48. 48-3+=45

The value obtained by doing the operation are ,a=0.11,b=11,c=(111/250)d=44,e=(1/8),f=(83/500),h=0.375,i=37.5%,j=(111/125),k=0.888,i=(777/1000),m=77%,n=0.2,o=20%,p=(777/1000),q=66.6%.

The quantity of anything is stated as though it were a fraction of a hundred. A fraction of 100 can be used to express the ratio.

The procedure of changing the basis of a number is known as number conversion. A binary number is made up of only two numbers: 1 and 0. A decimal number is made up of ten digits ranging from 0 to 9.

The values obtained from the table the decimal and the percentage form are solved as;

[tex]\frac{1}{9} \\\\ 0.3111 \\\\\ 0.1111 \times 100 \\\\ 11.11 \%[/tex]

The other missing values are;

a=0.11

b=11

c=(111/250)

d=44

e=(1/8)

f=(83/500)

h=0.375

i=37.5%

j=(111/125)

k=0.888

i=(777/1000)

m=77%

n=0.2

o=20%

p=(777/1000)

q=66.6%

Hence all the missing values are given above.

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