2 1/12 divided 2 5/8 fractions

Answer:

The student took the numbers for the factors 5, -4 for the zeros  instead of solving the equation

The zeros are -5, 4

Step-by-step explanation:

x^2 +x - 20=0

What 2 number multiply to -20 and add to 1

5*-4 = -20

5+-4 = 1

(x+5)(x-4) =0

Using the zero product property

x+5 = 0  x-4 =0

x=-5  x=4

Answer:

Solution given:

equation is:

x²+x-20=0

doing middle term factorisation

note:we need to get 1 while subtracting factor of the product of constant and coefficient of x².

20*1=20=2*5*2*1

we get 1 while subtracting 5-2*2=5-4

now substitute value 5-4 at coefficient of x

we get

x²+(5-4)x-20=0

now

distribute

x²+5x-4x-20=0

taking common from each two term

x(x+5)-4(x+5)=0

again taking common (x+5) and keeping remaining at another bracket

(x+5)(x-4)=0

either

x+5=0

x=-5

or

x-4=0

x=4

Error is:

x= 4 not -4

x=-5 not 5.

Answer:

40 min

Step-by-step explanation:

his sister can do it twice as fast so the boy does 1 and the sister does 2. he will do 1/3 of the work and it will take 1/3 of the time

Answer:

90 mins

Step-by-step explanation:

1 hour =60 mins

1/2 hour = 60/2=30

so, 60 +30=90

Answer:

A) U is called an upper triangular matrix because all entries below the principal diagonal element are zeros ( 0 )   since Uij = 0 if i >j  also

L is a lower triangular matrix because all entries above the principal diagonal element are zero ( 0 )

B) sum of two upper triangular matrices = upper triangular matrix.

C) product of two upper triangular matrices = upper triangular matrix

Step-by-step explanation:

A) U is called an upper triangular matrix because all entries below the principal diagonal element are zeros ( 0 )   since Uij = 0 if i >j  also

L is a lower triangular matrix because all entries above the principal diagonal element are zero ( 0 ) since Lij = 0  if i < j

B) To prove that sum of two upper triangular matrices

attached below

C) Prove or disprove that product of two upper triangular matrices is an upper triangular matrix

attached below

Answer:

Option B. M = Log 10000

Step-by-step explanation:

From the question given above, we were told that the intensity (I) is 10000 times that of the reference earthquake (I₀).

Thus, we can obtain the magnitude (M) of the earthquake as follow:

Let the reference earthquake (I₀) = A

Then, the intensity (I) = 10000 × A

M = Log(I/I₀)

M = Log(10000A / A)

M = Log 10000

Thus, option B gives the right answer to the question.

Answer:

[tex]sin\left(90\right)/x=sin\left(25\right)/16[/tex]

x = 37.85

Step-by-step explanation:

Answer:

a) 25/35

b) 30/42

Step-by-step explanation:

a)

Variable x = denominator if numerator is 25

5/7 = 25/x

5 × x = 7 × 25

5x = 175

x = 35

b)

Variable y = numerator if denominator is 42

5/7 = y/42

5 × 42 = 7 × y

210 = 7y

30 = y

25/35

30/42

To get 25/35 multiply by 5

To get 30/42 multiply by 6

Answer:

c. [0.25 0.75] ^T

Step-by-step explanation:

Bernoulli distribution is used to identify number of successes and failures in the selected sample. In the given problem Ber distribution trial is 0.25. There will be categorical distribution of 0.75 and the trial will be done on parameter vector.

Answer:

2x, a linear pair

Step-by-step explanation:

2/5x/3/4 = 7/4 = 1 3/4

Answer:

Step-by-step explanation:

Begin the solution by squaring both sides of the given equation.  We get:

(3x - 4)^2 = 2x^2 - 2x + 2, or:

9x^2 - 24x + 16 = 2x ^2 - 2x + 2

Combining like terms results in:

7x^2 - 22x + 14 = 0

and the coefficients are a = 7, b = -22, c = 14, so that the discriminant of the quadratic formula, b^2 - 4ac becomes (-22)^2 - 4(7)(14) = 92

According to the quadratic formula, the solutions are

       -b ± √discriminant           -(-22) ± √92             22 ± √92

x = ------------------------------- = ----------------------- = ------------------------

                   2a                                   14                            14

Solution :

Given data :

20.1     33.5     21.7      58.4     23.2     110.8     30.9

24.0    74.8     60.0

n = 10

Range : Arranging from lowest to highest.

20.1,   21.7,   23.2,    24.0,   30.9,    33.5,    58.4,    60.0,    74.8,   110.8

Range = low highest value - lowest value

           = 110.8 - 20.1

           = 90.7

Mean = [tex]$\frac{\sum x}{n}$[/tex]

         [tex]$=\frac{20.1+21.7+23.2+24.0+30.9+33.5+58.4+60.0+74.8+110.8}{10}$[/tex]

          [tex]$=\frac{457.4}{10}$[/tex]

         [tex]$=45.74$[/tex]

Sample standard deviation :

[tex]$S=\sqrt{\frac{1}{n-1}\sum(x-\mu)^2}$[/tex]

[tex]$S=\sqrt{\frac{1}{10-1}(20.1-45.74)^2+(21.7-45.74)^2+(23.2-45.74)^2+(24.0-45.74)^2+(30.9-45)^2}$[/tex]  

      [tex]\sqrt{(33.5-45.74)^2+(58.4-45.74)^2+(60.0-45.74)^2+(74.8-45.74)^2+(110.8-45.74)^2}[/tex]

[tex]$S=\sqrt{\frac{1}{9}(657.4+577.9+508.0+472.6+220.2+149.8+160.2+203.3+844.4+4232.8)}$[/tex][tex]$S=\sqrt{\frac{1}{9}(8026.96)}$[/tex]

[tex]$S=\sqrt{891.88}$[/tex]

S = 29.8644

Variance = [tex]S^2[/tex]

               [tex]=(29.8644)^2[/tex]

               = 891.8823

Answer:

m = 6

c = 0

General Formulas and Concepts:

Algebra I

Slope-Intercept Form: y = mx + c

Step-by-step explanation:

Step 1: Define

y = 6x

↓ Compare to Slope-Intercept Form

Slope m = 6

y-intercept c = 0

9514 1404 393

Answer:

  (-1, -3)

Step-by-step explanation:

Each coordinate is multiplied by the dilation factor when dilation is centered at the origin.

  (1/4)(-4, -12) = (-1, -3) . . . . the image of the given point

Answer:

Step-by-step explanation:

Functions are always relations, but not every relation is a function. This passes the vertical line test so it is a function. Since it's not a line, it's not linear. B is your choice.

Answer:

x = -2

Step-by-step explanation:

the line goes through -2 on the x-axis

the line also goes through -3 on the y-axis

============================================

Explanation for part (a)

All distances mentioned are in feet.

We'll have a right triangle which allows us to apply the pythagorean theorem. Refer to the diagram below.

a^2+b^2 = c^2

x^2+y^2 = z^2

Apply the derivative to both sides with respect to t. We'll use implicit differentiation and the chain rule.

[tex]x^2+y^2 = z^2\\\\\frac{d}{dt}[x^2+y^2] = \frac{d}{dt}[z^2]\\\\\frac{d}{dt}[x^2]+\frac{d}{dt}[y^2] = \frac{d}{dt}[z^2]\\\\2x*\frac{dx}{dt}+2y*\frac{dy}{dt}=2z*\frac{dz}{dt}\\\\x*\frac{dx}{dt}+y*\frac{dy}{dt}=z*\frac{dz}{dt}\\\\[/tex]

Now we'll plug in (x,y,z) = (4000,3000,5000). The x and y values are given. The z value is found by use of the pythagorean theorem. Ie, you solve 4000^2+3000^2 = z^2 to get z = 5000. Or you could note that this is a scaled copy of the 3-4-5 right triangle.

We know that dx/dt = 0 because the horizontal distance, the x distance, is not changing. The rocket is only changing in the y direction. Or you could say that the horizontal speed is zero.

The vertical speed is dy/dt = 800 ft/s and it's when y = 3000. It's likely that dy/dt isn't the same value through the rocket's journey; however, all we care about is the instant when y = 3000.

Let's plug all that in and isolate dz/dt

[tex]4000*0+3000*800=5000*\frac{dz}{dt}\\\\2,400,000=5000*\frac{dz}{dt}\\\\\frac{dz}{dt} = \frac{2,400,000}{5000}\\\\\frac{dz}{dt} = 480\\\\[/tex]

At the exact instant that the rocket is 3000 ft in the air, the distance between the camera and the rocket is changing by an instantaneous speed of 480 ft/s.

-----------------------------------------------------------------------

Explanation for part (b)

Again, refer to the diagram below.

We have theta (symbol [tex]\theta[/tex]) as the angle of elevation. As the rocket's height increases, so does the angle theta.

We can tie together the opposite side y with the adjacent side x with the tangent function of this angle.

[tex]\tan(\text{angle}) = \frac{\text{opposite}}{\text{adjacent}}\\\\\tan(\theta) = \frac{y}{x}[/tex]

Like before, we'll apply implicit differentiation. This time we'll use the quotient rule as well.

[tex]\tan(\theta) = \frac{y}{x}\\\\\frac{d}{dt}[\tan(\theta)] = \frac{d}{dt}\left[\frac{y}{x}\right]\\\\\sec^2(\theta)*\frac{d\theta}{dt} = \frac{\frac{dy}{dt}*x - y*\frac{dx}{dt}}{x^2}\\\\\sec^2(\theta)*\frac{d\theta}{dt} = \frac{800*4000 - 3000*0}{4000^2}\\\\\sec^2(\theta)*\frac{d\theta}{dt} = \frac{3,200,000}{16,000,000}\\\\\sec^2(\theta)*\frac{d\theta}{dt} = \frac{32}{160}\\\\\sec^2(\theta)*\frac{d\theta}{dt} = \frac{1}{5}\\\\[/tex]

Let's take a brief detour. We'll return to this later. Recall earlier that [tex]\tan(\theta) = \frac{y}{x}\\\\[/tex]

If we plug in y = 3000 and x = 4000, then we end up with [tex]\tan(\theta) = \frac{3}{4}\\\\[/tex] which becomes [tex]\tan^2(\theta) = \frac{9}{16}[/tex]

Apply this trig identity

[tex]\sec^2(\theta) = 1 + \tan^2(\theta)[/tex]

and you should end up with [tex]\sec^2(\theta) = 1+\frac{9}{16} = \frac{25}{16}[/tex]

So we can now return to the equation we want to solve

[tex]\sec^2(\theta)*\frac{d\theta}{dt} = \frac{1}{5}\\\\\frac{25}{16}*\frac{d\theta}{dt} = \frac{1}{5}\\\\\frac{d\theta}{dt} = \frac{1}{5}*\frac{16}{25}\\\\\frac{d\theta}{dt} = \frac{16}{125}\\\\\frac{d\theta}{dt} = 0.128\\\\[/tex]

This means that at the instant the rocket is 3000 ft in the air, the angle of elevation theta is increasing by 0.128 radians per second.

This is approximately 7.334 degrees per second.

The distance from the television camera to the rocket is changing at 480 ft/s while the camera's angle of elevation is changing at 0.128 rad/s

Let x represent the distance from the camera to the rocket and let h represent the height of the rocket.

a)

[tex]x^2=h^2+4000^2\\\\2x\frac{dx}{dt}=2h\frac{dh}{dt} \\\\x\frac{dx}{dt}=h\frac{dh}{dt} \\\\At \ h=3000\ ft, \frac{dh}{dt}=800\ ft/s;\\x^2=3000^{2} +4000^2\\x=5000\\\\\\x\frac{dx}{dt}=h\frac{dh}{dt} \\\\5000\frac{dx}{dt}=3000*800\\\\\frac{dx}{dt}=480\ ft/s[/tex]

b)

[tex]tan(\theta)=\frac{h}{4000} \\\\h=4000tan(\theta)\\\\\frac{dh}{dt}=4000sec^2(\theta)\frac{d\theta}{dt} \\\\\\At\ h=3000\ ft;\\\\tan\theta = \frac{3000}{4000}=\frac{3}{4} \\\\sec^2(\theta)=1+tan^2(\theta)=1+(\frac{3}{4})^2=\frac{25}{16} \\\\\\\frac{dh}{dt}=4000sec^2(\theta)\frac{d\theta}{dt}\\\\800=4000*\frac{25}{16}* \frac{d\theta}{dt}\\\\\frac{d\theta}{dt}=0.128\ rad/s[/tex]

Hence, The distance from the television camera to the rocket is changing at 480 ft/s while the camera's angle of elevation is changing at 0.128 rad/s

Find out more at: https://brainly.com/question/1306506

Step-by-step explanation:

the graph will be compressed vertically

Answer:

The zeros of this function would be: x = 4 and x = 6, assuming that option got caught off while you were taking a picture.

Step-by-step explanation:

When they're asking for the zeros of this type of function, where is forms this kind of U-shape or also known as a quadratic equation, they're asking what the x-value is when y = 0, or when the line of the function touches the x-axis. Notice that it happens when x = 4 and when x = 6.

In short, it's asking what the x-value is of the points of the function when it intersects the x-axis. Hopefully my explanation wasn't too confusing. Good luck on the rest of the quiz!

9514 1404 393

Answer:

  A = 108.78°

  a = 11 mi

  b = 7 mi

Step-by-step explanation:

The sum of angles in a triangle is 180°, so the third angle is ...

  A = 180° -38.71° -32.51°

  A = 108.78°

__

The remaining sides can be found from the law of sines.

  a/sin(A) = c/sin(C)

  a = sin(A)·c/sin(C) ≈ 0.946762 × 11.163896

  a ≈ 11 mi

  b = sin(B)·11.163896 ≈ 0.625379 × 11.163896

  b ≈ 7 mi

Answer:

Yes 7i is the answer

Step-by-step explanation:

they are equivalent.

Answer:

[tex]9(7x - 2) = 5(10x - 1) \\ \\ 63x - 18 = 50x - 5 \\ \\ 63x - 50x = - 5 + 18 \\ \\ 13x = - 5 + 18 \\ \\ 13x = 13 \\ \\ x = \frac{13}{13} \\ \\ x = 1[/tex]

Answer:

[tex]9(7x - 2) = 5(10x - 1) \\ 9 \times 7x - 9 \times 2 = 5 \times 10x - 5 \\ 63x - 18 = 50x - 5 \\ 63x - 50x = - 5 + 18 \\ 13x = 13 \\ x = 1 \\ thank \: you[/tex]

Answer:

(m-6+2root3)(m-6-2root3)

Step-by-step explanation:

m^2 - 12m +36 -12

= (m-6)^2 - 12

= (m-6+2root3)(m-6-2root3)[root 12 = 2root3]

Answer:

The answer is "21%".

Step-by-step explanation:

[tex]\to 26 + 20 + 29 + 49 = 124\\\\=\frac{26\ \text{brown times}}{124 \ \text{total times}} \\\\= \frac{26}{ 124} \\\\ = 0.209677 \\\\[/tex]

Calculating the percentage:

[tex]= 20.9677 \times 100\\\\=20.9677\% \approx 21\%[/tex]

1,4,9,16,25,36,49,........

7th pattern =49.....

Answer:

1, 4, 9, 16, 25, 36, 49…................the 7 the pattern is 49

Answer:

x=0,5/2

Step-by-step explanation:

Answer:

18 dollars

Step-by-step explanation:

sunglasses = 6 dollars

sandals = 3 * sunglasses

             = 3 * 6 dollars

             = 18 dollars

Step-by-step explanation:

14.99 - 11.53= 3.46+1.95 =4.41

Answer:

B. x = 2.77

Step-by-step explanation:

3^x = 21

You first look for a base for 21 that is 3 to the power of something.

21 = 3^2.77

So 3^x = 2^2.77

They have the same base so

x= 2.77

Answer:

$7800

Step-by-step explanation:

1. Principal x interest x rate

So: $7500 + 4% (0.04) x 1 year = $300

2. Interest + Principal

So: $7500 + $300

Mrs Lee had $7800 in the bank.

Answer:

the ratio of lengths of the two rods, Aluminum to Invar is 11.27

Step-by-step explanation:

coefficient of linear expansion of aluminum, [tex]\alpha _{Al} = 23 \times 10^{-6} /K[/tex]

Coefficient of linear expansion of Invar, [tex]\alpha _{Iv} = 1.2 \times 10^{-6}/K[/tex]

Linear thermal expansion is given as;

[tex]\Delta L = L_0 \times \alpha\times \Delta T\\\\where;\\\\L_0 \ is \ the \ original \ length \ of \ the \ metal\\\\\Delta L \ is \ the \ increase \ in \ length[/tex]

The increase in length of Invar is given as;

[tex]\Delta L_{Iv} = L_0_{Iv} \times \alpha _{Iv}\times \Delta T_{Iv}[/tex]

The increase in length of the Aluminum;

[tex]\Delta L_{ Al} = L_0_{Al} \times \alpha _{Al} \times \Delta T_{Al}\\\\from \ the\ given \ question, \ the \ relationship \ between \ the \ rods \ is \ given \ as\\\\ L_0_{Al} \times \alpha _{Al} \times \frac{1}{3} \Delta T_{Iv}= 2( L_0_{Iv} \times \alpha _{Iv} \times \Delta T_{Iv})\\\\ L_0_{Al} \times \alpha _{Al} \times \Delta T_{Iv}= 6( L_0_{Iv} \times \alpha _{Iv} \times \Delta T_{Iv})\\\\ L_0_{Al} \times \alpha _{Al} \times \Delta T_{Iv} = 6L_0_{Iv} \times 6\alpha _{Iv} \times 6 \Delta T_{Iv}\\\\[/tex]

[tex]\frac{L_0_{Al}}{6L_0_{Iv} } = \frac{6\alpha _{Iv} \ \times \ 6 \Delta T_{Iv}}{\alpha _{Al} \ \times \ \Delta T_{Iv}} \\\\\frac{L_0_{Al}}{6L_0_{Iv} } = \frac{6\alpha _{Iv} \ \times \ 6}{\alpha _{Al} \ } \\\\\frac{L_0_{Al}}{L_0_{Iv} } = 6^3(\frac{\alpha _{Iv} }{\alpha _{Al} } )\\\\\frac{L_0_{Al}}{L_0_{Iv} } = 6^3(\frac{1.2 \times 10^{-6} }{23\times 10^{-6} } )\\\\\frac{L_0_{Al}}{L_0_{Iv} } = 6^3(\frac{1.2}{23} )\\\\\frac{L_0_{Al}}{L_0_{Iv} } = \frac{259.2}{23} \\\\\frac{L_0_{Al}}{L_0_{Iv} } = 11.27[/tex]

Therefore, the ratio of lengths of the two rods, Aluminum to Invar is 11.27

The ratio of the lengths of the two rods which is length of aluminum to length of Invar rod is; 11.27

Formula for linear thermal expansion is;

ΔL = L × α × ΔT

Where;

ΔL is change in original length

L is original length

α is coefficient of linear expansion

ΔT is change in temperature

We are told that increase in length of aluminum rod is twice the increase in length of an Invar rod with only a third of the temperature increase.

Thus;

ΔL = 2ΔL

ΔT for the aluminum rod = ⅓ΔT for the Invar rod.

Thus, we have;

L_al × α_al × ⅓ΔT = 2L_in × 2α_in × 2ΔT

ΔT will cancel out to give;

⅓(L_al × α_al) = 2L_in × 2α_in × 2

Multiply both sides by 3 to get;

(L_al × α_al) = 6L_in × 6α_in × 6

From online tables, the linear coefficient of expansion of aluminum is 23 × 10^(-6) C¯¹

While the coefficient of thermal expansion for Invar rod is 1.2 × 10^(-6) K¯¹

Thus;

L_al × 23 × 10^(-6) = 6L_in × (6 × 1.2 × 10^(-6)) × 6

L_al/L_in = (6 × 6 × 1.2 × 10^(-6) × 6)/(23 × 10^(-6))

L_al/L_in = 11.27

Read more on coefficient of linear expansion at; https://brainly.com/question/6985348