What is the length of CG?

The missing angles are listed below:

sin 27.818° = 21 / 45

tan 26.565° = 8 / 16

cos 57.607° = 15 / 28

tan 38.956° = 38 / 47

sin 44.901° = 12 / 17

cos 52.398° = 36 / 59

tan 19.477° = 29 / 82

sin 16.601° = 8 / 28

tan 10.008° = 9 / 51

cos 36.870° = 16 / 20

sin 35.538° = 60 / 84

cos 53.576° = 19 / 32

tan 40.101° = 64 / 76

cos 39.571 = 37 / 48

cos 69.605° = 23 / 66

In this question we must find the measure of the angle associated to the rational numbers seen in the statement. There are two methods to estimate such measures: (i) Drawing an equivalent right triangle and estimate the measure of the angle by definition of trigonometric functions, (ii) Using inverse trigonometric functions and numerical methods since they are trascendent variables.

Herein we decided to use the second method by reasons of rapidness and effectivity:

sin 27.818° = 21 / 45

tan 26.565° = 8 / 16

cos 57.607° = 15 / 28

tan 38.956° = 38 / 47

sin 44.901° = 12 / 17

cos 52.398° = 36 / 59

tan 19.477° = 29 / 82

sin 16.601° = 8 / 28

tan 10.008° = 9 / 51

cos 36.870° = 16 / 20

sin 35.538° = 60 / 84

cos 53.576° = 19 / 32

tan 40.101° = 64 / 76

cos 39.571 = 37 / 48

cos 69.605° = 23 / 66

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Answer: 27°

Step-by-step explanation:

The easiest way to find the measure of angle Z is to know that in a rhombus, the diagonals are perpendicular bisectors of each other. This means that all the four angles in the center are 90°.

Since we know that all angles in a triangle add up to 180°, we can set the sum of z, 63, and 90 to be 180 and solve for z.

[tex]z+63+90=180\\z+153=180\\z=27[/tex]

Hence, z is 27°.

The percentages for each outcome are given as follows:

a) 34.4%.

b) 29.4%.

The percentage of an amount a over a total amount b is given by a multiplied by 100% and divided by b, that is:

P = a/b x 100%

For this problem, we already have the percentages, hence we just combine them.

For item a, he can be late in two ways, one that happens 24.4% of the time(waking up late), and the other 10%(stuck in traffic), hence the percentage is of 34.4%, as 24.4% + 10% = 34.4%.

For item b, the percentage stuck in traffic is of 5%, hence the percentage that he will be late is 24.4% + 5% = 29.4%.

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The probability that the battery life is  20 ± 2 hours (between 18 and 22 hours) for each of the phones is:

Using the normalcdf function on a calculator, the probability that Phone C's battery would last between 18 and 22 hours can be found by inputing the function:

normalcdf (20, IE99, 18, 2)

The result is 0.158655 or 15.9%.

For Phone T, the function is:

normalcdf (20, IE99, 20, 3)

The result is 0.5 or 50%.

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Answer:

13

Step-by-step explanation:

→ Substitute 3 into 2x

2 × 3

→ Evaluate

f ( x ) = 6

→ Substitute x = 3 into 3x - 2

3 × 3 - 2

→ Evaluate

7

→ Find the sum of the 2 results

13

When x = 3, the value of (f + g)(x) is 13.

To evaluate (f + g)(x) when f(x) = 2x and g(x) = 3x - 2, we substitute the given functions into the expression (f + g)(x).

(f + g)(x) = f(x) + g(x)

Substituting the given functions:

(f + g)(x) = 2x + (3x - 2)

Simplifying the expression:

(f + g)(x) = 2x + 3x - 2

Combining like terms:

(f + g)(x) = 5x - 2

Now, to find the value of (f + g)(x) when x = 3, we substitute x = 3 into the expression:

(f + g)(3) = 5(3) - 2

Simplifying further:

(f + g)(3) = 15 - 2

(f + g)(3) = 13

Therefore, when x = 3, the value of (f + g)(x) is 13.

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Answer:

f(–5)=–4

Step-by-step explanation:

as it is defined

Answer:

[tex]-8x^2 - 4x + 29 \\[/tex]

Step-by-step explanation:

Second expression evaluates to:

[tex](5 + x -2x)^2 = (5 -x)^2 = (-x+5)^2 = x^2 + 2(-x)(5) +5^2 = x^2 -10x + 25[/tex]  (1)

For (1) We are using the rule [tex](a+b)^2 = a^2 +2ab + b^2\\\\[/tex]

Here [tex]a = -x, b = 5\\\\[/tex]

First expression evaluates to

[tex](3x)^2 -6x - 4 = 9x^2 -6x -4[/tex]    (2)

Subtract (2) from (1)

[tex]x^2 - 10x +25 - (9x^2 -6x -4) = x^2-9x^2 -10x - (-6x) +25 -(-4)\\\\= -8x^2 - 4x + 29 \\[/tex]

Answer: B

Step-by-step explanation:

Answer:

B

Step-by-step explanation:

a dream like imageeeeee

Given the measure of side a, b and c of the triangle, the measure of angle C to the nearest tenth is 104.5°.

From the law of cosines;

cosC = ( a² + b² - c² ) / 2ab

Where C is the angle C and a, b, and c are the three sides of the triangle.

Given the data in the question;

Using law of cosine;

cosC = ( a² + b² - c² ) / 2ab

We substitute the values into the equation.

cosC = ( 2² + 3² - 4² ) / ( 2 × 2 × 3 )

cosC = ( 4 + 9 - 16 ) / ( 12 )

cosC = -3 / 12

cosC = -0.25

We find the inverse of cosine.

C = cos⁻¹( -0.25 )

C = 104.477°

C = 104.5°

Given the measure of side a, b and c of the triangle, the measure of angle C to the nearest tenth is 104.5°.

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a. The velocity t = [tex]v = Ce_{n} (\frac{mo}{mo - kt} ) - gt[/tex]

b. v60 = 7164

mdv/dt = ck - mg

dv/dt = ck/m - mg/m

= ck/m - g

dv = [tex](\frac{ck}{Mo-Kt} -g)dv[/tex]

Integrate the two sides of the equation to get

v [tex]-\frac{ck}{k} e_{n} (Mo- kt)-gt+c[/tex]

[tex]v = Ce_{n} (\frac{mo}{mo - kt} ) - gt[/tex]

b. fuel accounts for 55% of the mass

So final mass after fuel is burned out is = 0.45

c=2500

g=9.8

t=60

v = -2500ln0.45 - 9.8 x 60

= 7752 - 588

= 7164

A rocket, fired from rest at time t = 0, has an initial mass of m0 (including its fuel). Assuming that the fuel is consumed at a constant rate k, the mass m of the rocket, while fuel is being burned, will be given by m0 - kt. It can be shown that if air resistance is neglected and the fuel gases are expelled at a constant speed c relative to the rocket, then the velocity of the rocket will satisfy the equation where g is the acceleration due to gravity.

dv dt m =ck - mg

(a) Find v(t) keeping in mind that the mass m is a function of t.

v(t) =

m/sec

(b) Suppose that the fuel accounts for 55% of the initial mass of the rocket and that all of the fuel is consumed at 60 s. Find the velocity of the rocket in meters per second at the instant the fuel is exhausted. [Note: Take g = 9.8 m/s² and c = 2500 m/s.]

v(60) =

m/sec [Round to nearest whole number]

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Answer:x>1

Step-by-step explanation:

Subract 11 from both sides

Answer:

7, 9

Step-by-step explanation:

Let x and y represent the two positive integers.

We can write x in terms of y. The question states "one positive integer is 5 less than twice another". Let "one positive integer" represent x.

Therefore:

[tex]x=2y-5[/tex]

We also know that the sum of the squares of the two positive integers is 130. Using this information, we can write another equation.

[tex]x^2+y^2=130[/tex]

These two equations form a system of equations.

[tex]x=2y-5[/tex]

[tex]x^2+y^2=130[/tex]

We can solve this system of equations by the substitution method. Using this method, we substitute [tex]2y-5[/tex] for [tex]x[/tex].

[tex](2y-5)^2+y^2=130[/tex]

Expand [tex](2y-5)^2[/tex] . Notice that [tex](a-b)^2=a^2-2ab+b^2[/tex] :

[tex](2y-5)^2=4y^2-20y+25[/tex]

[tex]4y^2+y^2-20y+25=130[/tex]

Subtract 25 from both sides

[tex]4y^2+y^2-20y=105[/tex]

Add like terms

[tex]5y^2-20y=105[/tex]

Divide both sides by 5

[tex]y^2-4y=21[/tex]

Subtract 21 from both sides

[tex]y^2-4y-21 = 0[/tex]

Factor the equation

[tex]y^2-7y+3y-21=0\\y(y-7)+3(y-7)=0\\(y-7)(y+3)=0\\y=7\text{ or} \ -3[/tex]

Since the question states that the two integers are positive, one of the integers is 7.

We can use this information to find the other integer.

[tex]x=2y-5\\x=2(7)-5\\x=14-5\\x=9[/tex]

[tex]CHECK:\\x^2+y^2=7^2+9^2=49+81=130 \Rightarrow Correct![/tex]

Therefore the two integers are 7 and 9.

The graph is shown in the attached image.

The following classification of quadratic equations is presented below:

In this problem we have quadratic equations in factored form, whose form is presented below:

y = a · (x - r₁) · (x - r₂)      (1)

Where r₁ and r₂ are the roots of the equation and a is the leading coefficient. A value of x is a root if and only if y is zero. Besides, we must located all the quadratic equations according to their roots.

x = - 2 and x = 3

h(x) = (x + 2) · (x - 3)

k(x) = - 3 · (x + 2) · (x - 3)

x = 2 and x = - 3

g(x) = 8 · (x + 3) · (x - 2)

m(x) = (x + 3) · (x - 2)

Neither

f(x) = 3 · (x - 1) · (x + 2)

j(x) = (x - 2) · (x - 3)

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Answer:

n=11

Step-by-step explanation:

(n+4)/10 = (n-8)/2

We can solve using cross products

(n+4) * 2 = 10 * ( n-8)

Distribute

2n+8 = 10n -80

Subtract 2n from each side

2n+8-2n = 10n-80-2n

8 = 8n-80

Add 80 to each side

8+80= 8n-80+80

88 = 8n

Divide each side by 8

88/8 = 8n/8

11 = n

Answer: n=11

Step-by-step explanation:

(n+4)/10 = (n-8)/2

multiply both sides by 10

n+4 = (n-8)/2*10

cancel

n+4=(n-8)*5

n+4=5(n-8)

multiply

n+4=5n-40

subtract both sides by 5n

n-5n+4=-40

subtract both sides by 4

n-5n=-40-4

subtract the like terms

-4n=-44

cancel the negatives

4n=44

divide each side by 4

n=11

Considering it's discriminant, it is found that:

A. The classmate is wrong, as the discriminant is of zero, hence the equation has one solution.

B. The quadratic equation has 1 x-intercept.

A quadratic equation is modeled by:

y = ax^2 + bx + c

The discriminant is:

[tex]\Delta = b^2 - 4ac[/tex]

The solutions are as follows:

In this problem, the equation is:

y = 9x² - 6x + 1.

The coefficients are a = 9, b = -6 and c = 1, hence the discriminant is:

[tex]\Delta =(-6)^2 - 4(9)(1) = 36 - 36 = 0[/tex]

Since the discriminant is zero, the classmate is wrong, as it means that the equation has one solution = one x-intercept.

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Answer:

  order: 2, 3, 1

Step-by-step explanation:

Reasonableness checks and your knowledge of integers and fractions will help you solve this. The offered questions are intended to help you think this through.

For an output of -31, the machine (x -2)² cannot possibly be last. Its output can only be positive.

  machine 2, (x-2)², cannot be last

Also, machine 3 cannot be last. For the output to be -31, the input to machine 3 must be -1/31. Neither of the other machines can produce a fraction with the inputs they might receive.

For an input of x=0, the machine 1/x cannot possibly be first. 1/0 is undefined.

The other two machines will give the following outputs for an input of 0:

  machine 1: 4(0) -32 = -32

  machine 2: (0 -2)² = 4

It is unlikely that machine 1 will be first, because the other two machines cannot do anything useful with -32 as an input.

  machine 3, 1/x, cannot be first

The reasoning of part (a) tells you the last machine must be machine 1. The reasoning of part (b) tells you the first machine cannot be 3, so must be 2. The order of the machines must be ...

Using the addition rule of the Sine function and the Cosine function, we obtain [tex]\cos\dfrac{x+y}{2}=\dfrac{b}{\sqrt{a^2+b^2}}[/tex].

Given that

[tex]\sin x + \sin y = a\\\cos x + \cos y = b[/tex]

Then using the above formulas, we get:

[tex]2\sin\frac{x+y}{2}\cos\frac{x-y}{2}=a[/tex]       (1)

[tex]2\cos\frac{x+y}{2}\cos\frac{x-y}{2}=b[/tex]       (2)

Dividing the equation (1) by (2), we get:

[tex]\dfrac{\sin\dfrac{x+y}{2}}{\cos\frac{x-y}{2}}=\dfrac{a}{b}\\\Longrightarrow \tan\dfrac{x+y}{2}=\dfrac{a}{b}[/tex]             (3)

Now, we know that  [tex]\cos\theta=\dfrac{1}{\sqrt{1+\tan^2\theta}}[/tex].

Thus, using the above formula, we get from (3):

[tex]\cos\dfrac{x+y}{2}=\dfrac{1}{\sqrt{1+\tan^2\dfrac{x+y}{2}}}\\\Longrightarrow \cos\dfrac{x+y}{2}=\dfrac{1}{\sqrt{1+\dfrac{a^2}{b^2}}}\\\Longrightarrow \cos\dfrac{x+y}{2}=\dfrac{b}{\sqrt{a^2+b^2}}[/tex]

Therefore, using the addition rule of the Sine function and the Cosine function, we obtain [tex]\cos\dfrac{x+y}{2}=\dfrac{b}{\sqrt{a^2+b^2}}[/tex].

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Answer:

B

Step-by-step explanation:

• the opposite sides of a parallelogram are congruent , then

ZY = WX , that is

5x - 6 = 4x + 1 ( subtract 4x from both sides )

x - 6 = 1 ( add 6 to both sides )

x = 7

• the diagonals bisect each other , then

ZT = TX = 2x = 2 × 7 = 14

The adjusting entry to registering the Rent revenue shall be as pursues:

Unearned Rent Revenue Debit $6,000

Rent Revenue         Credit      $6,000

(being the adjustment made for Rent Revenue earned)

Adjusting entry to record the Rent Revenue are as follows:

It exists shown that on November 1, the company rented space to another tenant. A check in the amount of $9,000, symbolizing three months' rent in advance, stood obtained from the tenant on that date. The payment stood registered with a credit to the unearned rent revenue account.

Now on December 31, we must organize the adjusting access to record the Rent Revenue for the period (Nov. 1 to Dec. 31) which exists for two months. The Rent Revenue for two months shall be

9000 [tex]*[/tex] 2/3 = $6,000

Therefore the adjusting entry to registering the Rent revenue shall be as pursues:

Unearned Rent Revenue Debit $6,000

Rent Revenue         Credit      $6,000

(being the adjustment made for Rent Revenue earned)

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The complete question is:

On November 1, the company rented space to another tenant. A check for $9,000, representing three months' rent in advance, was received from the tenant on that date. The payment was recorded with a credit to the unearned rent revenue account. Complete the necessary adjusting entry for December 31 by selecting the account names and dollar amounts from the drop-down menus.

Answer:

$279

Step-by-step explanation:

100% = $930

30% = ???

(30/100) × 930

=$279

Since we know the sequence is quadratic, we expect the [tex]n[/tex]-th term to have the general form

[tex]x_n = an^2 + bn + c[/tex]

Plug in the first 3 known values of the sequence to form a system of equations.

[tex]x_1 = 3 = a + b + c[/tex]

[tex]x_2 = 12 = 4a + 2b + c[/tex]

[tex]x_3 = 25 = 9a + 3b + c[/tex]

Eliminating [tex]c[/tex] gives

[tex](4a + 2b + c) - (a + b + c) = 12 - 3 \implies 3a + b = 9[/tex]

[tex](9a + 3b + c) - (a + b + c) = 25 - 3 \implies 8a + 2b = 22 \implies 4a + b = 11[/tex]

Eliminating [tex]b[/tex] gives

[tex](4a + b) - (3a + b) = 11 - 9 \implies a = 2[/tex]

Solving for [tex]b,c[/tex], we get

[tex]4a + b = 11 \implies b = 11-4\cdot2 = 3[/tex]

[tex]a+b+c=3 \implies c=3-2-3 = -2[/tex]

So, the [tex]n[/tex]-th term of the sequence is

[tex]x_n = \boxed{2n^2 + 3n - 2}[/tex]

Answer: 6.25

Step-by-step explanation: Here, you want to find how much of 6,298 pounds of food each person got. You can do this by dividing. 6,298/1,007 should get you the answer because it shows how the 6,298 pounds were distributed amongst the 1,007 people.

Answer:

hey can you show me the pic because the question doesn't make any sense

Answer:

  4.346%

Step-by-step explanation:

The points and origination fee are assumed to be capitalized, so the loan payment amount is based on the total of the original loan amount and those fees. The APR is calculated as the rate that would be charged on the original loan value to produce that payment.

The loaned amount is the sum of the original loan value and the added points.

  P = $513,000×(1 +1.4% +0.65%) = $523,516.50

The monthly payment on this amount is given by the amortization formula ...

  A = P(r/12)/(1 -(1 +r/12)^(-12t))

where r=0.04175, the annual interest rate, t=30, the period in years

  A = $523,516.50(0.04175/12)/(1 -(1 +0.04175/12)^-360)) ≈ $2552.45

There is no formula for calculating the APR. It must be developed iteratively or graphically. The attached calculator images show a couple of different ways the value can be found.

The payment of $2552.45 on a loan value of $513,000 represents an APR of 4.346%.

Answer:

Step-by-step explanation:

Let r(y)=x.

[tex]x=\frac{2}{3}(4y-5)\\\\\frac{3}[2}x=4y-5\\\\\frac{3}{2}x+5=4y\\\\y=\frac{3}{8}x+\frac{5}{4}\\\\r^{-1}(x)=\frac{3}{8}x+\frac{5}{4}[/tex]

Using the monthly payment formula, we have that:

It is given by:

[tex]A = P\frac{\frac{r}{12}\left(1 + \frac{r}{12}\right)^n}{\left(1 + \frac{r}{12}\right)^n - 1}[/tex]

In which:

The down payment is 20% of $175,000, hence:

D = 0.2 x 175,000 = $35,000.

Hence the parameters are:

P = 175,000 - 35,000 = 140,000, r/12 = 0.055/12 = 0.004583, n = 30 x 12 = 360.

Then the monthly payment is found as follows:

[tex]A = P\frac{\frac{r}{12}\left(1 + \frac{r}{12}\right)^n}{\left(1 + \frac{r}{12}\right)^n - 1}[/tex]

[tex]A = 140,000\frac{0.004583(1.004583)^{360}}{(1.004583)^{360} - 1}[/tex]

A = $795.87.

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