Draw a line between the points representing wicket 3 and wicket 9 on the diagram. Then, find the
area of the rectangle. Show your work and round your answer to the nearest hundredth.

Answer:

[tex]\lim_{x\rightarrow +\infty } x-1+\frac{1+ln^{2}x}{x} = + \infty[/tex]

[tex]\lim_{x\rightarrow 0 } x-1+\frac{1+ln^{2}x}{x} = + \infty[/tex]

Step-by-step explanation:

[tex]\lim_{x\rightarrow +\infty } x-1+\frac{1+ln^{2}x}{x}[/tex]

[tex]= [\lim_{x\rightarrow +\infty } (x-1)]+[ \lim_{x\rightarrow +\infty } (\frac{1+ln^{2}x}{x})][/tex]

=          +∞             +          0

= +∞

[tex]\lim_{x\rightarrow +\infty } x-1+\frac{1+ln^{2}x}{x}[/tex]

[tex]= [\lim_{x\rightarrow 0 } (x-1)]+[ \lim_{x\rightarrow 0 } (\frac{1+ln^{2}x}{x})][/tex]

=            -1             +      +∞  

= +∞  

by solving a system of equations, we conclude that there are 7 students with blue eyes in the class.

We know that there are 40 students in the first-grade class, and we also know that:

Let's define the variables.

We know that:

H = 3*B

We also know that there are 3 students that have both blue eyes and blond hair.

And there are 15 students with none of these traits, so:

40 - 15 = 25

There are 25 students that have blue eyes, blond hair, or both.

Because we know that 3 students have both traits, we can write:

B + H - 3  =25

(Where we subtract 3 because we don't want to add these students twice).

Then we created a system of two equations:

H = 3*B

B + H - 3  =25

Replacing the first equation into the second one, we get:

B + (3*B) - 3 = 25

4*B = 25 + 3 = 28

B = 28/4 = 7

Then, by solving that system of equations, we conclude that there are 7 students with blue eyes in the class.

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The inequalities in the system of inequalities are y > x² – 2x – 3 and y > 3x + 4

The graph that completes the question is added as an attachment

The parabola

It has a vertex of (1,  -4)

So, we have:

y = a(x - 1)^2 - 4

It passes through (3, 0).

So, we have:

0 = a(3 - 1)^2 - 4

This gives

4a - 4 = 0

Solve for a

a = 1

Substitute a = 1 in y = a(x - 1)^2 - 4

y = (x - 1)^2 - 4

Expand

y = x^2 - 2x + 1 - 4

y = x^2 - 2x - 3

The inner part is shaded.

So, we have:

y > x^2 - 2x - 3

The linear equation

It passes through (0, 4).

So, we have:

y = mx + 4

It passes through (1, 7).

So, we have:

7 = m + 4

Solve for m

m = 3

Substitute m = 3 in y = mx + 4

y = 3x + 4

The upper part is shaded.

So, we have:

y > 3x + 4

Hence, the inequalities in the system of inequalities are y > x² – 2x – 3 and y > 3x + 4

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The probability of being dealt no pair is 0.5011 or 50.11% if the tandard deck of playing cards has 52 cards-4 suits.

It is defined as the ratio of the number of favourable outcomes to the total number of outcomes, in other words, the probability is the number that shows the happening of the event.

It is given that:

In 5-card poker, find the probability of being dealt the following hand. Refer to the table.

From the table:

Total number of outcomes = 2598960

Total number of favourable outcomes = 1302540

The probability of being dealt no pair:

P(no pair) = 1302540/2598960

P(no pair) = 0.5011

In percentage:

P(no pair) = 50.11%

Thus, the probability of being dealt no pair is 0.5011 or 50.11% if the tandard deck of playing cards has 52 cards-4 suits.

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Answer:

0.7

Step-by-step explanation:

Each increment is 0.1, so count the approximate number of increments needed to get from the orange circle on the left and the one on the right and multiply by 0.1 to get the difference.

Answer:

The ratio is 3 : 2.

Step-by-step explanation:

42 cm : 28 cm

= 21 : 14 (Divide both sides by 2)

= 3 : 2 (Divide both sides by 7)

The ratio between the length and width of the rectangle is 3:2.

We have,

The concept used here is the concept of ratio.

In mathematics, a ratio is a comparison of two quantities.

It expresses how many times one quantity contains another or how many times it is greater or smaller than the other quantity.

The ratio between the length and width of the rectangle can be found by dividing the length by the width:

Ratio = Length / Width

= 42 cm / 28 cm

= 7 x 6 / 7 x 4

Cancel out the common factor.

= 6/4

= 2 x 3 / 2 x 2

Cancel out the common factor.

= 3/2

Thus,

The ratio between the length and width of the rectangle is 3:2.

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Using the quadratic formula, we solve for [tex]z^2[/tex].

[tex]z^4 - 5(1+2i) z^2 + 24 - 10i = 0 \implies z^2 = \dfrac{5+10i \pm \sqrt{-171+140i}}2[/tex]

Taking square roots on both sides, we end up with

[tex]z = \pm \sqrt{\dfrac{5+10i \pm \sqrt{-171+140i}}2}[/tex]

Compute the square roots of -171 + 140i.

[tex]|-171+140i| = \sqrt{(-171)^2 + 140^2} = 221[/tex]

[tex]\arg(-171+140i) = \pi - \tan^{-1}\left(\dfrac{140}{171}\right)[/tex]

By de Moivre's theorem,

[tex]\sqrt{-171 + 140i} = \sqrt{221} \exp\left(i \left(\dfrac\pi2 - \dfrac12 \tan^{-1}\left(\dfrac{140}{171}\right)\right)\right) \\\\ ~~~~~~~~~~~~~~~~~~~~= \sqrt{221} i \left(\dfrac{14}{\sqrt{221}} + \dfrac5{\sqrt{221}}i\right) \\\\ ~~~~~~~~~~~~~~~~~~~~= 5+14i[/tex]

and the other root is its negative, -5 - 14i. We use the fact that (140, 171, 221) is a Pythagorean triple to quickly find

[tex]t = \tan^{-1}\left(\dfrac{140}{171}\right) \implies \cos(t) = \dfrac{171}{221}[/tex]

as well as the fact that

[tex]0<\tan(t)<1 \implies 0along with the half-angle identities to find

[tex]\cos\left(\dfrac t2\right) = \sqrt{\dfrac{1+\cos(t)}2} = \dfrac{14}{\sqrt{221}}[/tex]

[tex]\sin\left(\dfrac t2\right) = \sqrt{\dfrac{1-\cos(t)}2} = \dfrac5{\sqrt{221}}[/tex]

(whose signs are positive because of the domain of [tex]\frac t2[/tex]).

This leaves us with

[tex]z = \pm \sqrt{\dfrac{5+10i \pm (5 + 14i)}2} \implies z = \pm \sqrt{5 + 12i} \text{ or } z = \pm \sqrt{-2i}[/tex]

Compute the square roots of 5 + 12i.

[tex]|5 + 12i| = \sqrt{5^2 + 12^2} = 13[/tex]

[tex]\arg(5+12i) = \tan^{-1}\left(\dfrac{12}5\right)[/tex]

By de Moivre,

[tex]\sqrt{5 + 12i} = \sqrt{13} \exp\left(i \dfrac12 \tan^{-1}\left(\dfrac{12}5\right)\right) \\\\ ~~~~~~~~~~~~~= \sqrt{13} \left(\dfrac3{\sqrt{13}} + \dfrac2{\sqrt{13}}i\right) \\\\ ~~~~~~~~~~~~~= 3+2i[/tex]

and its negative, -3 - 2i. We use similar reasoning as before:

[tex]t = \tan^{-1}\left(\dfrac{12}5\right) \implies \cos(t) = \dfrac5{13}[/tex]

[tex]1 < \tan(t) < \infty \implies \dfrac\pi4 < t < \dfrac\pi2 \implies \dfrac\pi8 < \dfrac t2 < \dfrac\pi4[/tex]

[tex]\cos\left(\dfrac t2\right) = \dfrac3{\sqrt{13}}[/tex]

[tex]\sin\left(\dfrac t2\right) = \dfrac2{\sqrt{13}}[/tex]

Lastly, compute the roots of -2i.

[tex]|-2i| = 2[/tex]

[tex]\arg(-2i) = -\dfrac\pi2[/tex]

[tex]\implies \sqrt{-2i} = \sqrt2 \, \exp\left(-i\dfrac\pi4\right) = \sqrt2 \left(\dfrac1{\sqrt2} - \dfrac1{\sqrt2}i\right) = 1 - i[/tex]

as well as -1 + i.

So our simplified solutions to the quartic are

[tex]\boxed{z = 3+2i} \text{ or } \boxed{z = -3-2i} \text{ or } \boxed{z = 1-i} \text{ or } \boxed{z = -1+i}[/tex]

The measures of the angles are

Angle JPK

The given parameters are:

∠KPL = 46°

∠NPH = 25°

∠MPN = ∠NPH

By vertical angle theorem, we have:

∠JPK = ∠MPH

Where

∠MPH = ∠MPN + ∠NPH

This gives

∠MPH = ∠NPH + ∠NPH

So, we have:

∠MPH = 25° + 25°

Evaluate

∠MPH = 50°

Recall that:

∠JPK = ∠MPH

So, we have:

∠JPK = 50°

Hence, the measure of angle ∠JPK is 50°

Angle LPM

The angle on a straight line is 180°.

So, we have:

∠LPM +  ∠KPL + ∠MPH = 180°

This gives

∠LPM +  46° + 50° = 180°

Evaluate

∠LPM = 84°

Hence, the measure of angle ∠LPM is 84°

Angle NPM

In (a), we have:

∠NPM = ∠NPH

This gives

∠NPM = 50°

Hence, the measure of angle ∠NPM is 50°

Angle JPI

By vertical angle theorem, we have:

∠JPI = ∠LPM

So, we have:

∠JPI = 84°

Hence, the measure of angle ∠JPI is 84°

Angle HPI

By vertical angle theorem, we have:

∠HPI = ∠KPK

So, we have:

∠HPI = 46°

Hence, the measure of angle ∠HPI is 46°

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The length of the diagonal that he cut to the nearest whole inch is 36 inches

A red dotted line crosses the rectangle from the upper left corner to the lower right corner to form a triangle.

Length of the diagonal of a rectangle;

Hypotenuse² = adjacent² + opposite²

= 18² + 36²

= 324 + 1296

hyp² = 1620

hyp = √1620

hyp = 35.4964786985976

Approximately,

hypotenuse = 36 inches

Therefore, the length of the diagonal that he cut to the nearest whole inch is 36 inches.

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The total income and expenditure in September are "Income is R 106 360.00 and expenditure is R 4 850.00." Option C. This is further explained below.

Generally, the equation for  Insurance and investment  is  mathematically given as

II= 42% of the balance

II = 0.42*0.64A

II= 0.2688A

Generally, the equation for Total After All other exp is  mathematically given as

Texp = 0.64A – 0.2688A

Texp= 0.3712A

Total After All other exp = 111210

subbing ahead and equating we have

111210 = 0.3712 A

Therefore

A = 299596

The total income is 299596

Therefore, the calculation for the expense is given as

Expense = 0.25A + 0.11A + 0.2688A + 4850

Expense= 0.6288A + 4850 = (0.6288*299596) + 4850

Expense= 193236

The Expense is 299596

In conclusion,  At the end of September, Agri-Small Business's minimal spending on gasoline for their fleet amounted to R 4 850.00, and they still had a balance of R106 360.00 in their account. The majority of the remaining revenue was allocated to insurance and investments by the business, which accounted for 42% of the total. Wages and salaries accounted for 25% of the company's income in September. The overall revenue and expenses for the month of September were as follows: the income was R 106 360.00, and the expenses were R 4 850.00. Option C.

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What is ordinary number?

1 : a number designating the place (such as first, second, or third) occupied by an item in an ordered sequence — see Table of Numbers. 2 : a number assigned to an ordered set that designates both the order of its elements and its cardinal number.

The equation of the line that passes through the given points is

y = 1/3x + 5.

The formula for the equation of a line passing through two points (x1, y1) and (x2, y2) is

[tex](y-y1) = \frac{(y2-y1)}{(x2-x1)} (x-x1)[/tex]

Where the fraction (y2 - y1)/(x2 - x1) is the slope of the line denoted by 'm'.

It is given that, a line passes through the points (-3,4) and (6,7).

So, the equation of the line is

[tex](y-y1) = \frac{(y2-y1)}{(x2-x1)} (x-x1)[/tex]

On substituting x1 = -3, y1 = 4, x2 = 6, and y2 = 7

(y - 4) = [(7 - 4)/(6 + 3)](x + 3)

⇒ (y - 4) = (3/9)(x + 3)

⇒ (y - 4) = 1/3(x + 3)

⇒ y- 4 = 1/3x + 1

⇒ y = 1/3x + 1 + 4

∴ y = 1/3x + 5

Thus, the equation of the line passing through the points (-3,4) and (6,7) is y = 1/3x + 5. Where the slope m = 1/3.

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Answer: The slope is -2 and the y-intercept is (0,1)

Step-by-step explanation:

Answer:

A convenient formula to use is

S = ((1 + i)^n - 1) / i where S is the value of 1$ deposited for n periods at an interest rate of i

in this case n = 12 * 28 = 336 periods of deposit at an interest rate of

.0021 / 12 = .00175 = i

S = (1.00175^336 - 1) / .00175 = 456.8338  the value of 1$ after 336 periods

350 * 456.8338 = 159891.81   the value of 350 deposited monthly

Note that 350 * 336 would be 117,600

One must be careful to distinguish the above formula from

(1 - (1 + i)^-n) / i which gives the value of 1$ when the borrower is "paying" an interest rate of i - this would be the case for a mortgage - or what is  the value of 1$ paid for n periods when paying an interest rate of i

1a. 98 cm ^2

1b. 76. 98 cm^2

2a. 42cm

2b. 7. 19 cm

3. a + b/2 (h)

4. 24 cm ^2

5. πr + 2r

6. 13. 2m

7. 45cm^2

8. 252 cm ^ 2

9. 450 cm^ 2

1a. The shape given is a rectangle

The formula for area of a rectangle is given as;

Area = length × width

Area = 7 × 14

Area = 98 cm ^2

1b The shape given is a semi circle

The formula for area of a semicircle is given as;

Area = 1/2 π r^2

radius = diameter/2 = 14/2 = 7cm

Area = 1/2 × 3.142 × 7 × 7

Area = 76. 98 cm^2

2a. The shape given is a rectangle

The formula for perimeter of a rectangle is given as;

Perimeter = 2 ( length + width)

Perimeter = 2 ( 14 + 7) = 2( 21)

Perimeter = 42cm

2b. The shape is a semicircle

Perimeter = π r + 2r

r= 1.4cm; diameter divided by 2

Perimeter = 3. 142(1.4) + 2(1.4)

Perimeter = 7. 19 cm

3. The formula for area of a trapezium is given as

Area = a + b/2 (h)

4. The area of the trapezium is given as;

Area = 9 + 7/2 (3)

Area = 16/2 (3)

Area = 8 × 3

Area = 24 cm ^2

5. Area of semicircle = 1/2 πr^2

Perimeter of a semicircle = πr + 2r

6. From the information given, we have the following

Area = 480 m^2

a = 20m

b = unknown

h = 13. 2m

Area = a+b/2 (h)

Substitute the values

480 = 20+b/2 (13. 2)

480 = 10+ b (13. 2)

480/13. 2 = 10 + b

10+ b = 480/ 13. 2

10 + b = 36. 36

b = 36.36 - 10

b = 26. 36m

7. The formula for area of a rhombus is given as

Area = p × q/2

Where p and q are the diagonals

Area = 7. 5 × 12/2

Area = 90/2

Area = 45cm^2

8. The formula for area for a quadrilateral is given as;

Area of quadrilateral = (½) × diagonal length × sum of the length of the perpendiculars

sum of the length = 13 + 8 = 21cm

Diagonal A= 24cm

Area = 1/2 × 24 × 21

Area = 252 cm ^ 2

9. Area of a pentagonal park = 1/2 × sum of parallel sides × height

Sum of parallel sides = 15 + 15 = 30 cm

height = 30cm

Area = 1/2 × 30 × 30

Area = 450 cm^ 2

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Answer:15

Step-by-step explanation: 20 divided by 8=2.5 x 6=15

Answer:

40

Step-by-step explanation:

[tex]\frac{RW}{24}=\frac{30}{18} \\ \\ RW=40[/tex]

Answer:9[tex]a^{10}[/tex][tex]b^{-6}[/tex]=[tex]\frac{9a^{10} }{b^6}[/tex]

Step-by-step explanation:

Answer:

9*a^10*(1/(b^6))

Step-by-step explanation:

(-3a^5*b^-3)^2

(-3)^2 * (a^5)^2 * (b^-3)^2

9a^10*b^-6

9*a^10*(1/(b^6))

Answer:

4.12 miles

Step-by-step explanation:

Use distance formula

d^2 = (-8- -4)^2 + ( -9 - -10)^2

d^2 =   16   +   1

d^2 = 17

d = sqrt 17 = 4.12 miles

[tex]\boldsymbol{\sf{6-\dfrac{3}{4}x+\dfrac{1}{3}=\dfrac{1}{y}x+5 }}[/tex]

Convert 6 to the fraction 18/3.

                       [tex]\boldsymbol{\sf{\dfrac{18}{3} -\dfrac{3}{4}x+\dfrac{1}{3}=\dfrac{1}{y}x+5 }}[/tex]

Since the fractions 18/3 and 1/3 have the same denominator, we add their numerators to calculate them.

                       [tex]\boldsymbol{\sf{\dfrac{18+1}{3}-\dfrac{3}{4}x=\dfrac{1}{2}x+5 \ \longmapsto \ \ [Add \ 18+1] }}[/tex]

                              [tex]\boldsymbol{\sf{\dfrac{19}{3}-\dfrac{3}{4}x=\dfrac{1}{2}x+5 }}[/tex]

Subtract [tex]\bf{\frac{1}{2}x }[/tex] on both sides.

                               [tex]\boldsymbol{\sf{\dfrac{19}{3}-\dfrac{3}{4}x-\dfrac{1}{2}x=5 }}[/tex]

Combine [tex]\bf{-\frac{3}{4}x}[/tex] and [tex]\bf{-\frac{1}{2}x}[/tex] to get [tex]\bf{-\frac{5}{4}x}[/tex].

                                [tex]\boldsymbol{\sf{\dfrac{19}{3}-\dfrac{5}{4}x=5 }}[/tex]

Subtract 19x from both sides.

                                       [tex]\boldsymbol{\sf{-\dfrac{5}{4}x=5-\dfrac{19}{3} }}[/tex]

Convert 5 to the fraction 15/3.

                                      [tex]\boldsymbol{\sf{-\dfrac{4}{5}x=\dfrac{15}{3}-\dfrac{19}{3} }}[/tex]

Since the fractions 15/3 and 19/3 have the same denominator, we add their numerators to calculate them.

                             [tex]\boldsymbol{\sf{-\dfrac{5}{4}x=\dfrac{15-19}{3} \ \longmapsto \ \ [Subtract \ 15-19] }}[/tex]

                                 [tex]\boldsymbol{\sf{-\dfrac{5}{4}x=-\dfrac{4}{3} }}[/tex]

Multiply both sides by -4/3, the reciprocal of -4/3.

                                    [tex]\boldsymbol{\sf{x=-\dfrac{4}{5}\left(-\dfrac{4}{5}\right) }}[/tex]

Multiply -4/3 by -4/5 (to do this, multiply the numerator by the numerator and the denominator by the denominator).

                   [tex]\boldsymbol{\sf{x=\dfrac{-4(-4)}{3\times5} \ \ \longmapsto \ \ Multiply, \ numerator \ and \ denominator. }}[/tex]

                                 [tex]\red{\boxed{\boldsymbol{\sf{\blue{Answer \ \ \longmapsto \ \ \ \ x=\frac{16}{15} }}}}}[/tex]

Answer:

The Co-ordinate of C is (4/3, -1/2)

Step-by-step explanation:

We have two equation one is of straight line equation which is:

                                              y=2x-3           (i)

Other equation is of quadratic function which is:

                                         y=-3x^2+5         (ii)

Put the value of y from equation (i) in equation (ii)

So, we have:

                                     2x-3=-3x^2+5

                                      3x^2+2x-8=0

By factorization:

                                    3x^2+6x-4x-8=0

                                   3x(x+2)-4x(x+2)=0

                                        (x+2)(3x-4)=0

              x+2=0                             ;                      3x-4=0

              x=-2                                ;                       x=4/3

Put first x=-2 in equation (i)

                                                y=2(-2)-3

                                                  y=-4-3

                                                    y=-7

Now Put x=4/3 in equation (i)

                                             y=2(4/3)-3

                                                 y=8/3-3

                                                   y=-1/2

So, we have two Order pair One is (-2 , -7) and Second one is (4/3 , -1/2)

Hence the Co-ordinate of C is:

                                               C=(4/3 , -1/2)

Answer:

Point C:  (3, 3)

Point D:  (3, -22)

Step-by-step explanation:

If the distance between points C and D is 25 units, the y-value of point D will be 25 less than the y-value of point C.  The x-values of the two points are the same.

Therefore:

[tex]\textsf{Equation 1}: \quad y=2x-3[/tex]

[tex]\textsf{Equation 2}: \quad y-25=-3x^2+5[/tex]

As the x-values are the same, substitute the first equation into the second equation and solve for x to find the x-value of points C and D:

[tex]\implies 2x-3-25=-3x^2+5[/tex]

[tex]\implies 3x^2+2x-33=0[/tex]

[tex]\implies 3x^2-9x+11x-33=0[/tex]

[tex]\implies 3x(x-3)+11(x-3)=0[/tex]

[tex]\implies (x-3)(3x+11)=0[/tex]

[tex]\implies x=3, -\dfrac{11}{3}[/tex]

From inspection of the given graph, the x-value of points C and D is positive, therefore x = 3.

To find the y-value of points C and D, substitute the found value of x into the two original equations of the lines:

[tex]\begin{aligned} \textsf{Point C}: \quad 2x-3 & =y\\2(3)-3 & =3\\ \implies & (3, 3)\end{aligned}[/tex]

[tex]\begin{aligned} \textsf{Point D}: \quad -3x^2+5 & = y \\ -3(3)^2+5 & =-22\\ \implies & (3, -22)\end{aligned}[/tex]

Therefore, point C is (3, 3) and point D is (3, -22).

Usando la distribución hipergeométrica,  la probabilidad de que ambas estén numeradas con el valor 1, es: 0.0222 = 2.22%.

La fórmula es:

[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

Los parámetros son:

Los valores de los parámetros son:

N = 10, k = 2, n = 2.

La probabilidad de que ambas estén numeradas con el valor 1, es P(X = 2), entonces:

[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]

[tex]P(X = 2) = h(2,10,2,2) = \frac{C_{2,2}C_{8,0}}{C_{10,2}} = 0.0222[/tex]

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Total 8 triangles are formed in figure and 4 edges used to make triangles and The area of all triangle is 1 square unit.

According to the statement

we have given that a figure and we have to the number of triangle present in it and the area of each square and the number of edges used in all triangles.

So, For this purpose, we know that the

A triangle is a polygon with three edges and three vertices. It is one of the basic shapes in geometry.

So,

A. The number of triangles present in figure:

There is a rectangle provided with two diagonals and due to this structure

Firstly 4 triangles are formed with 2 diagonals (big size).

And 4 triangles are formed in diagonals because diagonals cut each other (small size).

So, Total 8 triangles are formed in figure.

B. Edges used to form triangle :

there are 4 edges used to form triangles because of presence of overall shape of rectangle because triangles are formed in a rectangle shape.

So, 4 edges used to make triangles.

C. Area of the triangle:

If the area is 1 square unit and area of all triangles are same

So, The area of all triangle is 1 square unit.

So, Total 8 triangles are formed in figure and 4 edges used to make triangles and The area of all triangle is 1 square unit.

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Answer: A

Step-by-step explanation:

The total value of the prizes is [tex]1000+500+2(50)=1600[/tex].

The total cost of the tickets is [tex]1000(4.00)=4000[/tex].

So, the total loss is $2400.

Dividing this by 1000 tickets gives $-2.40.

Project's cost at time 0 is $46,272.84

What is profitability index?

Profitability index is the present value of future cash flows divided by the project cost at time 0.

The present value of each future cash flow can be determined using the present value formula of a single cash flow as  below;

PV=FV/(1+r)^N

FV=future cash inflows in years 1-4

r=discount rate=8.5%

N=the year of cash flows, 1 for year 1 , 2 for year 2, 3 for year 3 and 4 for year 4.

PV of future cash flows=$12,150/(1+8.5%)^1+$12,600/(1+8.5%)^2+$15,700/(1+8.5%)^3+$10,200/(1+8.5%)^4

PV of future cash flow=$41,553.01

Profitability index=PV of future cash flows/Project cost

Profitability index= .898

Project cost=unknown(assume it is X)

.898=$41,553.01 /X

X=$41,553.01/ .898

X=project's cost=$46,272.84

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Answer:

no habla espanol

Step-by-step explanation:

Answer: 5 units

Step-by-step explanation:

x+2+2x-7=13

solve for x

3x=18

x=6

∴SR: 2x-7

=2(6)-7

=12-7

=5

Answer:

SR = 5

Step-by-step explanation:

[tex]TR=x+2+2x-7=3x-5[/tex]

[tex]3x-5=13[/tex]

[tex]3x=13+5[/tex]

[tex]x=18/3=6[/tex]

[tex]SR=2x-7=2(6)-7=12-7=5[/tex]

Hope this helps

5x-30y=-35

Divide both the side by 5 and we get

x-6y = -7

x = 6y -7

Answer:

x=-7+6y

You simply need to add 30y and then divide by 5 to isolate the variable.