Assume that the terminal side of thetaθ passes through the point (negative 12 comma 5 )(−12,5) and find the values of trigonometric ratios sec thetaθ and sin thetaθ.

Answer:

Range of the average number of tours is between 150 and 200 including 150 and 200.

Step-by-step explanation:

Given:

The profit function is modeled as:

[tex]p(x)=-2x^2+700x-10000[/tex]

The profit is at least $50,000.

So, as per question:

[tex]p(x)\geq50000\\-2x^2 + 700x-10000\geq 50000\\-2x^2+700x-10000-50000\geq 0\\-2x^2+700x-60000\geq 0\\\\\textrm{Dividing by 2 on both sides, we get}\\\\-x^2+350x-30000\geq 0[/tex]

Now, rewriting the above inequality in terms of its factors, we get:

[tex]-1(x-150)(x-200)\geq 0\\(x-150)(x-200)\leq 0[/tex]

Now,

[tex]x<150,(x-150)(x-200)>0\\x>200,(x-150)(x-200)>0\\For\ 150\leq x\leq200,(x-150)(x-200)\leq 0\\\therefore x=[150,200][/tex]

Therefore, the range of the average number of tours he must arrange per day to earn a monthly profit of at least $50,000 is between 150 and 200 including 150 and 200.

Answer:

x = 36

Step-by-step explanation:

[tex] x - 12\sqrt{x} + 36 = 0 [/tex]

Subtract x and 36 from both sides.

[tex] -12\sqrt{x} = -x - 36 [/tex]

Divide both sides by -1.

[tex] 12\sqrt{x} = x + 36 [/tex]

Square both sides.

[tex] 144x = x^2 + 72x + 1296 [/tex]

Subtract 144x from both sides.

[tex] 0 = x^2 - 72x + 1296 [/tex]

Factor the right side.

[tex] 0 = (x - 36)^2 [/tex]

[tex] x - 36 = 0 [/tex]

[tex] x = 36 [/tex]

Since the solution of the equation involved squaring both sides, we musty check the answer for possible extraneous solutions.

Check x = 36:

[tex] x - 12\sqrt{x} + 36 = 0 [/tex]

[tex] 36 - 12\sqrt{36} + 36 = 0 [/tex]

[tex] 36 - 12\times 6 + 36 = 0 [/tex]

[tex] 36 - 72 + 36 = 0 [/tex]

[tex] 0 = 0 [/tex]

Since 0 = 0 is a true statement, the solution x = 36 is a valid solution.

Answer:

(x-4)² - 11

Step-by-step explanation:

You find half of 8 which is 4 and half of x² which is x. this forms (x - 4).

However this would expand as

x²-8x+16 which isn't the expression. So to make it 5, you have to take away 11 leaving you with

(x-4)²-11

Answer:

(x - 4)^2 - 11.

Step-by-step explanation:

x^2 - 8x + 5

Note that x^2 - 8x = (x - 4)^2 - 16 so we have:

(x - 4)^2 - 16 + 5

= (x - 4)^2 - 11.

To get (x - 4)^2 - 16 I used the identity:

x^2 + ax = ( x + a/2)^2 - a^2/4    with a = -8.

Answer:

Option C) The symbol y represents the predicted value of price.

Step-by-step explanation:

We are given the following in the question:

We find a regression equation with x representing the ratings and y representing price.

The equation has a slope of 130 and a y-intercept of 350.

Comparing with the slope intercept form:

[tex]y = mx + c\\\text{where m is the slope and c is the y intercept}[/tex]

Thus, we can write the equation as:

[tex]y = 130x + 350[/tex]

Here, y is the predicted variable that is the price, c is the price of hotel when a rating of 0 is given.

Thus, symbol y represents:

C) The symbol y represents the predicted value of price.

The equation of the regression line is y = 130x + 350. The symbol 'y' in this equation represents the predicted price of a hotel based on its rating.

The equation of a line in slope-intercept form is given by y = mx + b, where m is the slope of the line and b is the y-intercept. In this case, we have been provided with a slope of 130 and a y-intercept of 350. Therefore, the equation of the regression line is y = 130x + 350. This equation is the model, created using regression analysis, predicting the price of hotels based on their ratings.

The symbol y in this situation refers to the predicted value of price for a hotel depending on its rating. Hence, the correct answer for part (b) is 'C) The symbol y represents the predicted value of price'.

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Answer: x = 2, y = -4 , z = 1

Step-by-step explanation:

-x + y + 3z = -3 - - - - - - - - - - 1

x - 2y - 2z = 8 - - - - - - -- - - - 2

3x - y - 4z = 6 - - - - - - - - - - - - - 3

Let us use the method of elimination

We would add equation 1 to equation 2. It becomes

-y+z= 5 - - - - - - - - - - - - - - -4

Multiply equation 2 by 3 and equation 3 by 1

3x - 6y -6z = 24- - - - - - - - - - 5

3x - y - 4z = 6 - - - - - - - - - - - - -6

Subtracting equation 6 from equation 5

-5y -2z = 18 - - - - - - - - - - 7

Substituting z = 5 + y into equation 7, it becomes

-5y -2(5+y) = 18

-5y -10-2y = 18

-5y -2y = 18+10

-7y = 28

y = 28/-7 = -4

z = 5 + y

z = 5 -4 = 1

We would substitute y = -4 and z = 1 into equation 2

It becomes

x - 2×-4 - 2×1 = 8

x+8-2 = 8

x +6 = 8

x = 8-6 = 2

x = 2, y = -4 , z = 1

Let us check by substituting the value into equation 1

-x + y + 3z = -3

-2-4+ 3= -3

-6 + 3 = -3

-3 = -3

For this case we have a function of the form [tex]y = f (x)[/tex], where:

[tex]f (x) = 10-x[/tex]

We must find the value of the function when:

[tex]x = -2,0,2[/tex]

[tex]f (-2) = 10 - (- 2) = 10 + 2 = 12[/tex]

[tex]f (0) = 10-0 = 10[/tex]

[tex]f (2) = 10-2 = 8[/tex]

Thus, we have that the function has a value of [tex]y = {12,10,8}[/tex] when [tex]x = {- 2,0,2}[/tex]

Answer:

[tex]y = {12,10,8}[/tex]

Answer: she runs 132 feets in each lap and 660 feets in 5 laps

Step-by-step explanation:

Stella runs laps around the edge of the yard. This means she runs round the entire shape of the yard.

Miss bridgeyard is 24 ft by 42 ft. This means that the length and width of Miss bridgeyard are not the same. Therefore, Miss bridgeyard has the shape of a rectangle. The distance that stella covers in one lap is the perimeter of the rectangular Miss bridgeyard.

Perimeter of a rectangle = 2( L + W )

If length,L = 42 feets and

Width ,W = 24 feets, the perimeter would be

2(42+24)/= 2×66 = 132 feets

She runs a distance of 132 feets in one lap.

Distance in 5 laps would be

132 × 5 = 660 feets

Answer:

a) No

b) No

c) No

d) No

Step-by-step explanation:

Remember, a set V wit the operations addition and scalar product is a vector space if the following conditions are valid for all u, v, w∈V and for all scalars c and d:

1. u+v∈V

2. u+v=v+u

3. (u+v)+w=u+(v+w).

4. Exist 0∈V such that u+0=u

5. For each u∈V exist −u∈V such that u+(−u)=0.

6. if c is an escalar and u∈V, then cu∈V

7. c(u+v)=cu+cv

8. (c+d)u=cu+du

9. c(du)=(cd)u

10. 1u=u

let's check each of the properties for the respective operations:

Let [tex]u=(u_1,u_2,u_3), v=(v_1,v_2,v_3)[/tex]

Observe that  

1. u+v∈V

2. u+v=v+u, because the adittion of reals is conmutative

3. (u+v)+w=u+(v+w). because the adittion of reals is associative

4. [tex](u_1,u_2,u_3)+(0,0,0)=(u_1+0,u_2+0,u_3+0)=(u_1,u_2,u_3)[/tex]

5. [tex](u_1,u_2,u_3)+(-u_1,-u_2,-u_3)=(0,0,0)[/tex]

then regardless of the escalar product, the first five properties are met for a), b), c) and d). Now let's verify that properties 6-10 are met.

a)

6. [tex]c(u_1,u_2,u_3)=(cu_1,u_2,cu_3)\in V[/tex]

7.

[tex]c(u+v)=c(u_1+v_1,u_2+v_2,u_3+v_3)=(c(u_1+v_1),u_2+v_2,c(u_3+v_3))\\=(cu_1+cv_1,u_2+v_2,cu_3+cv_3)=c(u_1,u_2,u_3)+c(v_1,v_2,v_3)=cu+cv[/tex]

8.

[tex](c+d)u=(c+d)(u_1,u_2,u_3)=((c+d)u_1,u_2,(c+d)u_3)=\\=(cu_1+du_1,u_2,cu_3+du_3)\neq (cu_1+du_1,2u_2,cu_3+du_3)=cu+du[/tex]

Since 8 isn't satify then V is not a vector space with the addition as in R^3 and the scalar product [tex]a(x,y,z)=(ax,y,az)[/tex]

b)  6. [tex]c(u_1,u_2,u_3)=(cu_1,0,cu_3)\in V[/tex]

7.

[tex]c(u+v)=c(u_1+v_1,u_2+v_2,u_3+v_3)=(c(u_1+v_1),0,c(u_3+v_3))\\=(cu_1+cv_1,0,cu_3+cv_3)=c(u_1,u_2,u_3)+c(v_1,v_2,v_3)=cu+cv[/tex]

8.

[tex](c+d)u=(c+d)(u_1,u_2,u_3)=((c+d)u_1,0,(c+d)u_3)=\\=(cu_1+du_1,0,cu_3+du_3)=(cu_1,0,cu_3)+(du_1,0,du_3) =cu+du[/tex]

9.

[tex]c(du)=c(d(u_,u_2,u_3))=c(du_1,0,du_3)=(cdu_1,0,cdu_3)=(cd)u[/tex]

10

[tex]1u=1(u_1,u_2,u3)=(1u_1,0,1u_3)=(u_1,0,u_3)\neq(u_1,u_2,u_3)[/tex]

Since 10 isn't satify then V is not a vector space with the addition as in R^3 and the scalar product [tex]a(x,y,z)=(ax,0,az)[/tex]

c) Observe that [tex]1u=1(u_1,u_2,u3)=(0,0,0)\neq(u_1,u_2,u_3)[/tex]

Since 10 isn't satify then V is not a vector space with the addition as in R^3 and the scalar product [tex]a(x,y,z)=(0,0,0)[/tex].

d)  Observe that [tex]1u=1(u_1,u_2,u3)=(2*1u_1,2*1u_2,2*1u_3)=(2u_1,2u_2,2u_3)\neq(u_1,u_2,u_3)=u[/tex]

Since 10 isn't satify then V is not a vector space with the addition as in R^3 and the scalar product [tex]a(x,y,z)=(2ax,2ay,2az)[/tex].

None of the given definitions make ( V ) a vector space because they fail to satisfy the necessary vector space axioms.

To determine whether ( V ) is a vector space under the given definitions of scalar multiplication, we need to check if each definition satisfies the vector space axioms.

Definition (a): [tex]\( a(x,y,z) = (ax,y,az) \)[/tex]

Additive Identity: Yes, [tex]\( 1(x,y,z) = (x,y,z) \)[/tex].

Scalar Distributive (over vectors): [tex]\( a((x_1,y_1,z_1)+(x_2,y_2,z_2)) = a(x_1+x_2, y_1+y_2, z_1+z_2) = (a(x_1+x_2), y_1+y_2, a(z_1+z_2)) \).[/tex]

Scalar Distributive (over scalars): [tex]\( (a+b)(x,y,z) = ((a+b)x,y,(a+b)z) = (ax+bx,y,az+bz) \).[/tex]

Associative: [tex]\( a(b(x,y,z)) = a(bx,y,bz) = (abx,y,abz) = (ab)(x,y,z) \).[/tex]

Conclusion: Does not satisfy scalar distributive over vectors.

Definition (b): [tex]\( a(x,y,z) = (ax,0,az) \)[/tex]

Additive Identity: Yes, \( 1(x,y,z) = (x,0,z) \).

Scalar Distributive (over vectors): [tex]\( a((x_1,y_1,z_1)+(x_2,y_2,z_2)) = a(x_1+x_2,y_1+y_2,z_1+z_2) = (a(x_1+x_2),0,a(z_1+z_2)) = (ax_1+ax_2,0,az_1+az_2) \)[/tex]

Scalar Distributive (over scalars): [tex]\( (a+b)(x,y,z) = ((a+b)x,0,(a+b)z) = (ax+bx,0,az+bz) \).[/tex]

Associative: [tex]\( a(b(x,y,z)) = a(bx,0,bz) = (abx,0,abz) = (ab)(x,y,z) \).[/tex]

Conclusion: Does not satisfy scalar distributive over vectors.

Definition (c): [tex]\( a(x,y,z) = (0,0,0) \)[/tex]

Additive Identity: Yes, [tex]\( 1(x,y,z) = (0,0,0) \).[/tex]

Scalar Distributive (over vectors): [tex]\( a((x_1,y_1,z_1)+(x_2,y_2,z_2)) = (0,0,0) \).[/tex]

Scalar Distributive (over scalars): [tex]\( (a+b)(x,y,z) = (0,0,0) \).[/tex]

Associative: [tex]\( a(b(x,y,z)) = (0,0,0) \).[/tex]

Conclusion: Does not satisfy any of the scalar distributive properties.

Definition (d): [tex]\( a(x,y,z) = (2ax,2ay,2az) \)[/tex]

Additive Identity: No, [tex]\( 1(x,y,z) = (2x,2y,2z) \).[/tex]

Scalar Distributive (over vectors): [tex]\( a((x_1,y_1,z_1)+(x_2,y_2,z_2)) = a(x_1+x_2, y_1+y_2, z_1+z_2) = (2a(x_1+x_2), 2a(y_1+y_2), 2a(z_1+z_2)) = (2ax_1+2ax_2, 2ay_1+2ay_2, 2az_1+2az_2) \).[/tex]

Scalar Distributive (over scalars): [tex]\( (a+b)(x,y,z) = (2(a+b)x, 2(a+b)y, 2(a+b)z) = (2ax+2bx, 2ay+2by, 2az+2bz) \).[/tex]

Associative: [tex]\( a(b(x,y,z)) = a(2bx,2by,2bz) = (4abx,4aby,4abz) \neq (2ab)(x,y,z) \).[/tex]

Conclusion: Does not satisfy scalar multiplication associativity.

Answer: the number of adult tickets is 210

The number if student tickets is 270

Step-by-step explanation:

Let x represent the number of adult tickets that were purchased.

Let y represent the number of student tickets that were purchased.

At the ritz, concert tickets for adults cost $6 and tickets for students cost $4. If the cost of total tickets purchased is $2340, then,

6x + 4y = 2340 - - - - - - - -1

Total number of tickets purchased is 480. This means that

x + y = 480

x = 480 - y

Substituting x = 480 - y into equation 1, it becomes

6(480 - y) + 4y = 2340

2880 - 6y + 4y = 2340

- 6y + 4y = 2340 - 2880

-2y = - 540

y = - 540/-2 = 270

x = 480 - 270

x = 210

Answer:

Let's x be the probability for 1, 3, 5 and 6.

The probability for 2 and 4 is going to be 3x.

The sum of the probabilities of all possible outcomes is always 1.

P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1

x + 3x + x + 3x + x + x = 1

10x = 1

x = 1/10

The probability of obtaining 1, 3, 5 or 6 is 1/10

The probability for 2 and 4 is 3/10

The probability of rolling a 2 or a 4 is [tex]$\frac{3}{14}$[/tex], and the probability of rolling any of the other numbers (1, 3, 5, or 6) is [tex]$\frac{1}{14}$.[/tex]

To solve this problem, we need to distribute the total probability of 1 (since the sum of all probabilities must equal 1) among the six outcomes of the die according to the given conditions.

 Let's denote the probability of rolling a 2 or a 4 as $p$. According to the problem, rolling a 2 or rolling a 4 is three times as likely as rolling each of the other four numbers. Therefore, the probability of rolling a 1, 3, 5, or 6 is [tex]$\frac{p}{3}$.[/tex]

 Since there are two outcomes with probability $p$ (rolling a 2 and rolling a 4) and four outcomes with probability $\frac{p}{3}$ (rolling a 1, 3, 5, or 6), we can set up the following equation to represent the total probability:

[tex]\[ 2p + 4\left(\frac{p}{3}\right) = 1 \][/tex]

 Now, let's solve for $p$:

[tex]\[ 2p + \frac{4p}{3} = 1 \][/tex]

[tex]\[ \frac{6p + 4p}{3} = 1 \][/tex]

[tex]\[ \frac{10p}{3} = 1 \][/tex]

[tex]\[ 10p = 3 \][/tex]

[tex]\[ p = \frac{3}{10} \][/tex]

 So, the probability of rolling a 2 or a 4 is[tex]$p = \frac{3}{10}$.[/tex]

 The probability of rolling a 1, 3, 5, or 6 is [tex]$\frac{p}{3} = \frac{3}{10} \times[/tex] [tex]\frac{1}{3} = \frac{1}{10}$.[/tex]

However, we must remember that the total probability must be distributed equally between rolling a 2 and rolling a 4. Since they are equally likely, each has a probability of half of $p$:

[tex]\[ p_{2} = p_{4} = \frac{p}{2} = \frac{3}{10} \times \frac{1}{2} = \frac{3}{20} \][/tex]

 Now, we can state the final probabilities for each outcome:

 - The probability of rolling a 1 is [tex]$\frac{1}{10}$.[/tex]

- The probability of rolling a 2 is [tex]$\frac{3}{20}$.[/tex]

- The probability of rolling a 3 is [tex]$\frac{1}{10}$.[/tex] - The probability of rolling a 4 is [tex]$\frac{3}{20}$.[/tex]

The probability of rolling a 5 is [tex]$\frac{1}{10}$.[/tex]

Answer: The correct answer is D). Between 150 and 200; exclusive

Step-by-step explanation:

Given profit function p(x) of a tour operator is modeled by

p(x)=[tex](-2)x^{2} +700x-10000[/tex]

Where, x is the average number of tours he arranges per day.

To find number of tours to arrange per day to get monthly profit of at least 50,000$:

Now, he should make at-least 50000$ profit.

we can write as p(x)>50000$

[tex](-2)x^{2} +700x-10000\geq50000[/tex]

[tex](-2)x^{2} +700x-60000\geq0[/tex]

Roots are x is 150 and 200

(x-150)(x-200)>0

Case 1 : x>150 and x>200

x>150 also satisfy the x>200.

Case2: x<100 and x<200

x<200 also satisfy the x<100

Thus, the common range is 150<x<200

The correct answer is D). Between 150 and 200; exclusive

Answer:  between 150 and 200; inclusive

Step-by-step explanation:

The answer is 'inclusive' NOT 'exclusive.'

Answer:

Step-by-step explanation:

Answer:

The answer is b.)  -5.2 degrees

Step-by-step explanation:

to find the mean of this problem you have to add all numbers and then divide it by how many numbers there is.

so you have to add  -42+ -17+14+-4+23 and that'll equal -26

so you take -26 and divide it by 5 because thats how many numbers their are to divide

-26 divided by 5 is (-5.2)

Answer: You would spend 160 hours in total of ten weeks.

Step-by-step explanation: Just add 12.5 + 3.5 which = 16. Then multiply 16 times 10 which is 160, and that is your answer.

(i) The length of the longer side is 3 cm

(ii) The length of the longer side is 6 cm

(iii) The length of the longer side is 15 cm

Step-by-step explanation:

A rectangle has sides in the ratio 1 : 3, we need to find the length of the longer side if:

Let us use the ratio method to solve the problem

(i)

∵ The ratio of the two sides of the rectangle is 1 : 3

∵ The length of the shorter side is 1 cm

→  Shorter    :    Longer

→  1               :    3

→  1               :    x

By using cross multiplication

∴ 1 × x = 1 × 3

∴ x = 3

∵ x represents the length of the longer side

∴ The length of the longer side = 3 cm

The length of the longer side is 3 cm

(ii)

∵ The ratio of the two sides of the rectangle is 1 : 3

∵ The length of the shorter side is 2 cm

→  Shorter    :    Longer

→  1               :    3

→  2               :    x

By using cross multiplication

∴ 1 × x = 2 × 3

∴ x = 6

∵ x represents the length of the longer side

∴ The length of the longer side = 6 cm

The length of the longer side is 6 cm

(iii)

∵ The ratio of the two sides of the rectangle is 1 : 3

∵ The length of the shorter side is 5 cm

→  Shorter    :    Longer

→  1               :    3

→  5               :    x

By using cross multiplication

∴ 1 × x = 5 × 3

∴ x = 15

∵ x represents the length of the longer side

∴ The length of the longer side = 15 cm

The length of the longer side is 15 cm

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Answer:

a) d(x)=[tex]\sqrt{17x^{2} -12x+5}[/tex]

b)d'(x)=[tex]\frac{17x-6}{\sqrt{17x^{2} -12x+5} }[/tex]

c)The critical point is x=[tex]\frac{6}{17}[/tex]

d)Closest point is ([tex]\frac{6}{17}[/tex],[tex]\frac{7}{17}[/tex]

Step-by-step explanation:

We are given the line

[tex]y=4x-1[/tex]

Let a point Q([tex]x,y[/tex]) lie on the line.

Point P is given as P(2,0)

By distance formula, we have the distance D between any two points

A([tex]x_{1},y_{1}[/tex]) and B([tex]x_{2},y_{2}[/tex]) as

D=[tex]\sqrt{(x_{1}-x_{2})^2 + (y_{1}-y_2)^2}[/tex]

Thus,

d(x)=[tex]\sqrt{(x-2)^2+(y-0)^2}[/tex]

But we have, [tex]y=4x-1[/tex]

So,

d(x)=[tex]\sqrt{(x-2)^2+(4x-1)^2}[/tex]

Expanding,

d(x)=[tex]\sqrt{17x^2-12x+5}[/tex]  - - - (a)

Now,

d'(x)= [tex]\frac{\frac{d}{dx} (17x^2-12x+5)}{2(\sqrt{17x^2-12x+5}) }[/tex]

i.e.

d'(x)=[tex]\frac{17x-6}{\sqrt{17x^{2} -12x+5} }[/tex] - - - (b)

Now, the critical point is where d'(x)=0

⇒ [tex]\frac{17x-6}{\sqrt{17x^{2} -12x+5} }[/tex] =0

⇒[tex]x=\frac{6}{17}[/tex]    - - - (c)

Now,

The closest point on the given line to point P is the one for which d(x) is minimum i.e. d'(x)=0

⇒[tex]x=\frac{6}{17}[/tex]

as [tex]y=4x-1[/tex]

⇒y=[tex]\frac{7}{17}[/tex]

So, closest point is ([tex]\frac{6}{17},\frac{7}{17}[/tex])   - - -(d)

Answer:

$7,661

Step-by-step explanation:

Closing balance = Opening balance + purchases - Issued items

Given

Office supplies account on January 1 = $6,791 - Opening balance

Purchases = $3,205

Supplies on hand on January 31 = $2,155 - Closing balance

Substituting into the formula above

2155 = 6791 + 3025 - Issued items

Issued items = 6791 + 3025 - 2155

                     = $7,661

The amount to be used for the appropriate adjusting entry is $7,661

Final answer:

The adjusting entry for the used office supplies for the month of January is $7,841, which is calculated by subtracting the supplies on hand at the month's end from the sum of the starting balance and purchases made during the month.

Explanation:

To calculate the adjusting entry for office supplies, you need to calculate the cost of supplies that were used during the month. Start with the balance of supplies on hand at the beginning of the month, add the purchases made during the month, and then subtract the balance of supplies on hand at the end of the month.

The calculation is as follows:

Starting balance on January 1: $6,791

Add purchases during January: $3,205

Subtract ending balance on January 31: $2,155

The adjusting entry for supplies used = (Starting balance + Purchases) - Ending balance
= ($6,791 + $3,205) - $2,155
= $9,996 - $2,155
= $7,841

Therefore, the adjusting entry to record the office supplies used would be for $7,841.

Answer:

x = 8; y = 2.5.

Step-by-step explanation:

As we know , when two complex numbers are equal their real as well as imaginary part are equal.

So comparing on both sides ,

2x = 16   and    5 = 2y

x = 8       and    y = 2.5.

So ,   x = 8; y = 2.5.

Answer:

(2.23, 7,57)

Step-by-step explanation:

equation of this circle is

x^2 + (y - 2)^2 = 36

y = 2.5x + 2

Substitute for y in the equation of the circle:-

x^2 + (2.5x + 2 - 2)^2 = 36

x^2 + 6.25x^2 = 36

x^2 = 36 / 7.25  

x = +/-   6  /  2.693  =  +/- 2.228

when x = 2.228 y = 2.5(2.228) + 2 =  7.57    to nearest hundredth

when x = -2.228 y = 2.5(-2.228) + 2 =   -3.57

So they intersect at 2 points but the intersect in the first quadrant is at (2.23, 7,57)        to nearest hundredth.

Answer:

About 1.65 meals per hour

Step-by-step explanation:

9 days of work in May * 8 hours per day = 72 hours of work in May per employee

72 hours * 13 employees = 936 hours worked for all employees in May

1540 meals in May/ 936 hours worked for all employees in May = about 1.65 meals per hour

Answer:

Option D.

Step-by-step explanation:

Given information: KLMN is parallelogram, K(7,7), L(5,3), M(1,1) and N(3,5).

Diagonals of a parallelogram bisect each other.

If diagonals of a parallelogram are perpendicular to each other then the parallelogram is a rhombus.

If a line passes through two points [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex], then the rate of change is

[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]

Slope of KM is

[tex]m_1=\frac{1-7}{1-7}=1[/tex]

Slope of LN is

[tex]m_2=\frac{5-3}{3-5}=-1[/tex]

The product of slopes of two perpendicular lines is -1.

Find the product of slopes.

[tex]m_1\cdot m_2=1\cdot (-1)=-1[/tex]

The product of slopes of KM and NL is -1. It means diagonals are perpendicular and KLMN is a rhombus.

Therefore, the correct option is D.

The total weight of the smaller and the bigger cat Alberto has is 26 1/12 pounds.

A fraction is written in the form of p/q, where q ≠ 0.

Fractions are of two types they are proper fractions in which the numerator is smaller than the denominator and improper fractions where the numerator is greater than the denominator.

Given, Alberto has 2 cats.

The smaller cat weighs 10 3/4 pounds and the larger cat weighs 15 1/3 pounds.

Therefore, The weights of the cats together is the sum of their individual

weights which is,

= (10 3/4 + 15 1/3) pounds.

= (43/4 + 46/3) pounds.

= [(3×43 + 4×46)/12] pounds.

= (129 + 184)/12 pounds.

= 313/12 pounds.

= 26 1/12 pounds.

So, Together the cats weigh 26 1/12 pounds.

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Answer: 85.333 or 256 over 3.

Answer: x = 2

Step-by-step explanation:

The diagram of the kite is shown in the attached photo.

The perimeter of the kite is the distance around the kite.

The kite has 2 pairs of equal sides.

This means that

Side ML = side MN and side NO =

side LO

If Side MN = 8x – 3 and side NO = 2x + 1, then The perimeter of the kite is ML + MN + NO + LO

The perimeter of kite LMNO is given as 36 feet.

Therefore

ML + MN + NO + LO = 36

8x – 3 + 2x + 1 +8x – 3 + 2x + 1 = 36

8x + 8x + 2x+ 2x -3 +1 - 3 + 1

20x -4 = 36

20x = 40

x = 40/20 = 2

Answer:

x  =  140  ft

w = 70 ft

A(max)  =  9800 ft²

Step-by-step explanation:

We have:

280 feet of fence to enclose three sides of a rectangular area

perimeter of the rectangle ( 3 sides ) is

p  =  L  =  x  +2w       w   = (L - x ) / 2       w   =  ( 280  -  x ) / 2

where:

x is the longer side

w is the width

A(x,w)  = x*w         ⇒   A(x)  =  x* ( 280 - x ) / 2  ⇒ A(x)  = (280x -x²)/2

Taking derivatives on bth sides of the equation

A´(x)  = ( 280 -2x)*2 /4          A´(x)  = 0      ( 280 -2x)  =  0

280 -2x  = 0     x = 280/2

x  =  140  ft

And   w  = ( 280 - x ) / 2  ⇒  w  =(  280  -140  )/ 2

w = 70 ft

A(max)  =  9800 ft²

Answer:

  m∠A = m∠D = 40°

Step-by-step explanation:

Angles A and D are corresponding angles in the congruent triangles, so have the same measure. You set one measure equal to the other:

  x + 20 = 2x

To solve this, subtract x from both sides:

  20 = x

Then both angle measures are 2x = 40°.

Answer: The distance between A and B is 300 miles.

Step-by-step explanation:

Hi, to solve this problem we have to analyze the information given.

We know that when he was 15 miles away from point B, he was traveling at 60mph. if we apply the formula : time= distance /speed;

So, he traveled that distance in 15 minutes.

That means that he returned to point A in 3.75 hours (4 hours -15minutes) at a speed of 80 mph.

Applying the formula again to calculate the distance:

Answer:

(Solution: X=16 and Y=-10)

Explanation:

》Total money that will be spent on flour and corn meal altogether(T)

=$25

》Since it is not mentioned that whether corn and flour are bought in same quantity or not, we will assume them of different quantity.

i.e.,Suppose

》X pound of flour is bought

&

》Y pound of corn is bought.

So,

》Cost of flour(F)=$1.50X

》Cost of corn(C)=$2.50Y

So total cost will be sum of cost of flour and corn altogether,

Writing it in equation(linear),

F+C=T

Also,

Total pounds=6

ie,

The system of linear equations is x + y = 6 and 1.5x + 2.5y = 25.

A connection between a number of variables results in a linear model when a graph is displayed. The variable will have a degree of one.

The linear equation is given as,

y = mx + c

Where m is the slope of the line and c is the y-intercept of the line.

Assume you purchase flour and corn dinner in mass to make flour tortillas and corn tortillas flour cost $1.50 per pound and corn feast cost $2.50 per pound would you like to spend veiling $25 on flour and corn feast yet you want somewhere around 6 pounds by and large

Let x be the number of pounds of flour and y be the number of pounds of corn meal. Then the system of linear equalities is given as,

x + y = 6

1.5x + 2.5y = 25

The system of linear equations is x + y = 6 and 1.5x + 2.5y = 25.

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Answer:

  11 miles

Step-by-step explanation:

After the first hour, they walk 3 more hours at 2 miles per hour. So, the total distance is ...

  5 mi + (3 h)(2 mi/h) = 5 mi + 6 mi = 11 mi

The blue team will walk 11 miles in 4 hours.

Answer:

The equation [tex]7\,\sqrt{x} =14[/tex] is a radical equation.

Step-by-step explanation:

If the equations given are (as I can read them from your typing):

a) [tex]x+\sqrt{5} =12[/tex]

b) [tex]x^2=16[/tex]

c) [tex]3+x\,\sqrt{7} =13[/tex]

d) [tex]7\,\sqrt{x} =14[/tex]

The only radical equation is the last one : [tex]7\,\sqrt{x} =14[/tex], because it is the only one where the unknown appears inside the root. The name "radical equations" is associated with the fact that the unknown is contained inside the root and therefore the process involved in solving for the unknown will need to include the elimination of the root via algebraic methods to free the unknown.

Notice that the options a) and c) have roots, but what appears inside them are numbers (5 and 7 respectively), and not an unknown like "x". Equation b) doesn't contain a root, and wouldn't classify as a radical equation.

A radical equation is one which contains roots in it, specially those which has root over variables or things whose values changes.

Thus, by above definition, we will have the fourth option: [tex]7\sqrt{x} = 14[/tex] as a radical equation.

A radical equation is one which contains roots in it, specially those which has root over variables or things whose values changes.

Since only in the fourth option we see there's root over x which is a variable here, thus the  fourth option: [tex]7\sqrt{x} = 14[/tex] is a radical equation.

Rest of the options, although containing roots, aren't having variables inside the root, thus they aren't classified as radical equations.

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