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3 years ago

9.

Mathematics

1 answer:

Alik [6]3 years ago

7 0

Answer:

B. y = 1

Step-by-step explanation:

Answer:

the asymptote of the graph y = 1

Step-by-step explanation:

The asymptote of the graph y = 1 (Horizontal asymptote)

As you can see a line that a graph is approạching to as it heads towards infinity but it does not intersect.

Moreover, the value of y is decreasing over its domain. as you can see in the graph.

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One troop member So tickets for 33 crab meals and 20 vegetarian meals with a total receipt of $2025 another sold tickets for 30

natima [27]

Answer:

Therefore the cost of a crab meal and a vegetarian are $40.41 and $1.94 respectively.

Step-by-step explanation:

Given that,tickets for 33 crab meals and 20 vegetarian meal with total receipt of $2025. Another sold ticket for 30 crab meal and 11 vegetarian meals with total receipt of $1827.

Let the cost of a crab meal and a vegetarian meal be x and y respectively.

The cost of a crab meal is x

The cost of 33 crab meal is $33x

The cost of 30 crab meal is $30x

The cost of a vegetarian meal is y.

The cost of 20 vegetarian meal is $20y.

The cost of 11 vegetarian meal is $11y

According to problem,

33x+20y=2025

33x+20y-2025=0........(1)

and

30x+11y=1827

30x+11y-1827=0 ..........(2)

Cross Multiplication Method,

$a_1x+b_1y+c_1=0$

$a_2x+b_2y+c_2=0$

$\frac{x}{b_1c_2-b_2c_2}=\frac{y}{a_2c_1-a_1c_2}=\frac{1}{a_1b_2-a_2b_1}$

${x}=\frac{b_1c_2-b_2c_2}{a_1b_2-a_2b_1}$

${y}=\frac{a_2c_1-a_1c_2}{a_1b_2-a_2b_1}$

Here $a_1=33$ , $b_1=20$ , $c_1=-2025$ , $a_2=30$ , $b_2=11$ , $c_2=-1827$

$\therefore x=\frac{20.(-1827)-11.(-2025)}{33.11-30.20}$ $y=\frac{30.(-2025)-33.(-1827)}{33.11-30.20}$

$\approx 40.41$ $\approx 1.94$

Therefore the cost of a crab meal and a vegetarian are $40.41 and $1.94 respectively.

8 0

3 years ago

Graph the image of this figure after a dilation with a scale factor of 13 centered at the point (4, −2) .

Sonja [21]

Answer:

From the given triangle figure;

Labelled the triangle as A , B and C.

The coordinates of this triangle ABC are;

A = (1, 10) ,

B = (-2, 4)

C = (7, 4).

Given : Scale factor(k) = $\frac{1}{3}$ and centered at point (4, -2).

The rule of dilation with $k= \frac{1}{3}$ and center at point (4,-2) is:

$(x, y) \rightarrow (\frac{1}{3}(x-4)+4 , \frac{1}{3}(y+2)-2)$

$(x, y) \rightarrow (\frac{1}{3}x-\frac{4}{3}+4 , \frac{1}{3}y+\frac{2}{3}-2)$

$(x, y) \rightarrow (\frac{1}{3}x+\frac{8}{3} , \frac{1}{3}y-\frac{4}{3})$

then, the dilation of the given figure are;

$A(1, 10) \rightarrow (\frac{1}{3}\cdot 1+\frac{8}{3} , \frac{1}{3} \cdot 10-\frac{4}{3})$ = $(\frac{1}{3}+\frac{8}{3} , \frac{10}{3}-\frac{4}{3})$ = $(\frac{9}{3} , \frac{6}{3})$ = A'(3 , 2)

$B(-2, 4) \rightarrow (\frac{1}{3}\cdot -2+\frac{8}{3} , \frac{1}{3} \cdot 4-\frac{4}{3})$ = $(-\frac{2}{3}+\frac{8}{3} , \frac{4}{3}-\frac{4}{3})$ = $(\frac{6}{3} , \frac{0}{3})$ =B'(2 , 0)

$C(7, 4) \rightarrow (\frac{1}{3}\cdot 7+\frac{8}{3} , \frac{1}{3} \cdot 4-\frac{4}{3})$ = $(\frac{7}{3}+\frac{8}{3} , \frac{4}{3}-\frac{4}{3})$ = $(\frac{15}{3} , \frac{0}{3})$ = C'(5 , 0)

The coordinates of dilation images are:

A' = (3,2) , B' = (2, 0) and C' = (5, 0)

You can see the graph of the dilated image as shown below:

3 0

3 years ago

Read 2 more answers

You’re repairing a large rectangular aquarium. The final step is to cover the glass with a protective film. The front and back o

Marianna [84]

Answer:

28 square metres

Step-by-step explanation:

Calculate the total surface area of the glass covered part of the rectangular aquarium

Surface area = 2(4 × 2) + 2(3 × 2)

= 16 + 12

Protective film needed = 28 square metres

6 0

3 years ago

Read 2 more answers

The probability distribution histogram shows the length distribution of lizards in the reptile collection at a zoo.

LuckyWell [14K]

Try looking at this and see if it helps http://openstudy.com/updates/580662f4e4b0d88e39fc9a0e

3 0

3 years ago

Given the equation (x-a)^2(x-2)=x^3+bx^2+12x-72 find a and b

ivann1987 [24]

Isolate the variable by dividing each side by factors that don't contain the variable.

a=- \frac{ x^{2}- \sqrt{( x^{3} + x^{2} b+12x-72(x-2)-2x} }{x-2} ,- \frac{ x^{2}+ \sqrt{( x^{3} + x^{2} b+12x-72(x-2)-2x} }{x-2}

Solve for b by simplifying both sides of the equation then isolating the variable.

b= \frac{12}{x}+ \frac{72}{ x^{2} }-2+2a- \frac{4a}{x}+ \frac{ a^{2} }{x}- \frac{2a^{z} }{ x^{2} }

Hopefully i helped ^.^ Mark brainly if possible. Lol once again i saw the same question so why not answer it again!

3 0

3 years ago