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2 years ago

$25 for a shirts $30 for a bag $40 get 10% ON WHAT YOU PURCHASE MARLIN BOUGHT 1 BAG 2 SHIRTS 2 PANTS

Mathematics

1 answer:

SSSSS [86.1K]2 years ago

7 0

Answer: The question is incomplete.

Step-by-step explanation: I cannot give you the answer because i do not know the price of the pants.

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2) Line segment MK has endpoints at (2, 3) and (5, ?4). Segment M'K' is the reflection of MK over the y-axis. Which statement de

Marianna [84]

Answer:

Option C -M'K' is the same length as MK

Step-by-step explanation:

Given : Line segment MK has endpoints at (2, 3) and (5,4)

M'K' is the reflection of MK over the y-axis

By definition of reflection: reflection of point (x,y) across the the y-axis is the point (-x,y)

which implies M'K' has end points (-2,3) and (-5,4)

Now, we find the length of MK

let $(x_1,y_1)=(2,3)\\\\(x_2,y_2)=(5,4)$

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

⇒ $d=\sqrt{(2-5)^2+(4-3)^2}$

⇒ $d=\sqrt{9+1}$

⇒ $d=\sqrt{10}$ ....(1)

Now, we find the length of M'K'

let $(x_1^{'},y_1^{'})=(-2,3)\\\\(x_2^{'},y_2^{'})=(-5,4)$

$d^{'}=\sqrt{(x_2^{'}-x_1^{'})^2+(y_2^{'}-y_1^{'})^2}$

⇒ $d^{'}=\sqrt{(-2+5)^2+(3-4)^2}$

⇒ $d^{'}=\sqrt{9+1}$

⇒ $d^{'}=\sqrt{10}$ .....(2)

from (1) and (2) we simply show that the length of MK and M'K' is equal

we can also refer the figure attached for reflection of MK and M'K'

therefore, Option C is correct

3 0

3 years ago

Read 2 more answers

What are the solutions of the quadratic equation?

ANEK [815]

In this proble, we are asked to determine the roots of the quadratic equation x^2 + 11 x = -24. The standard form is x^2 + 11 x + 24 = 0. the second term is the negative sum of the roots and the third terms is the equal to the product of the roots.Hence the roots should be -3 and -8.

8 0

2 years ago

Read 2 more answers

What is the equation of the line that is parallel to the line 2x − y = 3 and passes through the point (2, −1)?

Ivanshal [37]

Answer:

C. y = 2x − 5

Step-by-step explanation:

Given equation of line

2x − y = 3

rewriting the equation in slope intercept form as answer are given slope intercept form

y = 2x - 3

slope intercept form of equation is

y = mx + c

where m is the slope

c is y intercept

thus

slope for y = 2x - 3 when comparing with y = mx + c is 2

now we know slope of two parallel line is same

thus,

the slope of line that is parallel to the line 2x − y = 3

is 2

Let the equation of required line be y = mx + c

where m = 2

thus

y = 2x + c is the new required equation

it passes through the point (2, −1)

using y = -1 and x = 2 in y = 2x + c

-1 = 2*2 + c

c = -1 - 4

c = -5

thus,

equation of required line is option c

y = 2x - 5 ----answer

7 0

2 years ago

If sine theta equals negative square root of three over two and pi less than theta less than three times pi over two, what are t

Romashka-Z-Leto [24]

Cosine theta equals the negative square root of three over two; tangent theta equals the negative square root of three over three is the correct answer.

<h3>What is angle measurement?</h3>

An angle measure is the measurement of the angle created by two rays or arms at a shared vertex in geometry. A protractor is used to measure angles in degrees (°).

Given data;

sin(1/2) = π/6

The value of the trignometric function are;

cos(π/6) = (√3)/2

tan(π/6) = 1/√3 = (√3)/3

In the second quadrant, where the cosine and tangent signs are both negative, is the angle of interest.

θ = 5π/6

cos(5π/6) = -(√3)/2

tan(5π/6) = -(√3)/3

Hence. cosine theta equals the negative square root of three over two; tangent theta equals the negative square root of three over three is the correct answer.

To learn more about the angle measurement, refer to the link;

brainly.com/question/14684647

#SPJ1

7 0

1 year ago

Any volunteer please help

den301095 [7]

Answer:

Part A)

The equation in the point-slope form is:

$y-11=\frac{4}{3}\left(x-\left(-2\right)\right)$

Part B)

The graph of the equation is attached below.

Step-by-step explanation:

Part A)

Given

The point = (-2, 11)

m = 4/3

The point-slope form of the line equation is

$y-y_1=m\left(x-x_1\right)$

Here, m is the slope and (x₁, y₁) is the point

substituting the values m = 4/3 and the point (-2, 11) in the point-slope form of the line equation

$y-y_1=m\left(x-x_1\right)$

$y-11=\frac{4}{3}\left(x-\left(-2\right)\right)$

Thus, the equation in the point-slope form is:

$y-11=\frac{4}{3}\left(x-\left(-2\right)\right)$

Part B)

As we have determined the point-slope form which passes through the point (-2, 11) and has a slope m = 4/3

The graph of the equation is attached below.

5 0

2 years ago