All categories

3 years ago

Which point on the graph shows the price of 6 ounces of cheese? Use the formula y=x÷2 to find y when x is 6.

Mathematics

2 answers:

Gwar [14]3 years ago

8 0

Answer: The required point is A(6, 3) in the attached graph.

Step-by-step explanation: We are to find the point on the given graph that shows the price of 6 ounces of cheese.

The formula to be used is given by

$y=x\div 2\\\\\Rightarrow y=\dfrac{x}{2}~~~~~~~~~~~~~~~~(i).$

Given that,

when x = 2 then y = 1. That is, price of 2 ounces of cheese is $1.

when x = 7 then y = 3.5. That is, price of 7 ounces of cheese is $3.5.

when x = 14 then y = 7. That is, price of 14 ounces of cheese is $7.

Therefore, these points satisfy the given relation (i).

Hence, if x = 6, then from equation (i), we have

$y=\dfrac{x}{2}=\dfrac{6}{2}=3.$

Thus, the price of 6 ounces of cheese is &3. This gives that the required point is A(6, 3) on the graph attached below.

Scrat [10]3 years ago

8 0

Y=x divided by 2
y= (6) divided by 2
y=3

You might be interested in

Is 6200 ft greater or less than 1 mile 900 ft

Sphinxa [80]

1 mile is equal to 5280 feet

Add 900 to it:

5280 + 900 = 6180

6180 is less than 6200, so 1 mile and 900 feet is less than 6200 feet.

7 0

3 years ago

Read 2 more answers

What is the range of F(x)=72x-x^2

Marianna [84]

Answer:

Range of F(x) = (-∞ ,72).

Step-by-step explanation:

Range of the function F(x) means the set of values which are taken by the function along its domain.

F(x) = 72 - $x^{2}$

Since coefficient of $x^{2}$ is negative on the right side ,

Maximum value of F(x) will come at x = 0.

Which is, F(0) = 72.

Now, as x increases , F(x) decreases , when x will reach infinity ,

F(x) will be reaching negative of infinity .

Thus, F(x) is ranging from -∞ to 72 .

Range of F(x) = (-∞ ,72).

3 0

3 years ago

Solve: 1/2x - 7 = 1/3 (x-12)

IrinaVladis [17]

Sorry I don’t know the answer hope you find one bye

8 0

3 years ago

2. Calculate the perimeter of the shape below. 6cm 2cm 3cm 4cm

Brut [27]

22 cm is the answer

6 0

2 years ago

Part B

alisha [4.7K]

Question:

Consider ΔABC, whose vertices are A (2, 1), B (3, 3), and C (1, 6); let the line segment AC represent the base of the triangle.

(a) Find the equation of the line passing through B and perpendicular to the line AC

(b) Let the point of intersection of line AC with the line you found in part A be point D. Find the coordinates of point D.

Answer:

$y = \frac{1}{5}x + \frac{12}{5}$

$D = (\frac{43}{26},\frac{71}{26})$

Step-by-step explanation:

Given

$\triangle ABC$

$A = (2,1)$

$B = (3,3)$

$C = (1,6)$

Solving (a): Line that passes through B, perpendicular to AC.

First, calculate the slope of AC

$m = \frac{y_2 - y_1}{x_2 - x_1}$

Where:

$A = (2,1)$ --- $(x_1,y_1)$

$C = (1,6)$ --- $(x_2,y_2)$

The slope is:

$m = \frac{6- 1}{1 - 2}$

$m = \frac{5}{-1}$

$m = -5$

The slope of the line that passes through B is calculated as:

$m_2 = -\frac{1}{m}$ --- because it is perpendicular to AC.

So, we have:

$m_2 = -\frac{1}{-5}$

$m_2 = \frac{1}{5}$

The equation of the line is the calculated using:

$m_2 = \frac{y_2 - y_1}{x_2 - x_1}$

Where:

$m_2 = \frac{1}{5}$

$B = (3,3)$ --- $(x_1,y_1)$

$(x_2,y_2) = (x,y)$

So, we have:

$\frac{1}{5} = \frac{y - 3}{x - 3}$

Cross multiply

$5(y-3) = 1(x - 3)$

$5y - 15 = x - 3$

$5y = x - 3 + 15$

$5y = x +12$

Make y the subject

$y = \frac{1}{5}x + \frac{12}{5}$

Solving (b): Point of intersection between AC and $y = \frac{1}{5}x + \frac{12}{5}$

First, calculate the equation of AC using:

$y = m(x - x_1) + y_1$

Where:

$A = (2,1)$ --- $(x_1,y_1)$

$m = -5$

So:

$y=-5(x - 2) + 1$

$y=-5x + 10 + 1$

$y=-5x + 11$

So, we have:

$y=-5x + 11$ and $y = \frac{1}{5}x + \frac{12}{5}$

Equate both to solve for x

i.e.

$y = y$

$-5x + 11 = \frac{1}{5}x + \frac{12}{5}$

Collect like terms

$-5x -\frac{1}{5}x = \frac{12}{5} - 11$

Multiply through by 5

$-25x-x = 12 - 55$

Collect like terms

$-26x = -43$

Solve for x

$x = \frac{-43}{-26}$

$x = \frac{43}{26}$

Substitute $x = \frac{43}{26}$ in $y=-5x + 11$

$y = -5 * \frac{43}{26} + 11$

$y = \frac{-5 *43}{26} + 11$

Take LCM

$y = \frac{-5 *43+11 * 26}{26}$

$y = \frac{71}{26}$

Hence, the coordinates of D is:

$D = (\frac{43}{26},\frac{71}{26})$

3 0

3 years ago