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3 years ago

What is the equation of a horizontal line that has a y-intercept of 3?

Mathematics

1 answer:

Ira Lisetskai [31]3 years ago

4 0

Answer:

y = 3

Step-by-step explanation:

The equation of a horizontal line parallel to the x- axis is

y = c

where c is the value of the y- coordinates the line passes through.

here the line passes through (0, 3) with y- coordinate 3

y = 3 ← is the equation of the horizontal line

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A trapezoid has an area of 50 square units. The lengths of the bases are 10 units and 15 units. find the height.

Lera25 [3.4K]

Step-by-step explanation:

area = ( base 1 + base 2 ) ÷ 2 x height

50 = ( 10 + 15) ÷2 x height

50 ÷ 12.5 = height

height = 4

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3 years ago

Solve for y for both questions

vredina [299]

1) y= 4

2) y= -10

Those are the answers

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3 years ago

The running back for the Bulldogs football team carried the ball 5 times for a total loss of 11 1/4 yards. Find the average chan

Firdavs [7]

Answer:

$\displaystyle 2\ \frac{1}{4}$

Step-by-step explanation:

<u>Fractions</u>

They can be expressed as proper fractions, improper fractions or mixed numbers. Proper fractions are such that the numerator is less than the denominator, like 2/5, 7/11, -9/10. Improper fractions are those whose numerators are greater than the denominator, such as 5/3, 10/9, -21/8.

Mixed numbers are expressions made of whole numbers and a proper fraction, like 4 3/5, 1 1/2, -5 4/9. Mixed numbers can be transformed to improper fractions and vice-versa.

The question requires us to find the average change in field position on each run of the running back for the Bulldogs football team which carried the ball 5 times for a total loss of 11 1/4 yards.

The number 11 1/4 is mixed, to express it as an improper fraction, we add the numbers like

$\displaystyle 11+\frac{1}{4}=\frac{45}{4}$

This improper fraction will now be divided by 5 to find the average of 5 runs:

$\displaystyle \frac{\frac{45}{4}}{5}=\frac{9}{4}$

We now need to separate the improper fraction to a mixed number, let's just divide 9 by 4 to get 2 as the quotient and 1 for the remainder, thus

$\boxed{\displaystyle \frac{9}{4}=2\ \frac{1}{4}}$

6 0

3 years ago

Find the y-intercept and slope of the function graph and find the equation for the function

FrozenT [24]

Answer:

y=-x+5 the y intercept is 5 and the slope is -1

Step-by-step explanation:

7 0

2 years ago

Given below are the graphs of two lines, y=-0.5 + 5 and y=-1.25x + 8 and several regions and points are shown. Note that C is th

zalisa [80]

We have the following equations:

$(1) \ y=-0.5x+5 \\ (2) \ y=-1.25x+8$

So we are asked to write a system of equations or inequalities for each region and each point.

Part a)

Region Example A

$y \leq -0.5x+5 \\ y \leq -1.25x+8$

Region B.

Let's take a point that is in this region, that is:

$P(0,6)$

So let's find out the signs of each inequality by substituting this point in them:

$y \ (?)-0.5x+5 \\ 6 \ (?) -0.5(0)+5 \\ 6 \ (?) \ 5 \\ 6\ \textgreater \ 5 \\ \\ y \ (?) \ -1.25x+8 \\ 6 \ (?) -1.25(0)+8 \\ 6 \ (?) \ 8 \\ 6\ \textless \ 8$

So the inequalities are:

$(1) \ y \geq -0.5x+5 \\ (2) \ y \leq -1.25x+8$

Region C.

A point in this region is:

$P(0,10)$

So let's find out the signs of each inequality by substituting this point in them:

$y \ (?)-0.5x+5 \\ 10 \ (?) -0.5(0)+5 \\ 10 \ (?) \ 5 \\ 10\ \textgreater \ 5 \\ \\ y \ (?) \ -1.25x+8 \\ 10 \ (?) -1.25(0)+8 \\ 10 \ (?) \ 8 \\ 10 \ \ \textgreater \ \ 8$

So the inequalities are:

$(1) \ y \geq -0.5x+5 \\ (2) \ y \geq -1.25x+8$

Region D.

A point in this region is:

$P(8,0)$

So let's find out the signs of each inequality by substituting this point in them:

$y \ (?)-0.5x+5 \\ 0 \ (?) -0.5(8)+5 \\ 0 \ (?) \ 1 \\ 0 \ \ \textless \ \ 1 \\ \\ y \ (?) \ -1.25x+8 \\ 0 \ (?) -1.25(8)+8 \\ 0 \ (?) \ -2 \\ 0 \ \ \textgreater \ \ -2$

So the inequalities are:

$(1) \ y \leq -0.5x+5 \\ (2) \ y \geq -1.25x+8$

Point P:

This point is the intersection of the two lines. So let's solve the system of equations:

$(1) \ y=-0.5x+5 \\ (2) \ y=-1.25x+8 \\ \\ Subtracting \ these \ equations: \\ 0=0.75x-3 \\ \\ Solving \ for \ x: \\ x=4 \\ \\ Solving \ for \ y: \\ y=-0.5(4)+5=3$

Accordingly, the point is:

$\boxed{p(4,3)}$

Point q:

This point is the $x-intercept$ of the line:

$y=-0.5x+5$

So let:

$y=0$

Then

$x=\frac{5}{0.5}=10$

Therefore, the point is:

$\boxed{q(10,0)}$

Part b)

The coordinate of a point within a region must satisfy the corresponding system of inequalities. For each region we have taken a point to build up our inequalities. Now we will take other points and prove that these are the correct regions.

Region Example A

The origin is part of this region, therefore let's take the point:

$O(0,0)$

Substituting in the inequalities:

$y \leq -0.5x+5 \\ 0 \leq -0.5(0)+5 \\ \boxed{0 \leq 5} \\ \\ y \leq -1.25x+8 \\ 0 \leq -1.25(0)+8 \\ \boxed{0 \leq 8}$

It is true.

Region B.

Let's take a point that is in this region, that is:

$P(0,7)$

Substituting in the inequalities:

$y \geq -0.5x+5 \\ 7 \geq -0.5(0)+5 \\ \boxed{7 \geq \ 5} \\ \\ y \leq \ -1.25x+8 \\ 7 \ \leq -1.25(0)+8 \\ \boxed{7 \leq \ 8}$

It is true

Region C.

Let's take a point that is in this region, that is:

$P(0,11)$

Substituting in the inequalities:

$y \geq -0.5x+5 \\ 11 \geq -0.5(0)+5 \\ \boxed{11 \geq \ 5} \\ \\ y \geq \ -1.25x+8 \\ 11 \ \geq -1.25(0)+8 \\ \boxed{11 \geq \ 8}$

It is true

Region D.

Let's take a point that is in this region, that is:

$P(9,0)$

Substituting in the inequalities:

$y \leq -0.5x+5 \\ 0 \leq -0.5(9)+5 \\ \boxed{0 \leq \ 0.5} \\ \\ y \geq \ -1.25x+8 \\ 0 \geq -1.25(9)+8 \\ \boxed{0 \geq \ -3.25}$

It is true

7 0

4 years ago