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2 years ago

Can someone please help me out with this! Im struggling a lot with this! x=1 y=-5

Mathematics

2 answers:

Alla [95]2 years ago

4 0

Answer: See attached.

Step-by-step explanation:

When we have a simple x = a equation, the line will be vertical since all y values of that function/equation are equal to a. In this case, that is 1.

Similar thing for a y = b equation. The line will be horizontal since all x values of that function/equation are equal to b. In this case, that is -5.

Tems11 [23]2 years ago

4 0

Answer: Refer to below

Step-by-step explanation:

Given equations

x = 1

y = -5

Determine the line of x = 1

The slope is undefined for any line that is x = a, where a is a constant. This is because the line is vertical, which means that rise/run = ∞ / 0

Thus, x = 1 should be drawn vertically across the point of (1, 0)

Determine the line of y = -5

The slope is 0 for any line that is y = a, where a is a constant. This is because the line is horizontal, which means that rise/run = 0 / ∞

Thus, y = -5 should be drawn horizontally across the point of (0, -5)

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4. Find the standard from of the equation of a hyperbola whose foci are (-1,2), (5,2) and its vertices are end points of the dia

Ad libitum [116K]

The standard form of the equation of the hyperbola that satisfies all conditions is (x - 2)²/4 - (y - 2)²/5 = 1 .

<h3>How to find the standard equation of a hyperbola</h3>

In this problem we must determine the equation of the hyperbola in its standard form from the coordinates of the foci and a general equation of a circle. Based on the location of the foci, we see that the axis of symmetry of the hyperbola is parallel to the x-axis. Besides, the center of the hyperbola is the midpoint of the line segment with the foci as endpoints:

(h, k) = 0.5 · (- 1, 2) + 0.5 · (5, 2)

(h, k) = (2, 2)

To determine whether it is possible that the vertices are endpoints of the diameter of the circle, we proceed to modify the general equation of the circle into its standard form.

If the vertices of the hyperbola are endpoints of the diameter of the circle, then the center of the circle must be the midpoint of the line segment. By algebra we find that:

x² + y² - 4 · x - 4 · y + 4= 0

(x² - 4 · x + 4) + (y² - 4 · y + 4) = 4

(x - 2)² + (y - 2)² = 2²

The center of the circle is the midpoint of the line segment. Now we proceed to determine the vertices of the hyperbola:

V₁(x, y) = (0, 2), V₂(x, y) = (4, 2)

And the distance from the center to any of the vertices is 2 (semi-major distance, a) and the semi-minor distance is:

b = √(c² - a²)

b = √(3² - 2²)

b = √5

Therefore, the standard form of the equation of the hyperbola that satisfies all conditions is (x - 2)²/4 - (y - 2)²/5 = 1 .

To learn more on hyperbolae: brainly.com/question/27799190

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5 0

2 years ago

If he = 7x -27 find the value of x

Amiraneli [1.4K]

$\\ \sf\longmapsto 7x-27=1$

$\\ \sf\longmapsto 7x=1+27$

$\\ \sf\longmapsto 7x=28$

$\\ \sf\longmapsto x=\dfrac{28}{7}$

$\\ \sf\longmapsto x=4$

5 0

3 years ago

Need Help ASAP algebra 2

Anastasy [175]

Answer:

a = √10

Step-by-step explanation:

a² + b² = c²

a² + 9² = (√91)²

a² + 81 = 91

a² = 10

a = √10

hope this helps :)

7 0

4 years ago

Read 2 more answers

A= 3x3x3x3x3x5x7x7x7

olga nikolaevna [1]

Answer:

LCM(57624, 416745) = 23,337,720 or LCM = 23337720

Step-by-step explanation:

List all prime factors for each number.

Prime Factorization of 57624 is:

2 x 2 x 2 x 3 x 7 x 7 x 7 x 7 => 23 x 31 x 74

Prime Factorization of 416745 is:

3 x 3 x 3 x 3 x 3 x 5 x 7 x 7 x 7 => 35 x 51 x 73

For each prime factor, find where it occurs most often as a factor and write it that many times in a new list.

The new superset list is

2, 2, 2, 3, 3, 3, 3, 3, 5, 7, 7, 7, 7

Multiply these factors together to find the LCM.

LCM = 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 5 x 7 x 7 x 7 x 7 = 23337720

In exponential form:

LCM = 23 x 35 x 51 x 74 = 23337720

LCM = 23337720

Therefore,

LCM(57624, 416745) = 23,337,720

3 0

2 years ago

Plz help me!!!!

stich3 [128]

So we have the exponential equation: $y=18582(0.90)^x$
where
$y$ is the number of visitors
$x$ are the weekends since opening

1. Since the opening weekend is the day of the the park's opening, we just need to replace $x$ with $0$ in our exponential equation to get the number of visitors on the opening weekend:
$y=18582(0.90)^x$
$y=18582(0.90)^0$
$y=18582(1)$
$y=18582$

We can conclude that there were 18582 visitors on the opening weekend.

2. Just like before, to find the approximate number of visitors on the ninth weekend, we just need to replace $x$ with $9$ in our exponential equation:
$y=18582(0.90)^x$
$y=18582(0.90)^9$
$y=7199.0475$
And rounded to the nearest integer:
$y=7199$

We can conclude that there were approximately 7199 visitors on the $9^{th}$ weekend.

3. Here we know that the number of visitors is 15051. Since $y$ represents the number of visitors, we just need to replace $y$ with 15051 in our exponential equation and solve for $x$ :
$y=18582(0.90)^x$
$15051=18582(0.90)^x$
$\frac{15051}{18582} =0.90^x$
$0.90^x= \frac{5017}{6194}$
$ln(0.90^x)=ln(\frac{5017}{6194})$
$xln(0.90)=ln(\frac{5017}{6194})$
$x= \frac{ln(\frac{5017}{6194})}{ln(0.90)}$
$x=2.0002$
And rounded to the nearest whole number:
$x=2$

We can conclude that the number of visitors was about 15,051 after 2 weekends.

4 0

3 years ago