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docker41 [41]
3 years ago
15

645 plus 763 to the nearest hundredth

Mathematics
1 answer:
Vedmedyk [2.9K]3 years ago
8 0

Answer:

1408

What you have to do is 645 + 763 = 1408 that's how you doing I think so and sorry if I did it wrong

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Solve the system by elimination: <br> -2x + 2y + 3z = 0<br> -2x - y + z = -3 <br> 2x+ 3y + 3z = 5
Greeley [361]
First= -2x+2y+3z=0

= -2x+2y+3z +-2y=0+(-2y)

-2x+3z=-2y

-2x+3z=-2y+-3z

-2x=-2y-3z

Divide both by negative two
-2x/-2=-2y-3z/-2

x=y+3/2z that’s the answer.


Use this to help you with other equations

3 0
3 years ago
There are 4 grams of fiber in 1/2 cup of oats how many grams of fiber are in 3 1/2
Anna71 [15]

Answer:

28 grams

Step-by-step explanation:

In  cup of oats, amount of fiber is = 4 grams

In 1 cup of oats amount of fiber is =  = 8 grams

So, in  or  cups of oats, amount of fiber is = 28 grams

3 0
3 years ago
Read 2 more answers
–2x + y = 3 2x – y = 4 Using elimination to solve this system results in the equation .
tankabanditka [31]

Answer:

1 = x and -2 = y

I am fairly happy with this answer

6 0
3 years ago
What is the diameter of a 2m circle?
Varvara68 [4.7K]

Answer:

If 2m is the radius, the diameter is 4 meters.

2*2=4

6 0
3 years ago
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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 <em>standard</em> 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 <em>standard</em> form from the coordinates of the foci and a <em>general</em> 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 <em>general</em> equation of the circle into its <em>standard</em> 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 (<em>semi-major</em> distance, a) and the semi-minor distance is:

b = √(c² - a²)

b = √(3² - 2²)

b = √5

Therefore, the <em>standard</em> 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

#SPJ1

5 0
2 years ago
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