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Alla [95]
3 years ago
13

Write these equations in Standard Form: Ax + By=C 1) y=-5x

Mathematics
1 answer:
worty [1.4K]3 years ago
5 0
<h3>5x +y= 0 [Here , A =  5 , B= 1 and C = 0]</h3>

Step-by-step explanation:

The stander form is Ax + By = C

Therefore

y = -5x

⇔ y+ 5x = 0

⇔ 5x +y= 0

Here , A =  5 , B= 1 and C = 0

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A cylinder and a cone have the same volume. The cylinder has radius x and height y. The cone has radius 2x. Find the height of t
weqwewe [10]
V(cylinder)=πR²H
Radius of the cylinder R=x, height of the cylinder H=y.
We can write for the cylinder
V(cylinder)=πx²y

V(cone) =(1/3)πr²h
Radius of the cone r=2x.
We can write for the cone
V(cone)= (1/3)π(2x)²h=(1/3)π *4*x²h

V(cylinder) =V(cone)
πx²y=(1/3)π *4*x²h
y=(4/3)*h
h=(3/4)*y


7 0
3 years ago
Solve this system of equations.<br> y = 2x – 12<br> y = -x + 3
3241004551 [841]
X = 5, y = -2
-x +3 = 2x - 12
-3x = -15
3x = 15
x = 5
y = -5+3
y=-2
4 0
3 years ago
What is the awnser for 15+12x-5x+44-7
Allushta [10]
The answer is 52+7x I hope this helps
3 0
3 years ago
Describe the behavior of the function ppp around its vertical asymptote at x=-2x=−2x, equals, minus, 2. ​
insens350 [35]

Answer:

x->-2^{-}, p(x)->-\infty and as x->-2^{+}, p(x)->-\infty

Step-by-step explanation:

Given

p(x) = \frac{x^2-2x-3}{x+2} -- Missing from the question

Required

The behavior of the function around its vertical asymptote at x = -2

p(x) = \frac{x^2-2x-3}{x+2}

Expand the numerator

p(x) = \frac{x^2 + x -3x - 3}{x+2}

Factorize

p(x) = \frac{x(x + 1) -3(x + 1)}{x+2}

Factor out x + 1

p(x) = \frac{(x -3)(x + 1)}{x+2}

We test the function using values close to -2 (one value will be less than -2 while the other will be greater than -2)

We are only interested in the sign of the result

----------------------------------------------------------------------------------------------------------

As x approaches -2 implies that:

x -> -2^{-} Say x = -3

p(x) = \frac{(x -3)(x + 1)}{x+2}

p(-3) = \frac{(-3-3)(-3+1)}{-3+2} = \frac{-6 * -2}{-1} = \frac{+12}{-1} = -12

We have a negative value (-12); This will be called negative infinity

This implies that as x approaches -2, p(x) approaches negative infinity

x->-2^{-}, p(x)->-\infty

Take note of the superscript of 2 (this implies that, we approach 2 from a value less than 2)

As x leaves -2 implies that: x>-2

Say x = -2.1

p(-2.1) = \frac{(-2.1-3)(-2.1+1)}{-2.1+2} = \frac{-5.1 * -1.1}{-0.1} = \frac{+5.61}{-0.1} = -56.1

We have a negative value (-56.1); This will be called negative infinity

This implies that as x leaves -2, p(x) approaches negative infinity

x->-2^{+}, p(x)->-\infty

So, the behavior is:

x->-2^{-}, p(x)->-\infty and as x->-2^{+}, p(x)->-\infty

6 0
3 years ago
Write an algebraic rule to describe the translation c (5,-4) c'(-2,1)
Sauron [17]
We can solve this by using the formula:
(x, y) (x + a, y + b) = (5,-4) (-2,1)

So, plugging in the values and solving for a and b,
5 + a = -2
a = -8

-4 + b = 1
b = 5

Therefore, the translation is
(x,y) (x - 8, y +5)
5 0
2 years ago
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