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olga nikolaevna [1]
3 years ago
7

A triangle has one side that lies along the line y=1/4x and another that lies along the line y=-1/4x. Which of the following poi

nts could be a vertex of the triangle?
Mathematics
1 answer:
mestny [16]3 years ago
4 0

Answer:

We know that our triangle has one side along the line:

y = (1/4)*x

And other side along the line:

y = -(1/4)*x.

Now, we want to find the vertex.

And we know that the vertex is the point where the two sides conect, so the vertex must be a common point of both lines.

Then we have:

y = (1/4)*x = -(1/4)*x

x = -x

The only solution to that equation is x = 0.

now we evaluate our lines in x = 0 and get:

y = (1/4)*0 = 0

y = -(1/4)*0 = 0

Then the lines intersect in the point (0, 0)

Then the vertex must be in the point (0, 0)

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3 years ago
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Given a triangle with two sides that measure 6 m and 10 m and the included angle measuring 35°, what would be the area? Round an
kow [346]

Answer:

89 degrees

Step-by-step explanation:

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7 0
3 years ago
In the derivation of Newton’s method, to determine the formula for xi+1, the function f(x) is approximated using a first-order T
dimaraw [331]

Answer:

Part A.

Let f(x) = 0;

suppose x= a+h

such that f(x) =f(a+h) = 0

By second order Taylor approximation, we get

f(a) + hf'±(a) + \frac{h^{2} }{2!}f''(a) = 0

h = \frac{-f'(a) }{f''(a)} ± \frac{\sqrt[]{(f'(a))^{2}-2f(a)f''(a) } }{f''(a)}

So, we get the succeeding equation for Newton's method as

x_{i+1} = x_{i} + \frac{1}{f''x_{i}}  [-f'(x_{i}) ± \sqrt{f(x_{i})^{2}-2fx_{i}f''x_{i} } ]

Part B.

It is evident that Newton's method fails in two cases, as:

1.  if f''(x) = 0

2. if f'(x)² is less than 2f(x)f''(x)    

Part C.

In case  x_{i+1} is close to x_{i}, the choice that shouldbe made instead of ± in part A is:

f'(x) = \sqrt{f'(x)^{2} - 2f(x)f''(x)}  ⇔ x_{i+1} = x_{i}

Part D.

As given x_{i+1} = x_{i} = h

or                 h = x_{i+1} - x_{i}

We get,

f(a) + hf'(a) +(h²/2)f''(a) = 0

or h² = -hf(a)/f'(a)

Also,             (x_{i+1}-x_{i})² = -(x_{i+1}-x_{i})(f(x_{i})/f'(x_{i}))

So,                f(a) + hf'(a) - (f''(a)/2)(hf(a)/f'(a)) = 0

It becomes   h = -f(a)/f'(a) + (h/2)[f''(a)f(a)/(f(a))²]

Also,             x_{i+1} = x_{i} -f(x_{i})/f'(x_{i}) + [(x_{i+1} - x_{i})f''(x_{i})f(x_{i})]/[2(f'(x_{i}))²]

6 0
3 years ago
I need to write the equation in slope intercept form that passes through the points
VMariaS [17]
(x1, y1) (x2, y2)
(-4, 5)    (-1, 1)
x2-x1 
-1 - -4
-1+4
   3
y2-y1
1- -4
1+4
  5
 
answer : (3,5)
I believe this is right, hope this helps!


4 0
3 years ago
Will give Brainliest<br><br> Summations with Algebraic Expressions<br><br> Image Shown
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7 0
3 years ago
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