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fomenos
3 years ago
12

Help! I'll give brainliest to the first correct answer! ​

Mathematics
1 answer:
CaHeK987 [17]3 years ago
5 0

Answer:

I think it's D but I'm not sure

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Which ordered pairs in the form (x, y)(x, y) are solutions to the equation
kap26 [50]
The answer is (-1,-7) because you are pluging it in -1 is x and -7 is y. so 7×-1=-7 and 5×-7 = -35 so -7 minus -35 =28
6 0
3 years ago
What is the focus of the parabola? <img src="https://tex.z-dn.net/?f=y%3D%20%5Cfrac%7B1%7D%7B4%7D%20x%5E%7B2%7D%20-x%2B3" id="Te
OlgaM077 [116]
We know that
the equation of the parabola is of the form
y=ax²+bx+c
in this problem
y=1/4x²−x+3
where
a=1/4
b=-1
c=3
 the coordinates of the focus are
(-b/2a,(1-D)/4a)
where D is the discriminant b²-4ac
D=(-1)²-4*(1/4)*3-----> D=1-3---> D=-2
therefore
x coordinate of the focus
-b/2a----> 1/[2*(-1/4)]----> 2

y coordinate of the focus
(1-D)/4a------> (1+2)/(4/4)---> 3

the coordinates of the focus are (2,3)
3 0
3 years ago
Read 2 more answers
PLEASE HELP IM ON A TIME LIMIT :(
Mice21 [21]

Answer:

hey hope this helps

<h3 /><h3>Comparing sides AB and DE </h3>

AB =

\sqrt{ {1}^{2} +  {1}^{2}  }

=  \sqrt{2}

DE

= \sqrt{ {(3 - 5)}^{2} +  {(1 + 1}^{2}  }  \\   = \sqrt{ {( - 2)}^{2}  +  {(2)}^{2} }  \\    = \sqrt{4 + 4}  \\  =  \sqrt{8}  \\   = 2 \sqrt{2}

So DE = 2 × AB

and since the new triangle formed is similar to the original one, their side ratio will be same for all sides.

<u>scale factor</u> = AB/DE

= 2

It's been reflected across the Y-axis

<em>moved thru the translation of 3 units towards the right of positive x- axis </em>

for this let's compare the location of points B and D

For both the y coordinate is same while the x coordinate of B is 0 and that of D is 3

so the triangle has been shifted by 3 units across the positive x axis

7 0
2 years ago
Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that
kherson [118]

Answer:

The answer is

f(x) = {\displaystyle  8 + \frac{-28}{1}(x+2)+\frac{-36}{2!}(x+2)^2 + \frac{-18}{3!}(x+2)^2 }

Step-by-step explanation:

Remember that Taylor says that

f(x) = {\displaystyle \sum\limits_{k=0}^{\infty} \frac{f^{(k)}(a) }{k!}(x-a)^k }

For this case

f^{(0)} (-2) = 8(-2)-3(-2)^3 = 8\\f^{(1)} (-2) = 8-3(3)(-2)^2 = -28\\f^{(2)} (-2) = -3(3)2(-2) = -36\\f^{(2)} (-2) = -3(3)2 = -18

f(x) = {\displaystyle  8 + \frac{-28}{1}(x+2)+\frac{-36}{2!}(x+2)^2 + \frac{-18}{3!}(x+2)^2 }

5 0
3 years ago
Help me with this math question please I'm giving away brainliests​
anyanavicka [17]
The answer is A.All real numbers except -2.
7 0
3 years ago
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