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Licemer1 [7]
2 years ago
14

What are the coordinates of point H? ​

Mathematics
1 answer:
Blababa [14]2 years ago
3 0

Answer:

The coordinates of H are (-4,1).

Step-by-step explanation:

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Answer:

fraction 1 and 3 over 5

Step-by-step explanation:

2:3 is equal to 8 blocks

5:12 is equal to 5 blocks

if you divide them, the result is:

8/5 which is the same as 1 3/5

because 5 is contained 1 time in 8 and the rest is 3

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1. Solve log x=4
Helen [10]
1. A. 10,000

Explanation: 10^4 = 10,000

2. x = -12

Explanation: see photo
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3 years ago
Write a quadratic equation could some please help asap​
True [87]

Answer:

f(x) = x^2 + 3x + 2.

Step-by-step explanation:

The zeroes of the quadratic function are -1 and -2 so we can write it as:

f(x) = a(x + 1)(x + 2)    where  a is some constant.

Now, when x = 1,  f(x) = 6 so

6 = a(1+1)(1+2)

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a = 1.

So the function is f(x) = (x + 1)(x + 2) = x^2 + 3x + 2.

6 0
3 years ago
the Vertices of a hyperbola are at (-5, -2) and (-5, 12) and the point (-5, 30) is one of its foci. What is the equation of its
SpyIntel [72]
With the provided vertices and the focus point, the hyperbola will look like the one in the picture below.

notice the "c" distance from the center to the focus point.

since it's a vertical hyperbola, the positive fraction will be the one with the "y" in it, its center is clearly half-way between the vertices at -5, 5.

its major axis or traverse axis goes from -2 up to 12, so is 14 units long, therefore the "a" component is half that, or 7.

\bf \textit{hyperbolas, vertical traverse axis }
\\\\
\cfrac{(y- k)^2}{ a^2}-\cfrac{(x- h)^2}{ b^2}=1
\qquad 
\begin{cases}
center\ ( h, k)\\
vertices\ ( h,  k\pm a)\\
c=\textit{distance from}\\
\qquad \textit{center to foci}\\
\qquad \sqrt{ a ^2 + b ^2}
\end{cases}\\\\
-------------------------------

\bf \begin{cases}
h=-5\\
k=5\\
a=7\\
c=25
\end{cases}\implies \cfrac{(y-5)^2}{7^2}-\cfrac{[x-(-5)]^2}{b}=1
\\\\\\
\cfrac{(y-5)^2}{7^2}-\cfrac{(x+5)^2}{b}=1\\\\
-------------------------------\\\\
c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b\implies \sqrt{25^2-7^2}=b
\\\\\\
\sqrt{625-49}=b\implies \sqrt{576}=b\implies 24=b\\\\
-------------------------------\\\\
\cfrac{(y-5)^2}{7^2}-\cfrac{(x+5)^2}{24^2}=1\implies \cfrac{(y-5)^2}{49}-\cfrac{(x+5)^2}{576}=1

8 0
2 years ago
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