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Lady bird [3.3K]
3 years ago
9

On. f(3) = 2 +1; Find f (4)

Mathematics
1 answer:
Korolek [52]3 years ago
8 0

Answer:

f(3) = 2 +1; Find f (4)

Step-by-step explanation:

f(3) = 2 +1; Find f (4)

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Please hurry..........
Montano1993 [528]

Answer:

I think the answer is C

Step-by-step explanation:

5 0
3 years ago
I am 24 years old. My age is 2 less than twice my sister's age.
gavmur [86]

Answer: Therefore, the age of my sister is 13 years.

Step-by-step explanation:

Let the age of my sister is x yeas.

Then according to question..

24 = 2x -2

Or, 2x=26

or, x= 26/2

x= 13

4 0
2 years ago
Read 2 more answers
Find 4 consecutive even integers where the product of the two smaller numbers is 72 less than the product of the two larger numb
WARRIOR [948]

Answer: 6, 8, 10, 12

Step-by-step explanation:

Given that x is the number, the 4 numbers would be

x, x + 2, x + 4, x + 6

so the two smallest numbers would be x and x + 2

and the two largest numbers would be x+4 and x+6

now set up an equation

x(x+2) = (x+4)(x+6) - 72

now FOIL

x^2 + 2x = x^2 + 6x + 4x + 24 - 72

combine like terms

x^2 + 2x = x^2 + 10x -48

subtract x^2 from both sides

2x = 10x - 48

subtract 2x from both sides

0 = 8x - 48

add 48 to both sides

48 = 8x

divide both sides by 8

6 = x

so the four numbers, x, x+2, x+4, and x+6 when you plug in x are equal to

6, 8, 10, 12

5 0
3 years ago
Write the following fractions greater than 1as a sum of two products
cestrela7 [59]
What are the following fractions
7 0
3 years ago
Which equation represents a hyperbola with a center at (0, 0), a vertex at (−48, 0), and a focus at (50, 0)?
Lerok [7]

bearing in mind that "a" is the length of the traverse axis, and "c" is the distance from the center to either foci.

we know the center is at (0,0), we know there's a vertex at (-48,0), from the origin to -48, that's 48 units flat, meaning, the hyperbola is a horizontal one running over the x-axis whose a = 48.

we also know there's a focus point at (50,0), that's 50 units from the center, namely c = 50.


\bf \textit{hyperbolas, horizontal traverse axis } \\\\ \cfrac{(x- h)^2}{ a^2}-\cfrac{(y- k)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h\pm a, k)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2}\\ \textit{asymptotes}\quad y= k\pm \cfrac{b}{a}(x- h) \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}


\bf \begin{cases} h=0\\ k=0\\ a=48\\ c=50 \end{cases}\implies \cfrac{(x-0)^2}{48^2}-\cfrac{(y-0)^2}{b^2}=1 \\\\\\ c=\sqrt{a^2+b^2}\implies \sqrt{c^2-a^2}=b\implies \sqrt{50^2-48^2}=b \\\\\\ \sqrt{196}=b\implies 14=b~\hspace{3.5em}\cfrac{(x-0)^2}{48^2}-\cfrac{(y-0)^2}{14^2}=1\implies \cfrac{x^2}{48^2}-\cfrac{y^2}{14^2}=1

8 0
3 years ago
Read 2 more answers
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