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pochemuha
2 years ago
15

Help me find the value of f(-5)

Mathematics
2 answers:
deff fn [24]2 years ago
8 0

Answer:

F(-5) = 2

Step-by-step explanation:

go over -5 on the x axis, go up two and -5 directly crosses over positive 2

topjm [15]2 years ago
3 0
The answer to this problem is gonna be 2
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Lara randomly surveyed 15 students
Rina8888 [55]

Answer:

hii

Step-by-step explanation:

yes it represents school population

Hope this helps you

Thank you

Gudni8

6 0
3 years ago
How do you solve x^2-4X-12=0
Pepsi [2]
                                         x²  -  4x  - 12  =  0

Factor the left side:      (x + 2) · (x - 6)  =  0

The equation is true if either factor is zero.

If  (x + 2) = 0  then    x = -2 .

If  (x - 6) = 0  then     x = 6 .
3 0
3 years ago
[Geometry Basics] I'm just checking if this is correct:
Leno4ka [110]
\bf ~~~~~~~~~~~~\textit{middle point of 2 points }
\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&&(~ x &,& y~) 
%  (c,d)
&&(~ -9 &,& -1~)
\end{array}\qquad
%   coordinates of midpoint 
\left(\cfrac{ x_2 +  x_1}{2}\quad ,\quad \cfrac{ y_2 +  y_1}{2} \right)
\\\\\\
\left( \cfrac{-9+x}{2}~~,~~\cfrac{-1+y}{2} \right)=\stackrel{midpoint}{(8,14)}\implies 
\begin{cases}
\cfrac{-9+x}{2}=8\\\\
-9+x=16\\
\boxed{x=25}\\
-------\\
\cfrac{-1+y}{2} =14\\\\
-1+y=28\\
\boxed{y=29}
\end{cases}

\bf -------------------------------\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&&(~ x &,& y~) 
%  (c,d)
&&(~10 &,& 12~)
\end{array}\qquad
\\\\\\
\left( \cfrac{10+x}{2}~~,~~\cfrac{12+y}{2} \right)=\stackrel{midpoint}{(6,9)}\implies 
\begin{cases}
\cfrac{10+x}{2}=6\\\\
10+x=12\\
\boxed{x=2}\\
-------\\
\cfrac{12+y}{2} =9\\\\
12+y=18\\
\boxed{y=6}
\end{cases}
3 0
3 years ago
The number of fish in a lake is decreasing by 400 every year, as described by the function below.
FrozenT [24]

Answer:

D: 7800

Can you give me brainliest :>

Step-by-step explanation:

3 0
2 years ago
PLEASE HELP WILL MARK BRAINLIEST
Sholpan [36]

Answer: Choice D

b greater-than 3 and StartFraction 2 over 15 EndFraction

In other words,

b > 3 & 2/15

or

b > 3\frac{2}{15}\\\\

========================================================

Explanation:

Let's convert the mixed number 2 & 3/5 into an improper fraction.

We'll use the rule

a & b/c = (a*c + b)/c

In this case, a = 2, b = 3, c = 5

So,

a & b/c = (a*c + b)/c

2 & 3/5 = (2*5 + 3)/5

2 & 3/5 = (10 + 3)/5

2 & 3/5 = 13/5

The inequality 2 \frac{3}{5} < b - \frac{8}{15}\\\\ is the same as \frac{13}{5} < b - \frac{8}{15}\\\\

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

Let's multiply both sides by 15 to clear out the fractions

\frac{13}{5} < b - \frac{8}{15}\\\\15*\frac{13}{5} < 15*\left(b - \frac{8}{15}\right)\\\\39 < 15b-8\\\\

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

Now isolate the variable b

39 < 15b-8\\\\15b-8 > 39\\\\15b > 39+8\\\\15b > 47\\\\b > \frac{47}{15}\\\\b > \frac{45+2}{15}\\\\b > \frac{45}{15}+\frac{2}{15}\\\\b > 3+\frac{2}{15}\\\\b > 3\frac{2}{15}\\\\

Side note: Another way to go from 47/15 to 3 & 2/15 is to notice how

47/15 = 3 remainder 2

The 3 is the whole part while 2 helps form the fractional part. The denominator stays at 15 the whole time.

7 0
3 years ago
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