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VladimirAG [237]
3 years ago
6

Please fill out the function table.Questions are in the photo.

Mathematics
2 answers:
Alona [7]3 years ago
8 0
I hope this helps you

CaHeK987 [17]3 years ago
4 0
Y=2x+3
X-2 Y=1
X=-1 y=1
X=0y=3
X=1y=5
X=2y=7
------------
Y=-x+6
Y=12
Y=9
Y=4
Y=3
Y=-1
------------
Y=-11x
Y=33
Y=11
Y=-22
Y=-44
Y=-55
---------
Y=-4x-1
Y=19
Y=7
Y=3
Y=-13
Y=-17
---------
Y=5x-4
Y=-34
-19
-4
16
21
-----------
Y=3x-7
-25
-22
-10
-1
1
5
------- that is all
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Write it in an algebraic expression form!
Phoenix [80]
1.

(x/2) - 7 = 11


2.

2x + 7 = 27


3.

2x - 5 = 25




Hope this helped!! (:
7 0
2 years ago
The equation of a line that goes through the points (0,0) and (300000,365)
olga_2 [115]
\bf \begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&&(~ 0 &,& 0~) 
%  (c,d)
&&(~ 300000 &,& 365~)
\end{array}
\\\\\\
% slope  = m
slope =  m\implies 
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{365-0}{300000-0}\\\\\\ \cfrac{365}{300000}\implies \cfrac{73}{60000}

\bf \stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-0=\cfrac{73}{60000}(x-0)\implies y=\cfrac{73}{60000}x
4 0
3 years ago
If M is the midpoint of PQ and X is the midpoint of PM then PX = a PQ Which statement is True? a is equal to 1 a is equal to 1/2
garik1379 [7]

Answer:

D. a is equal to  \frac{1}{4}

Step-by-step explanation:

Let us assume that segment PQ is 8 units, so that its midpoint M is at 4 units to both P and Q.

Given that X is the midpoint of PM, this implies that X is at 2 units to both P and M.

Then;

PX = aPQ

⇒   = a x 8

      = 8a

Therefore, a must be equal to \frac{1}{4}, so that;

PX = 8a

     = 8 x  \frac{1}{4}

     = 2 units

Thus, a is equal to  \frac{1}{4} is the correct option.

6 0
4 years ago
Find f. f ″(x) = x^−2, x > 0, f(1) = 0, f(6) = 0
marin [14]

If you do in fact mean f(1)=f(6)=0 (as opposed to one of these being the derivative of f at some point), then integrating twice gives

f''(x) = -\dfrac1{x^2}

f'(x) = \displaystyle -\int \frac{dx}{x^2} = \frac1x + C_1

f(x) = \displaystyle \int \left(\frac1x + C_1\right) \, dx = \ln|x| + C_1x + C_2

From the initial conditions, we find

f(1) = \ln|1| + C_1 + C_2 = 0 \implies C_1 + C_2 = 0

f(6) = \ln|6| + 6C_1 + C_2 = 0 \implies 6C_1 + C_2 = -\ln(6)

Eliminating C_2, we get

(C_1 + C_2) - (6C_1 + C_2) = 0 - (-\ln(6))

-5C_1 = \ln(6)

C_1 = -\dfrac{\ln(6)}5 = -\ln\left(\sqrt[5]{6}\right) \implies C_2 = \ln\left(\sqrt[5]{6}\right)

Then

\boxed{f(x) = \ln|x| - \ln\left(\sqrt[5]{6}\right)\,x + \ln\left(\sqrt[5]{6}\right)}

3 0
2 years ago
What is 3/8 divided by 4/1
Leviafan [203]
3/32 hope this helps ig
5 0
3 years ago
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