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Brut [27]
3 years ago
7

Simply the equation 1 1/3

Mathematics
1 answer:
Dennis_Churaev [7]3 years ago
8 0

1.3 i think that is the answer

You might be interested in
The decimal expansion of the number 2 1 / 2​
Molodets [167]

Answer:

{ \tt{ = 2 \frac{1}{2} }} \\  \\   =  \frac{5}{2}  \\  \\ { \tt{ = 2.5}}

4 0
3 years ago
The area of a rectangular field is 2275 m2. The field is enclosed by 200 m of fence. What are the dimensions of the field?
AURORKA [14]
So it is enclosed by 200 m of  fence
perimiter=200 m
it is a rectangle to
p=legnth+legnth+width+width or
p=l+l+w+w or
p=2l+2w or
p=2(l+w)
so
p=200
200=2(l+w)
divide both sides by 2
100=l+w
area=l times w
area=2275
lw=2275
l+w=100
combine and solve
l+w=100
subtract w
l=100-w
subisutute 100-w for l in other eqution
(100-w)(w)=2275
distribute
-w^2+100w=2275
add (w^2-100w) to both sides
0=w^2-100w+2275
factor
find what 2 numbers add up to -100 and multiply to get 2275
guess (or factor 2275 and find factors that add up to -100)
figure out that they are -65 and -35
0=(w-65)(w-35)
set each to zero
0=w-65
0=w-35

solve for w
w=65 or 35
65>35 so
65 m=legnth
35 m=width
 
4 0
3 years ago
How to solve an expression with variables in the exponents?
gtnhenbr [62]

Answer:

  use logarithms

Step-by-step explanation:

Taking the logarithm of an expression with a variable in the exponent makes the exponent become a coefficient of the logarithm of the base.

__

You will note that this approach works well enough for ...

  a^(x+3) = b^(x-6) . . . . . . . . . . . variables in the exponents

  (x+3)log(a) = (x-6)log(b) . . . . . a linear equation after taking logs

but doesn't do anything to help you solve ...

  x +3 = b^(x -6)

There is no algebraic way to solve equations that are a mix of polynomial and exponential functions.

__

Some functions have been defined to help in certain situations. For example, the "product log" function (or its inverse) can be used to solve a certain class of equations with variables in the exponent. However, these functions and their use are not normally studied in algebra courses.

In any event, I find a graphing calculator to be an extremely useful tool for solving exponential equations.

8 0
3 years ago
Solve the proportions
DanielleElmas [232]

Do cross multiply or butterfly.   For example first one. 18/x=6/10  So cross multiply  18×10 =6x 180=6x Divide both sides by 6 to get x alone 180÷6=30 So x =30  Do the same to rest

4 0
3 years ago
I need help with these two answers please :/
4vir4ik [10]
Rise over run is -2/4 or simplified as -1/2
The second one is
\frac{1 - 3}{4 - 0}  =  \frac{ - 2}{4}

7 0
3 years ago
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