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Inessa [10]
3 years ago
6

Mallorie has 3$ in her wallet. If this is 10% of her monthly allowance, what is her monthly allowance

Mathematics
1 answer:
Mamont248 [21]3 years ago
8 0

Answer:

30

Step-by-step explanation:

Write 10% as a decimal then divide 3 by 0.1 (10% as a decimal)

hope this is good  owo

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How to prove this???
swat32
\cos^3 2A + 3 \cos 2A \\
\Rightarrow \cos 2A (\cos^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (\cos^2 2A + 3)  \\
\Rightarrow (\cos^2 A - \sin^2 A) (1 - \sin^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - \sin^2 2A) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - (2\sin A \cos A)(2\sin A \cos A)) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - 4\sin^2 A \cos^2 A) \\ 
\Rightarrow 4(\cos^2 A - \sin^2 A) (1 - \sin^2 A \cos^2 A) 


go to right side now

4( \cos^6 A - \sin^6 A)\\
\Rightarrow 4( \cos^3 A - \sin^3 A)(\cos^3 A + \sin^3 A)

use x^3 - y^3 = (x-y)(x^2 + xy + y^2) and x^3 + y^3 = x^2 - xy + y^2

4( \cos^6 A - \sin^6 A)\\ \Rightarrow 4( \cos^3 A - \sin^3 A)(\cos^3 A + \sin^3 A) \\
\Rightarrow  4(\cos A - \sin A)(\cos^2 A + \cos A \sin A + \sin^2 A) \\
~\quad  \quad\cdot ( \cos A + \sin A)(\cos^2 A - \cos A \sin A + \cos^2 A)

so \sin^2 A + \cos^2 A = 1

4( \cos^6 A - \sin^6 A)\\ \Rightarrow 4(\cos A - \sin A)(\cos^2 A + \cos A \sin A + \sin^2 A) \\ ~\quad \quad\cdot ( \cos A + \sin A)(\cos^2 A - \cos A \sin A + \cos^2 A) \\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 + \cos A \sin A )(1- \cos A \sin A ) \\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 - \cos^2 A \sin^2 A )\\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 - \sin^2 A \cos^2 A ) \\
 \Rightarrow Left hand side
4 0
3 years ago
Simplified form of this expression 5z-8-4z-3z+6
kvasek [131]

Answer:

-2z-2

Step-by-step explanation:

5z-8-4z-3z+6

5z-4z-3z-8+6

-2z-2

7 0
2 years ago
Read 2 more answers
Which of the following are roots of the polynomial function? Check all that apply.
lutik1710 [3]

You know 1 is not a root because the sum of the coeffcients of the equation is 14, not zero.

It is fairly easy to try 3 by synthetic division (see attachment), which tells you that 3 is a root and the remaining quadratic factor is x²-3x-5. The quadratic formula tells you the roots of that factor are

... x = (-b±√(b²-4ac))/(2a) = (3±√29)/2

The appropriate choices are

... C. (3-√29)/2

... D. (3+√29)/2

... F. 3

_____

The quadratic formula tells you the solution to

... ax²+bx+c=0

is x = (-b±√(b²-4ac))/(2a)

We have a=1, b=-3, c=-5.

8 0
2 years ago
A jet plane can carry up to 200,000 liters of fuel. It used to 130,000 liters of fuel during a flights. What percentage of the f
LekaFEV [45]
65%
Step-by-step explanation:
Given: Full capacity= 200000 litres
Amount of fuel used during flight= 130000 litres
Now, finding percentage of fuel used on the flight.
∴ Percentage of fuel used=
Solving it, we will get
∴ 65% of fuel capacity been used in this flight.
7 0
2 years ago
Question 5 and 6 please help me
sp2606 [1]

Problem 5

The function is continuous for the given domain x \ge 6

This is because y = (-5/6)x+5 is itself continuous, and any interval subset of this function is also continuous. We can plug in any real number that is equal to 6 or larger, and get some y output. If we plugged in x = 6, then we'd get

y = (-5/6)x+5

y = (-5/6)*6 + 5

y = -5+5

y = 0

This is the largest y value possible. Why? Because y = (-5/6)x+5 has a negative slope, so the graph is going downhill as you read it from left to right. As x gets bigger, y gets smaller. The smallest x value allowed in the domain produces the largest y value in the range. There is no smallest y value as the y values keep going down forever.

The range is therefore y \le 0

In interval notation, you can write the range as (-\infty, 0]. The square bracket indicates "include this endpoint as part of the range".

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

Problem 6

The function is discrete for this given domain. The domain itself is a discrete list of values. We cannot plug in values between say 0 and 2. We can only substitute one of those values from the list given. Consequently, the y values will also be a list, and not an interval like problem 5 had.

-----------

If you plugged in x = -4, then you should get...

y = (-1/2)*(-4)+2

y = 2+2

y = 4

So the input x = -4 lead the output y = 4

Repeat for x = -2

y = (-1/2)x+2

y = (-1/2)*(-2)+2

y = 1+2

y = 3

and the same for x = 0 as well

y = (-1/2)x+2

y = (-1/2)*0 + 2

y = 0 + 2

y = 2

and x = 2 also

y = (-1/2)x+2

y = (-1/2)*2 + 2

y = -1+2

y = 1

Finally, plug in x = 4

y = (-1/2)x+2

y = (-1/2)*4+2

y = -2+2

y = 0

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

If we plugged each of these x values {-4, -2, 0, 2, 4} one at a time into the equation y = (-1/2)x+2, then we get this list of values {4, 3, 2, 1, 0}

Sorting the values from smallest to largest, we have this range {0, 1, 2, 3, 4}

3 0
2 years ago
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