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Rufina [12.5K]
3 years ago
14

Drag the symbol to complete the solution to this inequality.

Mathematics
2 answers:
tankabanditka [31]3 years ago
6 0

Answer:

the 3 symbol on the right

Tomtit [17]3 years ago
6 0

Answer: see picture below

Step-by-step explanation: see the correct answer below

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HELP PLEASE!!! Which equation can be used to calculate the area of the shaded triangle in the figure below?
kykrilka [37]
It’s the first one 44 square feet
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If you were to solve the following system by substitution, what would be the best variable to solve for and from what equation?
fgiga [73]
X in the first equation

because 3x + 6y = 9 can be reduced by dividing by 3, thus, giving u 
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7 0
3 years ago
Boris choos three diffrent numbers.the sum of the three numbers is 36.One of trh numbers is a cube number.the other two number a
inna [77]

Answer:

4,5,27

Problem:

Boris chose three different numbers.

The sum of the three numbers is 36.

One of the numbers is a perfect cube.

The other two numbers are factors of 20.

Step-by-step explanation:

Let's pretend those numbers are:

a,b, \text{ and } c.

We are given the sum is 36: a+b+c=36.

One of our numbers is a perfect cube. a=n^3 where n is an integer.

The other two numbers are factors of 20. bk=20 and ci=20 where a,c,i, \text{ and } k \text{ are integers}.

n^3+\frac{20}{k}+\frac{20}{i}=36

From here I would just try to find numbers that satisfy the conditions using trial and error.

3^3+\frac{20}{2}+\frac{20}{2}

27+10+10

47

3^3+\frac{20}{4}+\frac{20}{5}

27+5+4

36

So I have found a triple that works:

27,5,4

The numbers in ascending order is:

4,5,27

4 0
3 years ago
Evaluate 9P2- need to make a good grade
Svetllana [295]
81, if you mean 9 Squared.
3 0
3 years ago
A small painting has an area of 400 cm^2. The length is 4 more than twice the width. Find the dimensions of the pool. Solve by c
anygoal [31]
A = l*w
l = 2w + 4
400 = (2w + 4)*w
0 = 2w^2 + 4w - 400

Then use the quadratic equation where a=2, b=4, and c=-400.

w = <span>13.2
l = 30.4
</span>


6 0
3 years ago
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