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jeka57 [31]
3 years ago
14

Help me please with my math

Mathematics
2 answers:
podryga [215]3 years ago
7 0
= 5 4/6
= 5 2/3 ....this is the answer
Mice21 [21]3 years ago
5 0
I can´t see much of it but you have to make the fractions mixed numbers and then find a common denomonator and the add or subtract them and you will get the answer.
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4x + 2(x − 3) = 3x + x − 12<br> find the answer for x
Nitella [24]
X=  -3...................
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3 years ago
A construction firm is building a football stadium. The building work has taken 499 days so far. It should take another 199 days
Gre4nikov [31]

499

199

+-------  the answer is 698 you have to add 499 and 199.

7 0
2 years ago
A given line has the equation 10x -2.2y What is the equation, in slope-intercept form, of the line that is parallel to given lin
Andre45 [30]
10x + 2y = -2

2y = -10x - 2

y = -(10/2)x - 2/2

y = -5x  - 1    y = mx + c,   slope m = -5.

If it is parallel, then the slope of the second line ought to be -5 as well.

So you should fill -5 in the bracket.
3 0
3 years ago
What’s 50/100 + 90/100 show your work
katen-ka-za [31]

Answer:

140/100 or 1 and 4/10

Step-by-step explanation:

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8 0
3 years ago
Suppose you choose a team of two people from a group of n &gt; 1 people, and your opponent does the same (your choices are allow
jonny [76]

Answer:

The number of possible choices of my team and the opponents team is

 \left\begin{array}{ccc}n-1\\E\\n=1\end{array}\right     i^{3}

Step-by-step explanation:

selecting the first team from n people we have \left(\begin{array}{ccc}n\\1\\\end{array}\right)  = n possibility and choosing second team from the rest of n-1 people we have \left(\begin{array}{ccc}n-1\\1\\\end{array}\right) = n-1

As { A, B} = {B , A}

Therefore, the total possibility is \frac{n(n-1)}{2}

Since our choices are allowed to overlap, the second team is \frac{n(n-1)}{2}

Possibility of choosing both teams will be

\frac{n(n-1)}{2}  *  \frac{n(n-1)}{2}  \\\\= [\frac{n(n-1)}{2}] ^{2}

We now have the formula

1³ + 2³ + ........... + n³ =[\frac{n(n+1)}{2}] ^{2}

1³ + 2³ + ............ + (n-1)³ = [x^{2} \frac{n(n-1)}{2}] ^{2}

=\left[\begin{array}{ccc}n-1\\E\\i=1\end{array}\right] =   [\frac{n(n-1)}{2}]^{3}

4 0
4 years ago
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