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bixtya [17]
2 years ago
10

I NEED HELP WITH MATH QUESTION PLEASE HELP ME NO LINKS !!!

Mathematics
2 answers:
lukranit [14]2 years ago
6 0

Answer:

21.8

Step-by-step explanation:

1.9 + 9 = 10.9

10.9 x 2 = 21.8

Margarita [4]2 years ago
3 0

Answer:

21.8

Step-by-step explanation:

2×1.9=3.8

2×9=18

total=21.8

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Help me please ^^ much appreciated
Reil [10]

Answer:

C

Step-by-step explanation:

Using the sine ratio in the right triangle.

sinC = \frac{opposite}{hypotenuse} = \frac{AB}{BC} = \frac{16}{28} , thus

∠ C = sin^{-1} (\frac{16}{28} ) ≈ 34.85° ( to 2 dec. places )

8 0
2 years ago
Find the equation of the line perpendicular to y=−2 that passes through the point (4, −2).
vfiekz [6]

Answer:

x = 4

Step-by-step explanation:

y = - 2 is the equation of a horizontal line parallel to the x- axis.

A perpendicular line is therefore a vertical line parallel to the y- axis with equation

x = c

where c is the value of the x- coordinates the line passes through.

The line passes through (4, - 2 ) with x- coordinate 4 , thus

x = 4 ← equation of perpendicular line

5 0
3 years ago
How do I enlarge a shape
Ulleksa [173]

By drawing?

If so, here: first outline the current shape as big as you want it... Then, erase the first shape!

6 0
3 years ago
Suppose you choose a team of two people from a group of n > 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
3 years ago
We will give brain list please help PS ignore the black stuff it’s private information
Alisiya [41]

Answer:

the last one? i think.

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
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