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3 years ago

What is 9 plus 10???

Mathematics

1 answer:

Anon25 [30]3 years ago

8 0

Answer:

Here’s legitimate proof that 9+10=21

(9 + 10) (base x) = 21 (base y)

9(1) + [1(x) + 0(1)] = 2(y) + 1

Simplify and solve for y:

2y = 8 + x

y = 4 + x/2

Since we have number bases, we want x and y to be positive integers. The term x/2 requires that x be a positive even number.

Also since 9 is in base x, we have x ≥ 10, as the digit 9 would not be used for a base 9 or smaller.

Thus we have the pairs of solutions:

x = 10, so y = 9

x = 12, so y = 10

x = 14, so y = 12

…

x, y = 4 + x/2 … Therefore 9+10=21!

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3 years ago

What is the quotient?

Feliz [49]

Answer:

Step-by-step explanation:

the 4 is already in all of the answer choices dividing exponents is subtracting the smaller numbers. since 8-(-4)=12 it is 4 x 10 to the twelfth power.

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4 years ago

If you can’t see the bottom part it’s an 8 ..so anybody ??! Help

jolli1 [7]

Multiply both sides by the reciprocal (-8/7). You would get 40/7 or 5.71428571

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3 years ago

Which equation can be used to solve ? HELPPP

cluponka [151]

Option D:

$\left[\begin{array}{l}x_{1} \\x_{2}\end{array}\right]=\left[\begin{array}{cc}0.5 & -3 \\0 & 1\end{array}\right]\left[\begin{array}{c}2 \\-3\end{array}\right]$

Solution:

Given equation:

$\left[\begin{array}{ll}2 & 6 \\0 & 1\end{array}\right]\left[\begin{array}{l}x_{1} \\x_{2}\end{array}\right]=\left[\begin{array}{l}2 \\-3\end{array}\right]$

where

$A=\left[\begin{array}{ll}2 & 6 \\0 & 1\end{array}\right]$ , $X=\left[\begin{array}{l}x_{1} \\x_{2}\end{array}\right]$ , $B =\left[\begin{array}{l}2 \\-3\end{array}\right]$

This is in the form of AX = B.

To solve this equation.

Multiply by $A^{-1}$ on both sides.

$A^{-1}AX=A^{-1}B$

$X=A^{-1} B$

To find $A^{-1}$ using matrix formula:

$$\left[\begin{array}{ll}a & b \\c & d\end{array}\right]^{-1}=\frac{1}{a d-b c}\left[\begin{array}{cc}d & -b \\-c & a\end{array}\right]$

$$\left[\begin{array}{ll}2 & 6 \\0 & 1\end{array}\right]^{-1}=\frac{1}{2\times1- 6\times0}\left[\begin{array}{cc}1 & -6 \\0 & 2\end{array}\right]$

$$=\frac{1}{2}\left[\begin{array}{cc}1 & -6 \\0 & 2\end{array}\right]$

Multiply $\frac{1}{2}$ into inside the matrix.

$$=\left[\begin{array}{cc}\frac{1}{2} & \frac{-6}{2} \\\frac{0}{2} & \frac{2}{2} \end{array}\right]$

$=\left[\begin{array}{cc}0.5 & -3 \\0 & 1\end{array}\right]$

Substitute into $X=A^{-1} B$ , we get

$\left[\begin{array}{l}x_{1} \\x_{2}\end{array}\right]=\left[\begin{array}{cc}0.5 & -3 \\0 & 1\end{array}\right]\left[\begin{array}{c}2 \\-3\end{array}\right]$

This equation can be used to solve the given matrix.

Option D is the correct answer.

3 0

3 years ago

Please answer correctly, will mark brainliest to best answer! Thanks<br><br> worth 50 points

Eva8 [605]

Answer:

0, -14, 0.8, -5/6, square root of 81, square root of 3, pi

Step-by-step explanation:

5 0

3 years ago

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What is 9 plus 10???​

What is 9 plus 10???