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3 years ago

Which of the following is the correct representation for a translation 8 units right and 4 units down?

Mathematics

1 answer:

Zolol [24]3 years ago

5 0

Answer:

A. (x + 8, y - 4)

Step-by-step explanation:

When something is translated to the right, x goes up, and when something is translated down, y goes down.

Alternatively, eft means down for x and up means up for y.

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How would the decimal point move to solve the equation 610 ÷ 103 = ________?

snow_tiger [21]

It moves to the left no matter what it is, but i’m not sure how many bec there’s not alr a decimal

7 0

3 years ago

Read 2 more answers

Prove that for any natural n the number 3^(n+4)−3^n is divisible by 16.

gtnhenbr [62]

Answer:

Proved (See Explanation)

Step-by-step explanation:

Show that 3ⁿ⁺⁴ - 3ⁿ is divisible by 16.

This is done as follows

$\frac{3^{n+4} - 3^n}{16}$

From laws of indices;

aᵐ⁺ⁿ = aᵐ * aⁿ.

So, 3ⁿ⁺⁴ can be written as 3ⁿ * 3⁴.

$\frac{3^{n+4} - 3^n}{16}$ becomes

$\frac{3^n * 3^4 - 3^n}{16}$

Factorize

$\frac{3^n(3^4 - 1)}{16}$

$\frac{3^n(81 - 1)}{16}$

$\frac{3^n(80)}{16}$

3ⁿ * 5

5(3ⁿ)

<em>The expression can not be further simplified.</em>

<em>However, we can conclude that when 3ⁿ⁺⁴ - 3ⁿ is divisible by 16, because 5(3ⁿ) is a natural whole number as long as n is a natural whole number.</em>

3 0

3 years ago

Chose parameters h and k such that the system has a) a unique solution, b) many solutions, and c) no solution. X1 + 3x2 = 4 2x1

AysviL [449]

Answer:

a) The system has a unique solution for $k\neq 6$ and any value of $h$ , and we say the system is consisted

b) The system has infinite solutions for $k=6$ and $h=8$

c) The system has no solution for $k=6$ and $h\neq 8$

Step-by-step explanation:

Since we need to base the solutions of the system on one of the independent terms ( $h$ ), the determinant method is not suitable and therefore we use the Gauss elimination method.

The first step is to write our system in the augmented matrix form:

$\left[\begin{array}{cc|c}1&3&4\\2&k&h\end{array}\right]$

The we can use the transformation $r_0\rightarrow r_0 -2r_1$ , obtaining:

$\left[\begin{array}{cc|c}1&3&4\\0&k-6&h-8\end{array}\right]$ .

Now we can start the analysis:

If $k\neq 6$ then, the system has a unique solution for any value of $k$ , meaning that the last row will transform back to the equation as:

$(k-6)x_2=h-8\\x_2=h-8/(k-6)$

from where we can see that only in the case of $k=6$ the value of $x_2$ can not be determined.

if $k=6$ and $h=8$ the system has infinite solutions: this is very simple to see by substituting these values in the equation resulting from the last row:

$(k-6)x_2=h-8\\0=0$ which means that the second equation is a linear combination of the first one. Therefore, we can solve the first equation to get $x_1$ as a function of $x_2$ o viceversa. Thus, $x_2$ ( $x_1$ ) is called a parameter since there are no constraints on what values they can take on.

if $k=6$ and $h\neq 8$ the system has no solution. Again by substituting in the equation resulting from the last row:

$(k-6)x_2=h-8\\0=h-8$ which is false for all values of $h\neq 8$ and since we have something that is not possible $(0\neq h-8,\ \forall \ h\neq 8)$ the system has no solution