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2 years ago

Help me with number 5 and number 7 please!!

Mathematics

1 answer:

evablogger [386]2 years ago

6 0

5. Because we don't have established rules, we have to make our own rules. We know that we have to translate to the left 9 units and down 1 unit. Going to the left 9 units is subtracting 9 from x, so our rule is (x-9). Going down 1 unit is subtracting 1 from y, so our rule is (y-1). A rule to use when you're not sure whether it's x, y, positive or negative is this:

When the problem says,

right: add to x

left: subtract from x

up: add to y

down: subtract from y

Now that we have our rules, we can solve. Substitute our original points into the rules to get our new points.

A: (11, 9) --> after substituting, (11-9), (9-1)

simplify: (2, 8)

B: (11, 3) --> after substituting, (11-9), (3-1)

simplify: (2, 2)

C: (5, 3) --> after substituting, (5-9), (3-1)

simplify: (-4, 2)

D: (5, 9) --> after substituting, (5-9), (9-1)

simplify: (-4, 8)

Final Coordinates:

A'= (2, 8)

B'= (2, 2)

C'= (-4, 2)

D'= (-4, 8)

7. Because there is already a rule that they gave us, we can apply it to each point to find the post image coordinates.

The rules are (x+7) and (y+2).

We can substitute each point's x and y coordinates into the x and y values in the rules.

A: (-5, 1) --> after substituting, (-5+7), (1+2)

simplify: (2, 3)

B: (3, -7) --> after substituting, (3+7), (-7+2)

simplify: (10, -5)

C: (-5, -15) --> after substituting, (-5+7), (-15+2)

simplify: (2, -13)

Final Coordinates:

A'= (2, 3)

B'= (10, -5)

C'= (2, -13)

hope this helped! comment if something doesn't make sense and i'll try to explain as best as i can. have a nice day!

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Answer:

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Step-by-step explanation:

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frutty [35]

<h2>Answer with explanation:</h2>

We are asked to prove by the method of mathematical induction that:

$2^n\leq 2^{n+1}-2^{n-1}-1$

where n is a positive integer.

Let us take n=1

then we have:

$2^1\leq 2^{1+1}-2^{1-1}-1\\\\i.e.\\\\2\leq 2^2-2^{0}-1\\\\i.e.\\2\leq 4-1-1\\\\i.e.\\\\2\leq 4-2\\\\i.e.\\\\2\leq 2$

Hence, the result is true for n=1.

Let us assume that the result is true for n=k

i.e.

$2^k\leq 2^{k+1}-2^{k-1}-1$

Now, we have to prove the result for n=k+1

i.e.

<u>To prove:</u> $2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Let us take n=k+1

Hence, we have:

$2^{k+1}=2^k\cdot 2\\\\i.e.\\\\2^{k+1}\leq 2\cdot (2^{k+1}-2^{k-1}-1)$

( Since, the result was true for n=k )

Hence, we have:

$2^{k+1}\leq 2^{k+1}\cdot 2-2^{k-1}\cdot 2-2\cdot 1\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{k-1+1}-2\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-2$

Also, we know that:

$-2$

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Since, for n=k+1 being a positive integer we have:

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Hence, the result holds true for n=k+1

Hence, we may infer that the result is true for all n belonging to positive integer.

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$2^n\leq 2^{n+1}-2^{n-1}-1$ where n is a positive integer.

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3 years ago

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iren2701 [21]

Answer:

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Step-by-step explanation:

Let's find the answer.

Because we have 3 equations and 3 variables (x1, x2, x3) a 3x3 matrix (A) can be constructed by using their respectively coefficients.

Equations:

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Eq. 2 : x1 + x2 + x3 = 6

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From eq. 1 : 1 ; 2 ; 5

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So matrix A is:

$\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]$

And the vector of vriables (X) is:

$\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]$

Now we can find the resulting vector (B) using the 'resulting values' from each equation:

$\left[\begin{array}{ccc}5\\6\\7\end{array}\right]$

In conclusion, AX=B is:

$\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]$

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