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2 years ago

What is the image point of (-8,-1) after a translation left 3 units and up 4 units

Mathematics

1 answer:

Sedaia [141]2 years ago

4 0

Answer:

(-11, 3)

Step-by-step explanation:

When you move -8 to the left its on the x-axis and it's on the negative side. Moving it to the left would mean the number is lower than -8. Moving it to the left 3 units would mean its -11. When you move -1 up 4 units it's on the y axis on the negative side in the beginning. Moving it up would mean the number is higher than -1. Moving -1 up 4 units would mean its 3.

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LUCKY_DIMON [66]

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3 years ago

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azamat

Answer:

$0.75

Step-by-step explanation:

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3 years ago

5. Area = 36cm<br>HM = 12 cm<br>OE = x

Tpy6a [65]

Answer:

I'm going to guess and say that if HM is one side and OE is the other side of a rectangle, then:

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2 years ago

Read 2 more answers

Prove or disprove (from i=0 to n) sum([2i]^4) <= (4n)^4. If true use induction, else give the smallest value of n that it doe

ddd [48]

Answer:

The statement is true for every n between 0 and 77 and it is false for $n\geq 78$

Step-by-step explanation:

First, observe that, for n=0 and n=1 the statement is true:

For n=0: $\sum^{n}_{i=0} (2i)^4=0 \leq 0=(4n)^4$

For n=1: $\sum^{n}_{i=0} (2i)^4=16 \leq 256=(4n)^4$

From this point we will assume that $n\geq 2$

As we can see, $\sum^{n}_{i=0} (2i)^4=\sum^{n}_{i=0} 16i^4=16\sum^{n}_{i=0} i^4$ and $(4n)^4=256n^4$ . Then,

$\sum^{n}_{i=0} (2i)^4 \leq(4n)^4 \iff \sum^{n}_{i=0} i^4 \leq 16n^4$

Now, we will use the formula for the sum of the first 4th powers:

$\sum^{n}_{i=0} i^4=\frac{n^5}{5} +\frac{n^4}{2} +\frac{n^3}{3}-\frac{n}{30}=\frac{6n^5+15n^4+10n^3-n}{30}$

Therefore:

$\sum^{n}_{i=0} i^4 \leq 16n^4 \iff \frac{6n^5+15n^4+10n^3-n}{30} \leq 16n^4 \\\\ \iff 6n^5+10n^3-n \leq 465n^4 \iff 465n^4-6n^5-10n^3+n\geq 0$

and, because $n \geq 0$ ,

$465n^4-6n^5-10n^3+n\geq 0 \iff n(465n^3-6n^4-10n^2+1)\geq 0 \\\iff 465n^3-6n^4-10n^2+1\geq 0 \iff 465n^3-6n^4-10n^2\geq -1\\\iff n^2(465n-6n^2-10)\geq -1$

Observe that, because $n \geq 2$ and is an integer,

$n^2(465n-6n^2-10)\geq -1 \iff 465n-6n^2-10 \geq 0 \iff n(465-6n) \geq 10\\\iff 465-6n \geq 0 \iff n \leq \frac{465}{6}=\frac{155}{2}=77.5$

In concusion, the statement is true if and only if n is a non negative integer such that $n\leq 77$

So, 78 is the smallest value of n that does not satisfy the inequality.

Note: If you compute $(4n)^4- \sum^{n}_{i=0} (2i)^4$ for 77 and 78 you will obtain:

$(4n)^4- \sum^{n}_{i=0} (2i)^4=53810064$

$(4n)^4- \sum^{n}_{i=0} (2i)^4=-61754992$

7 0

3 years ago

the length of AB_ is 9 cm a dilation with a scale factor of 2 is applied to AB_ what is the length of the image of AB after the

Minchanka [31]

The scale factor of two means you need to multiply the original length by 2:

9 x 2 = 18

The new length would be 18 cm.

6 0

2 years ago

What is the image point of (-8,-1) after a translation left 3 units and up 4 units​

What is the image point of (-8,-1) after a translation left 3 units and up 4 units