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Liono4ka [1.6K]
3 years ago
15

When a figure is translated on a coordinate grid, what conclusion can you draw from the pre-image and image? (2 points) Group of

answer choices The orientation changes, and the angles remain congruent. The shape changes, and the side lengths remain congruent The position changes, and the angles remain congruent. The size changes, and the sides remain congruent.
Mathematics
1 answer:
Nataly [62]3 years ago
8 0

Answer:

B.the size of the figures change

Step-by-step explanation:

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8 less than 7 is:<br><br> can i please have an answer
Damm [24]

Answer:

-1

Step-by-step explanation:

7-8=-1

4 0
2 years ago
Read 2 more answers
Prove or disprove (from i=0 to n) sum([2i]^4) &lt;= (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
Express 4 4/9 as an improper fraction. Example please.
miskamm [114]
Simple...

you have: 4 \frac{4}{9}

To make this into an improper fraction-->>

1.) Multiply whole number by denominator

2.)Add the number you got plus the numerator

3.) Use the original denominator to finish your improper fraction

Example: 4\frac{4}{9}

4*9=36

36+4=40

\frac{40}{9}

Thus, your answer.
3 0
3 years ago
A $500 television is on sale for 20% off. However, you must pay 6% in sales tax. What is the final price of the television after
GrogVix [38]

20% off means you are paying 80% of the original price ( 100% - 20% = 80%)

Multiply the price of the tv by 80% to find the sale price:

500 x 0.80 = 400

Now multiply the sales price by 1 plus the tax rate to find final cost:

400 x 1.06 = 424

Final price including tax is $424

5 0
3 years ago
Pls asap... use trigonometry to find the area of a regular 20- gon with each side = 6 in
irga5000 [103]

Answer:

Area of polygon[20 sides] = 1,140 inch² (Approx.)

Step-by-step explanation:

Given;

Number of sides = 20 sides

Length of each side = 6 inch

Find:

Area of polygon[20 sides]

Computation:

Perimeter of polygon[20sides] = 20 x Length of each side

Perimeter of polygon[20sides] = 20 x 6

Perimeter of polygon[20sides] = 120 inch

Apothem[Shortest distance from center] = S / 2tan[180/n]

Apothem[Shortest distance from center] = 6 / 2tan[180/20]

Apothem[Shortest distance from center] = 6 / 2tan[9]

Apothem[Shortest distance from center] = 18.94

Apothem[Shortest distance from center] = 19 inch (Approx.)

Area of polygon[20 sides] = [Perimeter x Apothem] / 2

Area of polygon[20 sides] = [120 x 19] / 2

Area of polygon[20 sides] = 1,140 inch² (Approx.)

6 0
2 years ago
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