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andrew11 [14]
2 years ago
13

I need help I failing

Mathematics
1 answer:
Liono4ka [1.6K]2 years ago
7 0

Answer:

A and C

Step-by-step explanation:

Both A and C are the same shape, same size, but C is flipped over. If you were to cut out the two shapes and put them over one another, they would be the same. You can also count the blocks that are in the shape itself to see if they match.

The longer side on both has 2.5 blocks in it, the side is two blocks tall, and the smaller line has 1.5 of the block shown.

I'll will implement a sketch to show you what I mean and submit it here. Look at it if you are still confused

I hope this helps!

-No one

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PLEASE HELP 17 points
Ksivusya [100]

Answer: I say B its a bit hard but i think i got it not  sure but Good Luck!

Step-by-step explanation:

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3 years ago
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What numeral in base 8 equivalent to 3325(denoting 332 base 5)
lakkis [162]
It can help to write each number in expanded form, where the digit is multiplied by a power of the base.

332 base 5 = 3*5^2 + 3*5^1 +2*5^0 = 75 +15 +2 = 92

In base 8, this is
92 = 64 + 24 + 4 = 1*8^2 + 3*8^1 + 4*8^0 = 134 base 8

_____
The conversion of an integer to base b is often done by finding the remainder when dividing by b.
92/8 = 11 remainder 4
11/8 = 1 remainder 3
1/8 = 0 remainder 1

The digits of the converted number (right to left) are the remainders at each step. Here: 92 = 134 (base 8).
5 0
3 years ago
17. Test Practice which represents the number that is ten more than two
AleksAgata [21]

Answer:

2,080

Step-by-step explanation:

Two thousand and seventy is written as 2,070.

Ten more than 2,070 would be 2,080 as you're just adding on 10.

Hope this helps :)

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3 years ago
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elena55 [62]

Answer:

Step-by-step explanation:

The correct answer is dotplot

8 0
3 years ago
Find the absolute minimum and absolute maximum values of f on the given interval. f(t) = t 25 − t2 , [−1, 5]
Zina [86]

Answer: Absolute minimum: f(-1) = -2\sqrt{6}

              Absolute maximum: f(\sqrt{12.5}) = 12.5

Step-by-step explanation: To determine minimum and maximum values in a function, take the first derivative of it and then calculate the points this new function equals 0:

f(t) = t\sqrt{25-t^{2}}

f'(t) = 1.\sqrt{25-t^{2}}+\frac{t}{2}.(25-t^{2})^{-1/2}(-2t)

f'(t) = \sqrt{25-t^{2}} -\frac{t^{2}}{\sqrt{25-t^{2}} }

f'(t) = \frac{25-2t^{2}}{\sqrt{25-t^{2}} } = 0

For this function to be zero, only denominator must be zero:

25-2t^{2} = 0

t = ±\sqrt{2.5}

\sqrt{25-t^{2}} ≠ 0

t = ± 5

Now, evaluate critical points in the given interval.

t = -\sqrt{2.5} and t = - 5 don't exist in the given interval, so their f(x) don't count.

f(t) = t\sqrt{25-t^{2}}

f(-1) = -1\sqrt{25-(-1)^{2}}

f(-1) = -\sqrt{24}

f(-1) = -2\sqrt{6}

f(\sqrt{12.5}) = \sqrt{12.5} \sqrt{25-(\sqrt{12.5} )^{2}}

f(\sqrt{12.5}) = 12.5

f(5) = 5\sqrt{25-5^{2}}

f(5) = 0

Therefore, absolute maximum is f(\sqrt{12.5}) = 12.5 and absolute minimum is

f(-1) = -2\sqrt{6}.

8 0
3 years ago
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