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Ne4ueva [31]
3 years ago
13

What is the value of the expression 1/7 ÷8 A 1/12 B 1/35 C 5/7 D 6/7

Mathematics
2 answers:
dsp733 years ago
7 0

Answer:

Step-by-step explanation:B and D I belive.

vesna_86 [32]3 years ago
3 0

Answer:

Looks like it's none.

Step-by-step explanation:

\frac{1}{7}*\frac{1}{8}=\frac{1}{56}

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Please help me with this homework
Nady [450]

Answer:

Rational

Step-by-step explanation:

The number comes to a stop (doesn't keep going), so it would be rational.

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3 years ago
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On the grid draw a scaled copy of quadrilateral ABCD with a scale factor 2/3
Tamiku [17]

Answer:

Attached diagram A'B'C'D'

Step-by-step explanation:

Given is a quadrilateral ABCD. It says to draw a dilated version with a scale factor 2/3.

We see that scale factor is less than 1 which means it shrinks the image to a smaller one.

To draw a scaled copy, we need to find the lengths of its sides.

To do so, we can draw the diagonals AC & BD, and they intersect at origin O(0,0) such that OA= -2, OB= 2, OC= 4, OD= -4.

Applying a scale factor of 2/3, we get OA' = -4/3, OB' = 4/3, OC' = 8/3, OD' = -8/3.

So we have attached a scaled copy A'B'C'D' of quadrilateral ABCD with a scale factor 2/3.

6 0
3 years ago
Find the slope of the line that passes through (4,9) and (1,7)
denis-greek [22]

Answer:

2/3

Step-by-step explanation:

Use the formula for finding gradient.

7 0
3 years ago
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Nastasia [14]

Answer:

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4 0
3 years ago
Use an induction proof to prove this statement:<br> For n≥1, 4^n+5 is divisible by 3.
Tpy6a [65]

Answer:

See below

Step-by-step explanation:

We shall prove that for all n\in\mathbb{N},3|(4^n+5). This tells us that 3 divides 4^n+5 with a remainder of zero.

If we let n=1, then we have 4^{1}+5=9, and evidently, 9|3.

Assume that 4^n+5 is divisible by 3 for n=k, k\in\mathbb{N}. Then, by this assumption, 3|(4^n+5)\Rightarrow4^k+5=3m,\: m\in\mathbb{Z}.

Now, let n=k+1. Then:

4^{k+1}+5=4^k\cdot4+5\\=4^k(3+1)+5\\=3\cdot4^k+4^k+5\\=3\cdot4^k+3m\\=3(4^k+m)

Since 3|(4^k+m), we may conclude, by the axiom of induction, that the property holds for all n\in\mathbb{N}.

3 0
2 years ago
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