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Readme [11.4K]
3 years ago
8

. Which axiom justifies the statement RS = RS?. . A. symmetric. B. reflexive. C. addition. D. transitive

Mathematics
2 answers:
lora16 [44]3 years ago
8 0
The axiom that justifies the statement RS=RS is reflective. The correct answer is B.
ale4655 [162]3 years ago
3 0

Answer:

reflective like a mirror

Step-by-step explanation:

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Idk what to do, can someone please help?
bekas [8.4K]
It's simple .....
just subtract R and C.....

so 50x+10-10x-275
= 40x-265

so the answer is B
6 0
3 years ago
(−8p ^3 +9p ^2q −5pq ^2) + (8p ^3 − pq + 5q ^2)
wariber [46]

Answer:

9p^2q-5pq^2-pq+5q^2      

Step-by-step explanation:

6 0
3 years ago
Brittany graphs the point (6,3) on a coordinate grid. Which describes where she
Daniel [21]

Answer:

D. 6 units to the right and 3 units up

Step-by-step explanation:

Given

Point (6,3)

Required

Position of the point.

The general format if a coordinated grid is (x,y)

Where x and y represent the x and y axis respectively

The following conditions are to be considered when positioning a point on a coordinate grid.

1. If x is positive, then it points to the right;

2. If x is negative, then it points to the left

3. If y is positive, then it points upwards

4. If y is negative, then it points downwards

Having said this;

By comparison

x = 6 and y = 3

x and y are both positive.

Condition 1 and 3 above are satisfied.

Hence, the grid coordinates is 6 units to the right and 3 units upwards.

Option D is correct and others are incorrect

6 0
4 years ago
Read 2 more answers
If you had 1052 toothpicks and were asked to group them in powers of 6, how many groups of each power of 6 would you have? Put t
sukhopar [10]

1052 toothpicks can be grouped into 4 groups of third power of 6 (6^{3}), 5 groups of second power of 6 (6^{2}), 1 group of first power of 6 (6^{1}) and 2 groups of zeroth power of 6 (6^{0}).

The number 1052, written as a base 6 number is 4512

Given: 1052 toothpicks

To do: The objective is to group the toothpicks in powers of 6 and to write the number 1052 as a base 6 number

First we note that, 6^{0}=1,6^{1}=6,6^{2}=36,6^{3}=216,6^{4}=1296

This implies that 6^{4} exceeds 1052 and thus the highest power of 6 that the toothpicks can be grouped into is 3.

Now, 6^{3}=216 and 216\times 5=1080, 216\times 4=864. This implies that 216\times 5 exceeds 1052 and thus there can be at most 4 groups of 6^{3}.

Then,

1052-4\times6^{3}

1052-4\times216

1052-864

188

So, after grouping the toothpicks into 4 groups of third power of 6, there are 188 toothpicks remaining.

Now, 6^{2}=36 and 36\times 5=180, 36\times 6=216. This implies that 36\times 6 exceeds 188 and thus there can be at most 5 groups of 6^{2}.

Then,

188-5\times6^{2}

188-5\times36

188-180

8

So, after grouping the remaining toothpicks into 5 groups of second power of 6, there are 8 toothpicks remaining.

Now, 6^{1}=6 and 6\times 1=6, 6\times 2=12. This implies that 6\times 2 exceeds 8 and thus there can be at most 1 group of 6^{1}.

Then,

8-1\times6^{1}

8-1\times6

8-6

2

So, after grouping the remaining toothpicks into 1 group of first power of 6, there are 2 toothpicks remaining.

Now, 6^{0}=1 and 1\times 2=2. This implies that the remaining toothpicks can be exactly grouped into 2 groups of zeroth power of 6.

This concludes the grouping.

Thus, it was obtained that 1052 toothpicks can be grouped into 4 groups of third power of 6 (6^{3}), 5 groups of second power of 6 (6^{2}), 1 group of first power of 6 (6^{1}) and 2 groups of zeroth power of 6 (6^{0}).

Then,

1052=4\times6^{3}+5\times6^{2}+1\times6^{1}+2\times6^{0}

So, the number 1052, written as a base 6 number is 4512.

Learn more about change of base of numbers here:

brainly.com/question/14291917

6 0
3 years ago
Someone help me please?
Lena [83]

32500

0.00604

2.4 \times 10^6

1.47 \times 10^{-3}

5 0
4 years ago
Read 2 more answers
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