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Ratling [72]
3 years ago
12

If U = {1, 3, 5, 7, 9, 11, 13}, then which of the following are subsets of U.

Mathematics
1 answer:
lana [24]3 years ago
6 0

Answer:

C

Step-by-step explanation:

It is C because members in the set C can be found in the universal set (U)

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7/12 x 1/8 Find the product of in simplified form​
Vanyuwa [196]

Answer:

7/96

Step-by-step explanation:

to find the answer you need to multiply both numerators together and put that over the product of the denominators, this is going to leave you with a fraction smaller than the original fractions. that is normal, because you are multiplying terms smaller than one together. same applies to decimals less than one

4 0
3 years ago
Examine A ABC below.
Roman55 [17]

Answer:

  • DE║AC

Step-by-step explanation:

<u>Given relationships:</u>

DE = 2 AC

  • Incorrect. Should be DE = 1/2 AC

DE║AC

  • Correct

m∠BCA = 2(m∠BED)

  • Incorrect. Should be m∠BCA = m∠BED

DE = AC

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8 0
3 years ago
Amelia competed in a race in which she ran at a constant speed. A graph that represents the relationship between the number of s
Zolol [24]

Answer:

B (3, 13.5)

Step-by-step explanation:

Using point (0,0) and (15, 67.5) we can find the slope(gradient).

(y - y¹) / (x - x¹) = (67.5 - 0) / (15 - 0)

= 67.5 / 15 = 4.5

slope = 4.5

using the point given in option A (0, 4.5) with point (15, 67.5) to calculate the slope it gives 4.2 which is not equal to what we calculated.

using the option B (3, 13.5) with (15, 67.5) gives a slope of 4.5 which is equal to the slope of the line.

3 0
3 years ago
How many times larger is 3x10^8 than 3x10^2
muminat
3x10^8 is 7 times larger than 3x10^2
3 0
3 years ago
ABCD is a parallelogram. AE ⊥ DC, A[] ⊥ BC, AD = 10, AB = 15, A[] = 12. Find the length of AE.
kaheart [24]

Answer:

AE= 8

Step-by-step explanation:

The parallelogram forms two right triangles and since it's a parallelogram the opposite angles are congruent meaning you can say the triangles are similar through AA~. The ratio is 2:3 and FA =12 making AE= 8

8 0
3 years ago
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