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3 years ago

Which of the following is always true about lines of symmetry?

Mathematics

1 answer:

madam [21]3 years ago

4 0

What are the answers being provided ?

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How do I do this problem

STatiana [176]

Answer:

shape one would have to move 8 units to the right and 5 units down and then flip

i hope this helped you to understand more

Step-by-step explanation:

3 0

3 years ago

Is the sequence arithmetic..... if so identify the common difference

const2013 [10]

Answer:

A) Yes, 7

Step-by-step explanation:

The given sequence An = {13, 20, 27, 34, ...}

Here the first term Ao = 13 and difference between the successive terms is (d) = 7

That is d = 20 - 13

d = 7

It is an arithmetic sequence. d = 7

Answer: Yes, it is arithmetic sequence and common difference (d) = 7

Thank you.

8 0

3 years ago

Read 2 more answers

Please help ASAP this is timed

EastWind [94]

Answer:

the answer is 30 with 75 kids voting for drinks in total

8 0

3 years ago

Read 2 more answers

Round 3292.53429 to the nearest hundred thousandth

kari74 [83]

Answer:

The answer is "0.00001"

Step-by-step explanation:

Given value:

$\to 3292.53429$

when we round a hundred thousandths (100,000) value. So, the given rounding 3292.53429 to the nearest 0.00001.

7 0

2 years ago

A drawer contains 3 white shirts, 2 blue shirts, and 5 gray shirts. A shirt is randomly

shutvik [7]

Answer:

Probability that first shirt is white and second shirt is gray if first shirt selected is set aside = $\frac{1}{4}$

Step-by-step explanation:

Given that

3 white, 2 blue and 5 gray shirts are there.

To find:

Probability that first shirt is white and second shirt is gray if first shirt selected is set aside = ?

Solution:

Here, total number of shirts = 3+2+5 = 10

First of all, let us learn about the formula of an event E:

$P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}$

$P(First\ White) = \dfrac{\text{Number of white shirts}}{\text {Total number of shirts left}}$

$P(First\ White) = \dfrac{3}{10}$

Now, this shirt is set aside.

So, total number of shirts left are 9 now.

$P(First\ White\ and\ second\ gray) = P(First White) \times P(Second\ Gray)\\\Rightarrow P(First\ White\ and\ second\ gray) = P(First White) \times \dfrac{\text{Number of gray shirts}}{\text{Total number of shirts left}}\\\\\Rightarrow P(First\ White\ and\ second\ gray) = \dfrac{3}{10} \times \dfrac{5}{9}\\\Rightarrow P(First\ White\ and\ second\ gray) = \dfrac{1}{2} \times \dfrac{1}{2}\\\Rightarrow P(First\ White\ and\ second\ gray) = \bold{\dfrac{1}{4} }$

So, the answer is:

Probability that first shirt is white and second shirt is gray if first shirt selected is set aside = $\frac{1}{4}$

4 0

3 years ago