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Zielflug [23.3K]
3 years ago
13

What is the area of the following triangle? A. 300 in B. 60 in C. 150 in D. 72 in

Mathematics
2 answers:
babymother [125]3 years ago
6 0
Answer: C. 150
Hopes this helps
NemiM [27]3 years ago
4 0

Answer:

C, 150 in

Step-by-step explanation:

12 in x 25 = 300

300/2

=150 in

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Explanation/proof why the lines are parallel and insights​
Delvig [45]

Answer:

The lines are parallel is the will never touch.

Step-by-step explanation:

5 0
2 years ago
Seamstress Jane has purchased 50 yards of red fabric and 110 yards of gold fabric to make red and gold curtains. What is the lar
Alisiya [41]

Answer: There are 10 largest number of curtains she can make if she wants all the curtains to be exactly the same length.

There are 5 red fabrics and 11 gold fabrics that would be used for each curtain.

Step-by-step explanation:

Since we have given that

Number of yards of red fabric purchased by Jane = 50

Number of yards of gold fabric purchased by Jane = 110

We need to find the largest number of curtains she can make if she wants all the curtains to be exactly the same length.

For this we will find H.C.F. of 50 and 110.

Factors of 50 = 1, 2, 5, 10, 25, 50.

Factors of 110 = 1, 2, 5, 10, 11, 22, 55, 110.

So, Highest common factor of 50 and 110 is 10.

So, there are 10 largest number of curtains she can make if she wants all the curtains to be exactly the same length.

Number of yards of red fabric is given by

\frac{50}{10}=5

Number of yards of gold fabric is given by

\frac{110}{10}=11

Hence, there are 5 red fabrics and 11 gold fabrics that would be used for each curtain.

4 0
3 years ago
Lin ran 4 laps around the track in 5 minutes, if lin ran 21 laps, how long does it take her
lyudmila [28]

Answer:

26.25 min

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
How do i solve this.
jasenka [17]

Answer:

here is the answer to your question

5 0
1 year ago
The width of each of five continuous classes in a frequency distribution is 5 and the lower class-limit of the lowest class is 1
Helga [31]

Let x and y be the upper and lower class limit of frequency distribution.

Given, width of the class = 5

⇒ x-y= 5 …

Also, given lower class (y) = 10 On putting y = 10, we get

x – 10= 5 ⇒ x = 15 So, the upper class limit of the lowest class is 15

Hence, the upper class limit of the highest class

=(Number of continuous classes x Class width + Lower class limit of the lowest class)

= 5 x 5+10 = 25+10=35

Hence,’the upper class limit of the highest class is 35.

<em><u>Alternate Method</u></em><em><u>.</u></em>

After finding the upper class limit of the lowest class, the five continuous classes in a frequency distribution with width 5 are 10-15,15-20, 20-25, 25-30 and 30-35.

Thus, the highest class is 30-35..

<h3>Hence, the upper limit of this class is 35.</h3>

<h2>_____________________</h2>

4 0
2 years ago
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