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IceJOKER [234]
4 years ago
15

Maria has 20 metre of ribbon. She needs 4/5 of metre to tie a bow for wrapping birthday present. How many bows can Maria make fr

om the ribbon?
Mathematics
2 answers:
AlekseyPX4 years ago
6 0
She can make 25 bows
Dahasolnce [82]4 years ago
6 0

Answer:

25 bows

Step-by-step explanation:

4/5 metre = 0.8m

20m/0.8m = 25 bows

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The three trigonometric functions can be used to calculate unknown side lengths and angles in which type of triangle(s)?
malfutka [58]
Using the three trigonometric functions can be used to calculate unknown side lengths and angle in a right-angle triangle only.

But, if ever you we're given to find a non right-angle triangle, you will have to use sine rule or cosine rule.
3 0
3 years ago
If m // n, what is m∠1?
vampirchik [111]
The answer is c. alternate interior angles are congruent..

brainliest is appreciated..
8 0
4 years ago
Plzzzzz help me on this and explain
Galina-37 [17]
The Answers are B and C
Its B because 12 is MORE than twice as much as M. Meaning 140 is larger than 2*m
Its C because again it is MORE than twice as much as M. Meaning that 140 is larger than 2m.

Answers;B,C
Hope this helped!
7 0
3 years ago
Find percent of each number <br> 1) 5% of 40<br> 2) 250% of 44<br> 3) 0.5% of 13.5<br> 4) 32% of 54
Alex
1.) 2
2.) 110
3.) <span>0.0675
4.) 17.28</span>
8 0
3 years ago
Angles α and β are angles in standard position such that: α terminates in Quadrant II and sinα = 3/5, β terminates in Quadrant I
777dan777 [17]

We are given

Angles α and β are angles in standard position

and

α terminates in Quadrant II

β terminates in Quadrant I

and we have

sin(\alpha)=\frac{3}{5}

we can use triangle and find cos(α)

we get

cos(\alpha)=-\frac{4}{5}

and we have

cos(\beta)=\frac{4}{5}

we can draw triangle

sin(\beta)=\frac{3}{5}

now, we can use formula

cos(\alpha+\beta)=cos(\alpha)cos(\beta)-sin(\alpha)sin(\beta)

now, we can plug values

cos(\alpha+\beta)=-\frac{4}{5}\times \frac{4}{5}-\frac{3}{5}\times \frac{3}{5}

now, we can simplify it

cos(\alpha+\beta)=-\frac{16}{25}-\frac{9}{25}

cos(\alpha+\beta)=-\frac{(16+9)}{25}

cos(\alpha+\beta)=-\frac{(16+9)}{25}

cos(\alpha+\beta)=-\frac{25}{25}

cos(\alpha+\beta)=-1...............Answer

3 0
3 years ago
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