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saw5 [17]
3 years ago
14

He is 187cm tall how many metres is this

Mathematics
1 answer:
k0ka [10]3 years ago
6 0
He is one meter and 87 centimeters tall. 
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Mary is preparing for her college entrance exams in a practice test she answered 12 problems in 30 minutes at this rate how many
kap26 [50]

Answer:

40 problems in 100 minutes

Step-by-step explanation:

This is a ratio problem.

You first divide 30 minutes by three to get how many problems she can do in 10 minutes. What you do to one side you do to the other side. So then, you should have 4 problems in 10 minute. Then, you multiply 10 by 10 and 4 by 10 to get 40 problems in 100 minutes.

6 0
3 years ago
Is -29/8 greater than -3.62
Darina [25.2K]
-3.62 is greater than -29/8
5 0
3 years ago
Robert went to the grocery store and purchased cans of soup and frozen dinners. each can of soup has 300 mg of sodium and each f
Paraphin [41]

There are 9 cans of soup and 5 cans of frozen dinner.

The variables used for soup and frozen dinner are x and y respectively.

Step-by-step explanation:

Let,

Soup = x

Frozen Dinner = y

According to given statement of sodium;

300x + 500y = 5200 eqn 1

And,

x + y = 14 eqn 2

Multiplying\ equation\ 2\ with\ 300\\300(x+y=14)\\300x+300y=4200\ Eqn 3\\

Subtracting Eqn 3 from Eqn 1;

(300x+500y)-(300x+300y)=5200-4200\\300x+500y-300x-300y=1000\\200y=1000\\y=\frac{1000}{200} \\y=5

Putting value of y in Eqn 1;

300x+500(5)=5200\\300x+2500=5200\\\\Subtracting\ 2500\ from\ both\ sides\ \\300x+2500-2500=5200-2500\\300x=2700\\x=\frac{2700}{300} \\x=9

There are 9 cans of soup and 5 cans of frozen dinner.

The variables used for soup and frozen dinner are x and y respectively.

Keywords: Variables, linear equations, subtraction.

Learn more about linear equations at;

  • brainly.com/question/5047646
  • brainly.com/question/5048216

#LearnwithBrainly

5 0
3 years ago
Fill in the table for side length and area of different squares.
Gnoma [55]

Answer:

0.09

Step-by-step explanation:

0.03 divided by 100

7 0
3 years ago
the figure below shows a circle centre of radius 10 cm the chord PQ=16cm calculate the area of the shaded region​
m_a_m_a [10]

Step-by-step explanation:

OPQ originally forms a sector, the formula for sector is

\frac{x}{360} \pi {r}^{2}

where x is the degrees of rotation between the two radii.

We know three sides length and is trying to find an angle between the radii so we can use law of cosines which states that

16 =  \sqrt{10 {}^{2} + 10 {}^{2}   - 2(100) \times  \cos(o) }

This isn't the standard formula, it's for this problem

16 =  \sqrt{200 - 200 \times  \cos(o) }

16 =  \sqrt{200  - 200 \cos(o) }

256 = 200 - 200 \cos(o)

56 =  - 200 \cos(o)

-  \frac{7}{25}  =  \cos(o)

\cos {}^{ - 1} (  - \frac{7}{25} )  =  \cos {}^{ - 1} ( \cos(o) )

106.26 = o

So we found our angle of rotation, which is 106.26

Now, we do the sector formula.

\frac{106.26}{360} (100)\pi

\frac{10626}{360} \pi

is the area of sector.

Now let find the area of triangle, we can use Heron formula,

The area of a triangle is

\sqrt{s(s - a)(s - b)(s - c)}

where s is the semi-perimeter.

To find s, add all the side lengths up, 10,10,16 and divide it by 2.

Which is

\frac{36}{2}  = 18

\sqrt{18(18 - 10)(18 - 10)(18 - 16)}

\sqrt{18(8)(8)(2)}

\sqrt{(36)(64)}

6  \times 8 = 48

So our area of the triangle is 48. Now, to find the shaded area subtract the main area,( the sector of the circle) by the area of the triangle so we get

\frac{10626}{360}  - 48

Which is an approximate or

44.73

6 0
2 years ago
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