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Dmitry_Shevchenko [17]
2 years ago
5

4. Janet has 30 classmates in her classroom. 18 of those

Mathematics
1 answer:
boyakko [2]2 years ago
6 0

Answer:

300

Step-by-step explanation:

Let x represent number of girls.

18 girls / 30 students

x girls / 500 students

18/30 = x/500

x = 18/30 x 500

x= 300 girls

Therefore there are approximately 300 girls in the school.

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Tom and Sophie travel from home to school and back once a day. How many kilometres more does Sophie travel in 5days than Tom?
exis [7]

Answer:

1000 m

Step-by-step explanation:

Tom's distance = 400 m

Sophie's distance = 600 m

Hence. Distance per day = 600 - 400 = 200 m per days

In 5 days ;

200 m * 5 = 1000 m

Sophie travels 1000 m more

6 0
3 years ago
Please Help Me!
alex41 [277]

Answer:

✔️2 sets of corresponding angles

<D and <S

<R and <L

✔️2 sets of corresponding sides

DR and SL

RM and LT

Step-by-step explanation:

When two polygons are congruent, it implies that they have the same shape and size. Therefore, their corresponding angles and sides are congruent to each other.

When naming congruent polygons, the arrangement of the vertices are kept in a definite order of arrangement.

Therefore, Given that polygon DRMF is congruent to SLTO, the following angles and sides correspond to each other:

<D corresponds to <S

<R corresponds to <L

<M corresponds to <T

<F corresponds to <O

For the sides, we have:

DR corresponds to SL

RM corresponds to LT

MF corresponds to TO

FD corresponds to OS.

We can select any two out of these sets of corresponding angles and sides as our answer. Thus:

✔️2 sets of corresponding angles

<D and <S

<R and <L

✔️2 sets of corresponding sides

DR and SL

RM and LT

4 0
3 years ago
Describe the graph of y = negative x squared and compare it with the graph of y = x squared.
levacccp [35]

Answer:

The graph of the first equation opens downward while the graph of the second equation opens upward.  Both graphs are parabolas that pass through 0,0.

5 0
3 years ago
Use the principle of inclusion and exclusion to find the number of positive integers less than 1,000,000 that are not divisable
wel
This is a simple problem based on combinatorics which can be easily tackled by using inclusion-exclusion principle.
We are asked to find number of positive integers less than 1,000,000 that are not divisible by 6 or 4.
let n be the number of positive integers.
∴ 1≤n≤999,999
Let c₁ be the set of numbers divisible by 6 and c₂ be the set of numbers divisible by 4.
Let N(c₁) be the number of elements in set c₁ and N(c₂) be the number of elements in set c₂.

∴N(c₁) = \frac{999,999}{6} = 166666
N(c₂) = \frac{999,999}{4} = 250000
∴N(c₁c₂) = \frac{999,999}{24} = 41667
∴ Number of positive integers that are not divisible by 4 or 6,

N(c₁`c₂`) = 999,999 - (166666+250000) + 41667 = 625000
Therefore, 625000 integers are not divisible by 6 or 4
8 0
4 years ago
Use base ten blocks to find 182 ÷14 describe the steps you took to find your answer
docker41 [41]
It would be 
one 10 block 
and 3 single blocks 
4 0
3 years ago
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