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Dominik [7]
3 years ago
10

If 45 crackers will serve 15 students, how many crackers are needed for 30 students? Please help I'm not smart enough to solve t

his-
Mathematics
1 answer:
alexandr402 [8]3 years ago
8 0
Since 45 crackers feeds 15 students to get 30 you need to do 15+15, meaning you need 45+45 which equals 90

You need 90 crackers
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5x-(x+3)=1/3 (9x+18)-5<br> Please show me step by step
zzz [600]

Step-by-step explanation:

5x-(x+3)=1/3(9x+18)-5

5x-x+3=3x+6-5

1/3 multiplied by 9/1 is three because if you take away the ones and put your problem like this: 9/3, you'll get 3.

1/3 multiplied by 18/1 is six because if you take away the ones and switch to division, it'll look like this: 18/3. 18/3 is 6.

5x-1x+3=3x+6-5

If a variable is alone, you should - in this case - put a one before it, but if the variable is alone you should know that it's automatically a one.

4x+3=3x-1

6-5=1 Simple math problem, probably self explanatory

4x+3=3x-1

Subtract 3x from 4x and you'll get 1x because 4-3 is 1.

1x+3=-1

Then you add 3 to -1 which is 2.

The answer is 1x=2

Or you could've done:

4x+3=3x-1

-4x    -4x

    3=-1x-1

 +-1    +-1

2=1x

It's the same answer, but all I did was subtract the 4x from the 4x and 3x.

3 0
3 years ago
Triangle A'B'C' is a rotation of triangle ABC by 60 degrees using Q as the center.
butalik [34]

Answer:

The correct options are;

D. Triangles ABC and A'B'C' are congruent

E. Angle ABC is congruent to angle A'B'C'

F. Segment BC is congruent to segment B'C'

H. Segment AQ is congruent to segment A'Q'

Step-by-step explanation:

The given information are;

The angle of rotation of triangle ABC = 60°

Therefore, given that a rotation of a geometric figure about a point on the coordinate plane is a form of rigid transformation, we have;

1) The length of the sides of the figure of the preimage and the image are congruent

Therefore;

BC ≅ B'C'

2) The angles formed by the sides of the preimage are congruent to the angles formed by the corresponding sides of the image

Therefore;

∠ABC ≅ ∠A'B'C'

3) The distances of the points on the figure of the preimage from the coordinates of the point of rotation are equal to the distances of the points on the figure of the image from the coordinates of the point of rotation

Therefore;

Segment AQ ≅ A'Q'.

4 0
3 years ago
Find the sum. (−a^3+4a−3)+(5a^3−a)
aev [14]

Answer:

4a³+3a-3

Step-by-step explanation:

You can solve this by combining like terms. Add the numbers with the correct exponents together.

-a³+5a³

4a-a

Thus, the answer is

4a³+3a-3

5 0
3 years ago
First answer get brainliest
Tasya [4]

Answer:

A. 63 × 9 = ?

plz mark brainliest

6 0
3 years ago
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Ratling [72]
The radius of a circle should be placed in the Z section making the answer A.
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3 years ago
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