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andreyandreev [35.5K]
2 years ago
12

Rachel multiplied each side of x ≥ 2 by 3. She wrote the result as 3x ≤ 6. Explain the error Rachel made.

Mathematics
1 answer:
andrey2020 [161]2 years ago
6 0

Answer:

Her mistake was to change the inequality sign from ≥ to ≤.

The sign only changes when you multiply or divide both sides of an inequality by a negative number.

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2) (a^3-2a)-(3a^2-4a^3
mylen [45]

Answer:

a x (a2-2)

Step-by-step explanation:

the 2 is supposed to be above the a I don't know how to that while typing.

5 0
3 years ago
Part of a tiling design is shown. The center is a regular hexagon. A square is on each side of the hexagon, and an equilateral t
VMariaS [17]

1,119.6 in. the answer is c my diddle.


Find the area of the hexagon which is 259.81

Find the area of the squares which is (6)100

Find the area of the triangles which was (6)43.3

259.81 + 600 + 259.8

519.6 + 600 = 1119.6

8 0
3 years ago
In isosceles △ABC the segment BD (with D∈ AC ) is the median to the base AC . Find BD, if the perimeter of △ABC is 50m, and the
Morgarella [4.7K]

Let the equal sides of the isosceles Δ ABC be x.

Given that the perimeter of Δ ABC = 50m.

Therefore, 2x + AC = 50 --- (1)

It is also given that the perimeter of Δ ABD = 40m.

Therefore, x + BD + AD = 40

BD is the median of the Δ ABC. Therefore, D is the midpoint of AC.

So AD = CD.

Or, AD = \frac{1}{2} AC

Therefore, x + BD + \frac{1}{2} AC = 40

Multiply both sides by 2.

2x + 2BD + AC = 80

From (1), 2x + AC = 50.

Therefore, 2BD + 50 = 80

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BD = 15m.

6 0
3 years ago
Read 2 more answers
P2 + 14p + 40 written as a binomial
BartSMP [9]
P² + 14p + 40
= p² + 10p + 4p + 40
= p(p+10) + 4(p+10)
= (p+10)(p+4)


7 0
3 years ago
Find the mode of the data set: 9 6 7 3 10 9
Phoenix [80]

Answer:

9

Step-by-step explanation:

Math definition Mode means anything that repeats more than anything else.

Eg.(4, 5, 6, 2, 2, 4, 4, 4, 4, 5, 3, 5)

The mode in the bracket is 4. Some of the numbers occured more than once but 4 occured the most that's why it's 4

4 0
3 years ago
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