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Vinil7 [7]
3 years ago
14

Inequalities

Mathematics
1 answer:
Verdich [7]3 years ago
7 0

Answer:

Im so sorry I couldnt find u since I didnt save the question!

Step-by-step explanation:

I will solve this right now and I answered right now before Arthur takes ur points

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There are 448 students at west over middle school. a total of 3/8 of the students participate in after-school activities. Of tho
iragen [17]

Answer:

56

Step-by-step explanation:

Hope it helped <3

7 0
3 years ago
Solve each inequality. Graph the solution on a on a number line.<br> p-7/12&gt;3/10
Dovator [93]

Answer:

so to solve an inequality you just treat the inequaltity sign like an equals sign except if you multiply by negative 1 you flip the direction of the sign.

p-7/12>3/10

add 7/12 to both sides

p>7/12+3/10

p>35/60+18/60

p>53/60

so if you had to graph this on a number line you would have an open circle on 53/60 and an arrow coming off pointing to the right

Step-by-step explanation:

3 0
2 years ago
Solve y = x + 6 for x.<br><br> A.x = y + 6<br> B.x = −y + 6<br> C.x = y − 6<br> D.x = −y − 6
Ivan

Answer:

C) x = y - 6

Step-by-step explanation:

you have to make it in terms of x, or isolate x:

y = x + 6

- 6 from both sides

y - 6 = x + 6 - 6

combine 6 and -6

y - 6 = x

so, x = y - 6

4 0
3 years ago
Read 2 more answers
Why does the quotient of 8 ÷ 1 not change when we add a place value in the dividend and the divisor to make 80 ÷ 10?
Alex Ar [27]

Answer:

Basically this is because division can be thought of as how many times does the divisor have to be multiplied in order to produce the dividend.

So you would need to multiply 1 8 times in order to produce the dividend,

Similarly, 10 goes into 80 8 times. The zeros are simply cancelled out in the division.

4 0
3 years ago
The perpendicular bisectors of two sides of a triangle meet at point that belongs to the third side. Prove that this is a right
Bond [772]

Statement : If the perpendicular bisectors of two sides of a triangle meet at point that belongs to the third side then this is a right angle triangle.

Prove:

Let ABC is triangle,

In which DF and EF are the perpendicular bisectors of legs AB and BC respectively.

That is, ∠ FDB = ∠BEF=  90°

Where, F\in AC

Then By the property of the circumcenter,

F must be mid point of leg AC.

That is, DF ║ BC

⇒  ∠ FDB + ∠ DBC = 180° ( the sum of two adjacent angles on the parallel lines by the same transversal is supplementary )

But,  ∠ FDB = 90°

⇒∠ DBC = 90°

⇒ ∠ ABC = 90°

That is, Δ ABC is a right angle triangle.



4 0
3 years ago
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