MATHEMATICS MIDDLE SCHOOL

Find the distance between (150, -27) and (58, -67)

Answers

Answer 1

Answer:

Answer:

100.31949

Step-by-step explanation:

Answer 2

Answer:

The distance between (150, -27) & (58, -67) is about 26.

I hope this helps!

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kimi is playing hide-and-seek with Tommy and Manuel. Tommy is hiding 12 feet south of Kimi, and Manuel is hiding due east of Tommy. If Kimi is 20 feet from Manuel, how far apart are Tommy and Manuel?

Answers

Answer:

Tommy and Manuel are 16 ft apart

Step-by-step explanation:

The locations of all three players are shown in the image below

They form a right triangle where the hypotenuse is 20 ft, and one of the legs is 12 ft. We must find the other leg.

We must use Pythagoras's theorem. Being a and b the legs of a right triangle and c its hypotenuse, then

$c^2=a^2+b^2$

Knowing c and one of the legs, say b:

$a^2=c^2-b^2$

Using the values c=20, b=12 we find

$a^2=20^2-12^2=400-144=256$

$a=√(256)=16$

So, Tommy and Manuel are 16 ft apart

Solve |3x + 9| + 6 = 6

Answers

Answer:

x = -3

Step-by-step explanation:

|3x + 9| + 6 = 6

|3x + 9| = 6 - 6

|3x + 9| = 0

3x + 9 = 0

3x = - 9

x = -3

Solve x+1>10, x+11>20, and x+21>30

Answers

x+1>10
x>10-1
x>9

x+11>20
x>20-11
x>9

x+21>30
x>30-21
x>9

$\framebox[1.1\width]{Godspeed!} \par$

$a) \n x+1>10 \n x>10-1 \n x>9 \n \n \n b) \n x+11>20 \n x>20-11 \n x>9 \n \n \n c) \n x+21>30 \n x>30-21 \n x>9 \n \n \n \framebox[1.1\width]{Good Luck !} \par$

Which algebraic expression is equivalent to this expression 8(4x+2)+12A. 32x+4
B. 48x
C. 32x+28
D. 8x+6

Answers

c i hope that helps:)

Find the first four terms of the multiplication pattern given by the formula.10 • 4n – 1
_, _, _, _

plz give explanation along with answer thx!

Answers

You just need to plug in 1, 2, 3, and 4.
10*4(1)-1=39
10*4(2)-1=79
10*4(3)-1=119
10*4(4)-1=159

Equivalent to log3(x+4)

Answers

the equivalent expression of this would be log3 + log(x+4) because we know log(AB) = log(A) + log(B). to see why this is true, check out the attachment.

Let me know if you have any questions!

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