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3 years ago

Solve for x. Please helpppp!

Mathematics

1 answer:

Anni [7]3 years ago

4 0

Answer:

Pythagoreas Theorem

Step-by-step explanation:

a^2+b^2=c^2

4^2+b^2=15^2

16+b^2=225

b^2= 209

14.45=b

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Convert 10 1/2 to fractional notation

Phoenix [80]

10 1/2 =
10/1 + 1/2=
2/2 (10/1) + 1/2=
20/2 + 1/2=
21/2

3 0

3 years ago

Isosceles triangle LMN is graphed with vertices L(0, 1), M(3, 5), and N(6,1). What is the slope of side LN?

Vaselesa [24]

L(0, 1)
N(6, 1)

slope formula;
m = y2 - y1/ x2 - x1

m = 1 - 1/ 6 - 1

m = 0/5

m = 0

hope this helped, God bless!

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3 years ago

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Find the greatest common factor. 8j, 6j

VLD [36.1K]

Answer:

Step-by-step explanation:

3 0

3 years ago

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The circumference of a circle is 32π. What is the diameter of this circle?

lawyer [7]

C = 2 (pi) r

32 (pi) = c

32 (pi)= 2 (pi) r

r= 16

2r = d
d= 32

hope that helps

4 0

3 years ago

Security A rounds every 2 minutes and 20 seconds in the gate 1. Security B rounds every 1 minute and 40 seconds in the gate 2. I

Elden [556K]

Answer:

$Time = 11\frac{2}{3}\ min$

Step-by-step explanation:

Given

$Security\ A = 2\ mins, 20\ secs$

$Security\ B = 1\ mins, 40\ secs$

Required

After how many minutes, will they round together

First, convert the given time to minutes

$Security\ A = 2\ mins, 20\ secs$

$Security\ A = 2\ mins+ 20\ secs$

$Security\ A = 2\ mins+ \frac{20}{60}\ min$

$Security\ A = 2\ mins+ \frac{1}{3}\ min$

$Security\ A = 2\frac{1}{3}\ min$

$Security\ B = 1\ mins, 40\ secs$

$Security\ B = 1\ mins+ 40\ secs$

$Security\ B = 1\ mins+ \frac{40}{60}\ min$

$Security\ B = 1\frac{2}{3}\ min$

So, we have:

$Security\ A = 2\frac{1}{3}\ min$

$Security\ B = 1\frac{2}{3}\ min$

List out the multiples of the time of both security personnel take round.

$Security\ A = 2\frac{1}{3}min,\ 4\frac{2}{3}min,\ 7min,\ 9\frac{1}{3}min,\ 11\frac{2}{3}min...$

$Security\ B = 1\frac{2}{3}min,\ 3\frac{1}{3}min,\ 5min,\ 6\frac{2}{3}min,\ 8\frac{1}{3}min,\ 10min\ ,11\frac{2}{3}min,...$

In the above lists, the common time is:

$Time = 11\frac{2}{3}\ min$

This implies that they go on round after $11\frac{2}{3}\ min$

4 0

3 years ago