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REY [17]
3 years ago
11

The stripe is about. Cm long.

Mathematics
1 answer:
kodGreya [7K]3 years ago
4 0

Replace S with the given side length, then solve for d.

d=√2(37)^2

d= √2(1369)

d= √2738

d=52.33 cm long.


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Solve this using Substitution. <br> {y=x+1 <br> {2x+y=7
Elina [12.6K]
Y = x + 1....so we sub in x + 1 for y in the other equation

2x + y = 7
2x + x + 1 = 7 ....combine like terms
3x + 1 = 7....subtract 1 from both sides
3x = 7 - 1
3x = 6 ....divide both sides by 3
x = 6/3
x = 2

y = x + 1
y = 2 + 1
y = 3

so ur solution is : (2,3)
7 0
4 years ago
Tameron is driving 540 miles to college. If he drives at an average rate of 45 miles per hour, how many hours will it take him t
andreev551 [17]
Given:
540 miles to college
average rate of 45 miles per hour

How many hours will it take him to get 3/4 of the way there?

540 miles * 3/4 = (540 * 3) / 4 = 1620/4 = 405 miles

405 miles / 45 mph = 9 hours.

It would take Tameron 9 hours to travel 3/4 of the way there.
5 0
3 years ago
Indica el exponente de X
Veseljchak [2.6K]

Answer:

x=56

Step-by-step explanation:

8 0
3 years ago
5. Find the point P along the directed line segment from X(-3, 3) to Y(6,-3) that divides
ASHA 777 [7]
Consider point P(x,y) such that P, X and Y are collinear,

As vectors

XP = XO + OZ where O(0,0)
XP = OZ - OX
XP= (x,y) - (-3,3)
XP = (x+3, y-3)

Similarly,

PY = (6-x, -3-y)

But XP= 2^PY

[x+3, y-3] = [2(6-x), 2(-3-y)]

Given both vectors are equal, as they go in the same direction, Solve for x and y accordingly:

x+3 = 12 - 2x

x = 3

y-3 = -6-2y

y = -1

Therefore, P(3,-1)

6 0
3 years ago
Fine the exact distance between the line 6x-y=-3 and the point (6,2). Show your work. Explain your answer.
AVprozaik [17]

Answer:

\sqrt{37} units

Step-by-step explanation:

If we draw a perpendicular from point (6,2) on the line 6x - y = - 3, then we have to find the length of the perpendicular.  

We know the formula of length of perpendicular from a point (x_{1}, y_{1} ) to the straight line ax + by + c = 0 is given by  

\frac{|ax_{1} + by_{1} +c |}{\sqrt{a^{2}+b^{2}}}

Therefore, in our case the perpendicular distance is  

\frac{|6(6)-2+3|}{\sqrt{6^{2} +(-1)^{2} } } = \frac{37}{\sqrt{37} } = \sqrt{37}  units. (Answer)

6 0
4 years ago
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