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gayaneshka [121]
3 years ago
12

Help plslsllsl its due tmr

Mathematics
1 answer:
olga55 [171]3 years ago
6 0
X=12.5
If the perimeter is 90
90=3x+3x+15
Combine like terms
90=6x+15
Subtracts 15 from both sides
75=6x
Divide by six to get x by itself
12.5=x
Does this help?
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In a 30°- 60° - 90° right triangle, the longer leg has a length of 10 square root of 3. what is the length of the shorter leg?
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In a 30°-60°-90° triangle, the longer leg is √3 times the shorter leg. In your triangle, ...
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Which angle is an alternate interior angle to 8?
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For the function f(x)=3x−4 , what is the ordered pair for the point on the graph when x=6 ? Show work for full credit.
Ksivusya [100]

To find the ordered pair, we simply set the x value equal to 6.

We already know the first coordinate will be 6.

f(6) = 3(6) - 4

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<h3><u>The y value is 14, and so our coordinate pair is (6, 14)</u></h3>
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What is (4c)^2d ?<br> c = 5 and d = 8
Arte-miy333 [17]

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Walk fifty meters at 30o north or east from the old oak tree. (2) Turn 45o to your left (you should now be facing 75o north of e
saw5 [17]

Answer:

A straight line of approximately 75 meters, 1.4º north

Step-by-step explanation:

Hi, let's make it step by step to make it clearer

1) If we walk 50 meters in 30º angle Northeast, assuming the Old Oak tree is the point 0,0 and we're dealing with vectors in R^{2}. To say 30º Northeast is 30º clockwise (or 60º counter clockwise).

2) Then there was a the turning point to the left. If I turn to the left, on my compass 45º , I'll face 75º northeast.

3) Finally, the last vector leads to the treasure from the Old Oak Tree, i.e. the resultant.

So, let's calculate the norm which is the length of the each vector.

1) Graphing them we can find the points, then the components and then calculate the norm, the length of each vector.  

Since the Oak Tree is on (0,0). The turning point (50,86.61) and the Rock (R=(1.4,74,85) we can write the following vectors:

\vec{u}=\left \langle 50,86.61 \right \rangle\\\vec{v}=\left \langle -48.6,-11.76\right \rangle\\\vec{w}=\left \langle 1.4,74.85 \right \rangle

Now, let's calculate each vector length by calculating the norm.

\left \| \vec{u} \right \|=\sqrt{50^{2}+86.6^2}=100\\\left \| \vec{v} \right \|=\sqrt{(-48.6)^2+(-11.76)^2}=50\\\left \| \vec{u} \right \|=\sqrt{(1.4)^2+(74.85)^2}=74.86

The path is almost 75 meters. And since it is less than 15º degrees to the left of the North (or to the right) its direction is still north of the Old Oak Tree.

8 0
3 years ago
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