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vodomira [7]
3 years ago
5

I Need Help Solving For X And Y.

Mathematics
1 answer:
Margaret [11]3 years ago
5 0
2x-25=x+5
x=30
30+5=35
180-35=145
9y+28=145
9y=117
y=13
I hope help you this question
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Multiply. a^5 . a . a^0 . a^-3
stiks02 [169]
a^5 \cdot a \cdot a^0 \cdot a^{-3}=a^{5+1-3}\cdot 1 = a^3
6 0
4 years ago
A car is priced at $12,000. The monthly payment is 3% of the price. How much is the monthly payment?
Yuliya22 [10]

Answer: $ 360

Step-by-step explanation:

since the monthly payment is 3% of the price , this means that the monthly payment is 3% of 12000

= 0.03 x 12000

= $ 360

5 0
3 years ago
Please help. Thank you!
Ann [662]
<h2 />

<em><u>WRITE</u></em><em><u> </u></em><em><u>FORMULA</u></em>

  • <u>B</u><u>×</u><u>H</u><u>=</u><u> </u><u>Area</u><u> </u><u>of</u><u> </u><u>Parallelogram</u>

<em><u>SUBSTITUTE</u></em>

  • <em><u>2.7cm</u></em><em><u> </u></em><em><u>×</u></em><em><u> </u></em><em><u> </u></em><em><u>3.1cm</u></em>

<em><u>SOLUTION</u></em>

  • <u>b</u><u>×</u><u>h</u>=Area of Parallelogram

  • 2.7cm×3.1cm

  • 8.37cm=Area of Parallelogram

<em>Therefore</em><em> </em><em>the</em><em> </em><em>Area</em><em> </em><em>of</em><em> </em><em>Parallelogram</em><em> </em><em>is</em><em> </em><em>8.37cm</em>

<em><u>im</u></em><em><u> </u></em><em><u>Filipino</u></em>

7 0
3 years ago
Read 2 more answers
Find a vector ü normal to the plane 2x + 2y + 2 = 3
eduard

Answer:

ü=2i+2j+0k

Step-by-step explanation:

The given plane 2x + 2y + 2 = 3 can also be written as:

2x+2y=3-2

2x+2y=1

The general equation for a plane is Ax+By+Cz=D and by definition the normal vector of that plane is n=Ai+Bj+Ck

Where i,j,k are the unit vectors

In order to demostrate that the vector n is normal to the plane, let R1=(a1,b1,c1) and R2=(a2,b2,c2) be two vectors that are in the plane.

If R1 ∈ Ax+By+Cz=D then Aa1+Bb1+Cc1=D

If R2 ∈ Ax+By+Cz=D then Aa2+Bb2+Cc2=D

Therefore, the vector R1R2=R2-R1=(a2-a1)i+(b2-b1)j+(c2-c1)k

You can apply the dot product. <em>If the dot product of the two vectors is zero then the vectors are normal.</em>

R1R2_{o}n= [(a2-a1)i+(b2-b1)j+(c2-c1)k]_{o}(Ai+Bj+Ck)\\R1R2_{o}n = A(a2-a1) + B(b2-b1) + C(c2-c1)\\R1R2_{o}n = Aa2 + Bb2 +Cc2 - (Aa1+Bb1+Cc1)\\R1R2_{o}n = D - D\\R1R2_{o}n = 0

So, the vector which components are A,B,C is normal to the plane becase it is normal to any vector contained in the plane.

In this case:

A=2, B=2, C=0

ü=2i+2j+0k

8 0
3 years ago
I need help with this
kondor19780726 [428]
C^2/f^2

hope that helps
4 0
3 years ago
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