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AnnyKZ [126]
3 years ago
8

I NEED HELP ON DOING THIS

Mathematics
2 answers:
-BARSIC- [3]3 years ago
5 0
1. Write out an equation to solve this.
2x + 7x - 9 = 90 degrees

2. Add 2x to 7x
9x - 9 = 90 degrees

3. Add 9 to both sides of the equal sign
9x = 99 degrees

4. Divide 9x by 9 to get x
x = 11 degrees
stira [4]3 years ago
4 0
If all the angles add up to 90 degrees then you can set the equation 7x-9*2x=90
Simplify
9x-9=90
Plus 9 both sides
9x=99
Divide by 9 on both sides
X=11
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It takes 2 1/2 posterboard to display half of your project how many poster boards will you need for your entire project
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If 2 1/2 posterboard is enough for 1/2 of the project, it will take another 2 1/2 posterboard to display the other half of the project.

2 1/2 + 2 1/2 = 4 + 1/2 + 1/2 = 4 + 1 = 5

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KengaRu [80]

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Step-by-step explanation:

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3 years ago
Let T be the plane-2x-2y+z =-13. Find the shortest distance d from the point Po=(-5,-5,-3) to T, and the point Q in T that is cl
GaryK [48]

Answer:

d=10u

Q(5/3,5/3,-19/3)

Step-by-step explanation:

The shortest distance between the plane and Po is also the distance between Po and Q. To find that distance and the point Q you need the perpendicular line x to the plane that intersects Po, this line will have the direction of the normal of the plane n=(-2,-2,1), then r will have the next parametric equations:

x=-5-2\lambda\\y=-5-2\lambda\\z=-3+\lambda

To find Q, the intersection between r and the plane T, substitute the parametric equations of r in T

-2x-2y+z =-13\\-2(-5-2\lambda)-2(-5-2\lambda)+(-3+\lambda) =-13\\10+4\lambda+10+4\lambda-3+\lambda=-13\\9\lambda+17=-13\\9\lambda=-13-17\\\lambda=-30/9=-10/3

Substitute the value of \lambda in the parametric equations:

x=-5-2(-10/3)=-5+20/3=5/3\\y=-5-2(-10/3)=5/3\\z=-3+(-10/3)=-19/3\\

Those values are the coordinates of Q

Q(5/3,5/3,-19/3)

The distance from Po to the plane

d=\left| {\to} \atop {PoQ}} \right|=\sqrt{(\frac{5}{3}-(-5))^2+(\frac{5}{3}-(-5))^2+(\frac{-19}{3}-(-3))^2} \\d=\sqrt{(\frac{5}{3}+5))^2+(\frac{5}{3}+5)^2+(\frac{-19}{3}+3)^2} \\d=\sqrt{(\frac{20}{3})^2+(\frac{20}{3})^2+(\frac{-10}{3})^2}\\d=\sqrt{\frac{400}{9}+\frac{400}{9}+\frac{100}{9}}\\d=\sqrt{\frac{900}{9}}=\sqrt{100}\\d=10u

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