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3 years ago

Find the perimeter of the figure. ( Its a square, and one side says 20 ft while another says 18ft. I am confused. )

Mathematics

1 answer:

Luba_88 [7]3 years ago

3 0

It might look like a square but it might actually be a rectangle but the answer is 76ft

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4 - 2x = 10 plz solve

posledela

Answer:

x=7

Step-by-step explanation:

Opposite of subtraction is adding so you do 4+10=14 so that means that x equals seven because 2x7 equals 14.

4 0

3 years ago

Find the gradient of the line Joining the points p(-2,6) and Q(3,-2)

Rudiy27

Answer:

$- \frac{8}{5}$

Step-by-step explanation:

$\boxed{gradient = \frac{y1 - y2}{x1 - x2} }$

(x₁, y₁) is the 1st coordinate while (x₂, y₂) is the 2nd coordinate.

Gradient of the line

$= \frac{6 - ( - 2)}{ - 2 - 3}$

$= \frac{6 + 2}{ - 5}$

$= - \frac{8}{5}$

7 0

3 years ago

Pls help me and thank you!

Rudik [331]

Answer : -3/4 ..............

6 0

3 years ago

What is the simplified expression for the expression below? 4(x + 8) + 5(x – 3)Which expression is equivalent to 6(x – 4)

Mashutka [201]

Answer:4x+32+5x-15

Step-by-step explanation: that’s the simplified equation, you would just have to distribute the 4 to its own equation and the 5 into its own equation.. but i think the second part of ur question is missing..

8 0

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

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

7 0

3 years ago