What is the answer to 2 + -1/3 =

Please help me on this problem

Orlov [11]

Answer:

The Proof with Figure, Statement and Reasons is given below.

Step-by-step explanation:

Given:

∠DAF ≅ ∠EBF,

DF ≅ FE

Prove:

Δ ADF ≅ Δ BFE

Proof:

Statements Reasons

a. ∠DAF ≅ ∠EBF ...........................Given

48. ∠DFA ≅ ∠EFB ..........................Vertical Angles are congruent

DF ≅ FE ...........................49. Given

50. Δ ADF ≅ Δ BFE .........................By Angle-Angle-Side test

6 0

3 years ago

I need the answer for number 6 please:)

Goshia [24]

Answer: 12.5

25 / 2 = 12.5

and so does

25 x .50 = 12.5

We are just cutting the price in half, or 50%.

Hope this helps you!

5 0

3 years ago

Who know the answer of this? And can u pls explain for me?

anyanavicka [17]

Answer:

1.5 kg

Step-by-step explanation:

17*92 g = 1564

1564/1000= 1.564 kg

4 0

3 years ago

Read 2 more answers

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

show me how to work this problem, point A(7,9)is dilated from the origin by scale factor r was= 6. what are the coordinates of p

frutty [35]

If the point is dilated from the origin by a scale factor of 6

Then the coordinates of point A becomes

A( 6x7 , 6x9)

$A_{0,6}(7,9)=(42,\text{ 54)}$

6 0

1 year ago