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3 years ago

Given the information below, match the following items.

1 answer:

7 0

Let this be the guide in dealing with conditional statement.

Statement = if p, then q
Converse = if q, then p
Inverse = if NOT p, then NOT q
Contrapositive = if NOT q, then NOT P.

Statement:
If two angles add to 180°, then they are supplementary.

Converse:
If two angles are supplementary, then they add to 180°

Inverse:
If two angles do not add to 180°, then they are not supplementary.

Contrapositive:
If two angles are not supplementary, then they do not add to 180°

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In rectangle ABCD, AB = 10 cm and BC = 14 cm. If rectangle EFGH is the image after a dilation, which could be its side lengths?

lana [24]

Answer is D

After dilation the image remain same but it is stretched or shrinked to the original size.

That means the ratio of the sides and the angles between the sides remain same .

Here we did dilation of ABCD which made it to EFGH.

Hence the ratio of the corresponding sides of the original rectangle ABCD should remain same even after dilation.

the corresponding sides are : AB and EF

BC and FG

CD and GH

DA and HE

* Let us find ratio of the sides AB and BC

given that AB= 10 and BC= 14

AB/BC= 10/14 = 5 /7 ( 10 and 14 both are multiple of 2 so we reduced them by a factor of 2 )

* the raio of the corresponding sides EF and FG should be same ( 5/7)

in the option D EF= 25 and FG= 35

so EF/FG= 25/35 = 5 /7 ( both are multiple of 5 so we reduced them by the factor of 5 )

Since ratio of the corresponding sides are coming out to be same for the EFGH given in option D it should be the dilation of the ABCD

4 0

2 years ago

Read 2 more answers

What is the length of arc S shown below? Enter an exact expression.

aleksandrvk [35]

Answer:

S = 15π = 47.12

length of arc S = 15π

Attached is the image of the arc.

Step-by-step explanation:

Given;

Radius arc r = 10

Angle at the center of arc ⍉ = 3π/2

The length of the arc S can be derived using the formula;

S = r × ⍉

Substituting the values;

S = 10 × 3π/2

S = 30π/2

S = 15π = 47.12

length of arc S = 15π

6 0

3 years ago

Use lagrange multipliers to find the shortest distance, d, from the point (4, 0, −5 to the plane x y z = 1

Varvara68 [4.7K]

I assume there are some plus signs that aren't rendering for some reason, so that the plane should be $x+y+z=1$ .

You're minimizing $d(x,y,z)=\sqrt{(x-4)^2+y^2+(z+5)^2}$ subject to the constraint $f(x,y,z)=x+y+z=1$ . Note that $d(x,y,z)$ and $d(x,y,z)^2$ attain their extrema at the same values of $x,y,z$ , so we'll be working with the squared distance to avoid working out some slightly more complicated partial derivatives later.

The Lagrangian is

$L(x,y,z,\lambda)=(x-4)^2+y^2+(z+5)^2+\lambda(x+y+z-1)$

Take your partial derivatives and set them equal to 0:

$\begin{cases}\dfrac{\partial L}{\partial x}=2(x-4)+\lambda=0\\\\\dfrac{\partial L}{\partial y}=2y+\lambda=0\\\\\dfrac{\partial L}{\partial z}=2(z+5)+\lambda=0\\\\\dfrac{\partial L}{\partial\lambda}=x+y+z-1=0\end{cases}\implies\begin{cases}2x+\lambda=8\\2y+\lambda=0\\2z+\lambda=-10\\x+y+z=1\end{cases}$

Adding the first three equations together yields

$2x+2y+2z+3\lambda=2(x+y+z)+3\lambda=2+3\lambda=-2\implies \lambda=-\dfrac43$

and plugging this into the first three equations, you find a critical point at $(x,y,z)=\left(\dfrac{14}3,\dfrac23,-\dfrac{13}3\right)$ .

The squared distance is then $d\left(\dfrac{14}3,\dfrac23,-\dfrac{13}3\right)^2=\dfrac43$ , which means the shortest distance must be $\sqrt{\dfrac43}=\dfrac2{\sqrt3}$ .