Write the multiplicative comparison statement as an equation. <br><br> 18 is 3 times as many as 6.

Find x and y. <br><br> pls pls help

Anna [14]

Answer:

x = 65 , y = 64

Step-by-step explanation:

y - 16 and 2y + 4 are same- side interior angles and sum to 180° , that is

y - 16 + 2y + 4 = 180

3y - 12 = 180 ( add 12 to both sides )

3y = 192 ( divide both sides by 3 )

y = 64

then

2y + 4 = 2(64) + 4 = 128 + 4 = 132

the exterior angle of a triangle is equal to the sum of the 2 opposite interior angles.

2y + 4 is an exterior angle of the triangle , then

x + 67 = 132 ( subtract 67 from both sides )

x = 65

7 0

2 years ago

Find the measure of angle BCD in the following parallelogram. what's does BCD equal

DiKsa [7]

Answer:

120 degrees

Step-by-step explanation:

This is a quadrilateral, and by definition, the sum of all the interior angles is 360. So, we can just sum up all the expressions and set them equal to 360:

2x + x + 2x + x = 360

Combine like terms and solve for x:

6x = 360

x = 360/6 = 60

<BCD = 2x = 2 * 60 = 120

Thus, the answer is 120 degrees.

Hope this helps!

4 0

4 years ago

Triangle JKL has vertices J(2,5), K(1,1), and L(5,2). Triangle QNP has vertices Q(-4,4), N(-3,0), and P(-7,1). Is (triangle)JKL

Tems11 [23]

Answer:

Yes they are

Step-by-step explanation:

In the triangle JKL, the sides can be calculated as following:

J(2;5); K(1;1)

=> JK = $\sqrt{(1-2)^{2} + (1-5)^{2} } = \sqrt{(-1)^{2}+(-4)^{2} } = \sqrt{1+16}=\sqrt{17}$

J(2;5); L(5;2)

=> JL = $\sqrt{(5-2)^{2} + (2-5)^{2} } = \sqrt{3^{2}+(-3)^{2} } = \sqrt{9+9}=\sqrt{18} = 3\sqrt{2}$

K(1;1); L(5;2)

=> KL = $\sqrt{(5-1)^{2} + (2-1)^{2} } = \sqrt{4^{2}+1^{2} } = \sqrt{1+16}=\sqrt{17}$

In the triangle QNP, the sides can be calculate as following:

Q(-4;4); N(-3;0)

=> QN = $\sqrt{[-3-(-4)]^{2} + (0-4)^{2} } = \sqrt{1^{2}+(-4)^{2} } = \sqrt{1+16}=\sqrt{17}$

Q (-4;4); P(-7;1)

=> QP = $\sqrt{[-7-(-4)]^{2} + (1-4)^{2} } = \sqrt{(-3)^{2}+(-3)^{2} } = \sqrt{9+9}=\sqrt{18} = 3\sqrt{2}$

N(-3;0); P(-7;1)

=> NP = $\sqrt{[-7-(-3)]^{2} + (1-0)^{2} } = \sqrt{(-4)^{2}+1^{2} } = \sqrt{16+1}=\sqrt{17}$

It can be seen that QPN and JKL have: JK = QN; JL = QP; KL = NP

=> They are congruent triangles

7 0

3 years ago

Read 2 more answers

100 points! Please help asap! Look at the picture attached. My teacher said that both 1 and 2 are neither can someone explain wh

Neko [114]

Answer:

Your teacher is right, there is not enough info

Step-by-step explanation:

<h3>Question 1</h3>

We can see that RS is divided by half

The PQ is not indicated as perpendicular to RS or RQ is not indicates same as QS

So P is not on the perpendicular bisector of RS

<h3>Question 2</h3>

We can see that PD⊥DE and PF⊥FE

There is no indication that PD = PF or ∠DEP ≅ FEP

So PE is not the angle bisector of ∠DEF

6 0

3 years ago

Which ordered pair makes both inequalities true? y < 3x – 1 y > –x + 4 On a coordinate plane, 2 straight lines are shown.

Katen [24]

Answer:

(4,0)

Step-by-step explanation:

we have

$y< 3x-1$ ----> inequality A

$y \geq -x+4$ ----> inequality B

we know that

If a ordered pair is a solution of the system of inequalities, then the ordered pair must satisfy both inequalities (makes true both inequalities)

Verify each ordered pair

case 1) (4,0)

<em>Inequality A</em>

$0< 3(4)-1$

$0< 11$ ----> is true

<em>Inequality B</em>

$0 \geq -(4)+4$

$0 \geq 0$ ----> is true

so

the ordered pair makes both inequalities true

case 2) (1,2)

<em>Inequality A</em>

$2< 3(1)-1$

$2< 2$ ----> is not true

so

the ordered pair not makes both inequalities true

case 3) (0,4)

<em>Inequality A</em>

$4< 3(0)-1$

$4< -1$ ----> is not true

so

the ordered pair not makes both inequalities true

case 4) (2,1)

<em>Inequality A</em>

$1< 3(2)-1$

$1< 5$ ----> is true

<em>Inequality B</em>

$1 \geq -(2)+4$

$1 \geq 2$ ----> is not true

so

the ordered pair not makes both inequalities true

5 0

3 years ago

Read 2 more answers

Write the multiplicative comparison statement as an equation. 18 is 3 times as many as 6.