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3 years ago

ABCD is a parallelogram. What are the measures of angles B and C? b=(3n+20) D= (6n-25)

Mathematics

1 answer:

dybincka [34]3 years ago

8 0

Answer:

B = 65

C = 115

Step-by-step explanation:

Givens

ABCD is a parallelogram

It is lettered in such a way that B is opposite D

B = 3n+20

D = 6n-25

Solution

<B = <D Property of Parallelogram

3n + 20 = 6n - 25 Add 25 to both sides

3n + 20 + 25 = 6n Combine

3n +45 = 6n Subtract 3n from both sides

3n +45 = 6n-3n Combine

45 = 3n Divide by 3

45/3=3n/3

n = 15 Value of n

B = 3n + 20 Find B. Let n = 15

B= 3*15+20 Combine

B = 45 + 20 Combine

B = 65

==================

B and C are supplementary. They add up to 180

B + C = 180

65 + C = 180

C = 180-65

C = 115

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A jetski rental company charges a $50 deposit, plus $25 per hour to rent a jetski. This can be represented by the formula T = 25

loris [4]

Answer:

None, the answer will be the same

Step-by-step explanation:

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3 years ago

Find the distance between a point (– 2, 3 – 4) and its image on the plane x+y+z=3 measured across a line (x + 2)/3 = (2y + 3)/4

dimaraw [331]

Answer:

Distance of the point from its image = 8.56 units

Step-by-step explanation:

Given,

Co-ordinates of point is (-2, 3,-4)

Let's say

$x_1\ =\ -2$

$y_1\ =\ 3$

$z_1\ =\ -4$

Distance is measure across the line

$\dfrac{x+2}{3}\ =\ \dfrac{2y+3}{4}\ =\ \dfrac{3z+4}{5}$

So, we can write

$\dfrac{x-x_1+2}{3}\ =\ \dfrac{2(y-y_1)+3}{4}\ =\ \dfrac{3(z-z_1)+4}{5}\ =\ k$

$=>\ \dfrac{x-(-2)+2}{3}\ =\ \dfrac{2(y-3)+3}{4}\ =\ \dfrac{3(z-(-4))+4}{5}\ =\ k$

$=>\ \dfrac{x+4}{3}\ =\ \dfrac{2y-3}{4}\ =\ \dfrac{3z+16}{5}\ =\ k$

$=>\ x\ =\ 3k-4,\ y\ =\ \dfrac{4k+3}{2},\ z\ =\ \dfrac{5k-16}{3}$

Since, the equation of plane is given by

x+y+z=3

The point which intersect the point will satisfy the equation of plane.

So, we can write

$3k-4+\dfrac{4k+3}{2}+\dfrac{5k-16}{3}\ =\ 3$

$=>6(3k-4)+3(4k+3)+2(5k-16)\ =\ 18$

$=>18k-24+12k+9+10k-32\ =\ 18$

$=>\ k\ =\dfrac{13}{8}$

So,

$x\ =\ 3k-4$

$=\ 3\times \dfrac{13}{8}-4$

$=\ \dfrac{7}{4}$

$y\ =\ \dfrac{4k+3}{2}$

$=\ \dfrac{4\times \dfrac{13}{8}+3}{2}$

$=\ \dfrac{19}{4}$

$z\ =\ \dfrac{5k-16}{3}$

$=\ \dfrac{5\times \dfrac{13}{8}-16}{3}$

$=\ \dfrac{-21}{8}$

Now, the distance of point from the plane is given by,

$d\ =\ \sqrt{(x-x_1)^2+(y-y_1)^2+(z-z_1)^2}$

$=\ \sqrt{(-2-\dfrac{7}{4})^2+(3-\dfrac{19}{4})^2+(-4+\dfrac{21}{8})^2}$

$=\ \sqrt{(\dfrac{-15}{4})^2+(\dfrac{-7}{4})^2+(\dfrac{9}{8})^2}$

$=\ \sqrt{\dfrac{225}{16}+\dfrac{49}{16}+\dfrac{81}{64}}$

$=\ \sqrt{\dfrac{1177}{64}}$

$=\ 4.28$

So, the distance of the point from its image can be given by,

D = 2d = 2 x 4.28

= 8.56 unit

So, the distance of a point from it's image is 8.56 units.

4 0

3 years ago

Can someone please help me with these math problems

yuradex [85]

Answer: 138

Step-by-step explanation:

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3 years ago

Zalmon walks three fourhs of a mile in three tenths of an hour. what is his speed in miles per hour?

nataly862011 [7]

2.5 miles per hour
Because 3/4 divided by 3/10 is equal to 2.5

4 0

3 years ago

Read 2 more answers

PLEASE HELP ASAP!!! CORRECT ANSWERS ONLY PLEASE!!!

MAXImum [283]

The last option is the answer. You can double check by substituting the -3 for the x values and the 17 for y values and the statements would be true.

4 0

3 years ago