All categories

3 years ago

What is the largest integer value an angle of a parallelogram can have?

Mathematics

1 answer:

zloy xaker [14]3 years ago

4 0

Only 2 or more 180 degree angles can be considered parallel.

You might be interested in

A line passes through the points (1, –6) and (4, 3).

Marina CMI [18]

Answer:

y = 3x - 9

Step-by-step explanation:

We first need to find the slope (m)

m = $\frac{y_2-y_1}{x_2-x_1}$ = $\frac{3 - (-6)}{4-1}$ = $\frac{9}{3}$ = 3

y = mx + b

y = 3x + b

(4, 3)

y = 3x + b

3 = 3(4) + b

3 = 12 + b

-12 -12

-9 = b

(1, –6)

y = 3x + b

-6 = 3(1) + b

-6 = 3 + b

-3 -3

-9 = b

y = 3x - 9

To check: (1, –6)

y = 3x - 9

-6 = 3(1) - 9

-6 = 3 - 9

-6 = -6

Hope this helps!

6 0

2 years ago

Solve for m: 18d + 2m + 9k < 5k + 10

BigorU [14]

Use the photomath app

7 0

3 years ago

Using the slope formula, find the slope of the line through the given points ( 9, 4 ) and ( 4, 4 )

irakobra [83]

Answer: 0

Step-by-step explanation: it’s undefined

8 0

3 years ago

Read 2 more answers

HELP PLEASE WILL MARK BRAINLIEST!!

Arte-miy333 [17]

Answer:

Answer is option d)346.4 ft

Step-by-step explanation:

$from \: the \: figure \\ AB = 300 \: ft \: \: \: angle \: C = 60 \:( theta)\: degress \\ in \: triangle \: ABC \: using \: tan \: theta\\ \tan(60) = \frac{opposite \: side}{adjacent \: side} \\ \tan(60) = \frac{300}{BC} \\ \sqrt{3} = \frac{300}{BC} \\ BC = \frac{300}{ \sqrt{3} } \\ BC = 100 \sqrt{3} ft \\ then \: in \: triangle \: ABD \: using \: tan \: theta \\ angle \: D = 30 \: degrees \: (theta) \\ \tan(30 ) =\frac{opposite \: side}{adjacent \: side} \\ \tan(30) = \frac{300}{BD} \\ \frac{1}{ \sqrt{ 3} } = \frac{300}{BD} \\ BD = 300 \sqrt{3} \\ CD = BD - BC \\ = 300 \sqrt{3 } - 100 \sqrt{3} \\ = (300 - 100) \sqrt{3} \\ = 200 \sqrt{3 } \\ = 200 \times 1.732 \\ = 346.4 \: ft$