x + 2y + 3 = 0
Subtract x from both sides.
2y + 3 = -x
Subtract 3 from both sides.
2y = -x - 3
Divide by 2 on both sides
y = -(x+3)/2
x + y + 4 = 0
Subtract x and 4 from both sides
y = -x - 4
3x - 2y + 4 = 0
Subtract 3x and 4 from both sides.
-2y = -3x -4
Divide by -2 from both sides.
y = -(3x + 4) / 2
The answer is the graph that contains these slopes and lines on the graph, which was not provided.
Answer:
Step-by-step explanation:
ABC have sides: 5, 7 and 10
5^2 + 7^2 = 25+47 = 72 < 10^2
so triangle ABC is obtuse
JKL has sides: 12, 35 and 37
12^2 + 35^2 = 144 + 1225 = 1369 = 37^2
so triangle JKL is right-angled
PQR has sides 12, 10 and 16
12^2 + 10^2 = 144 + 100 = 244 > 16^2
so triangle PQR is acute
Step-by-step explanation:
cot x / (1 + csc x)
Multiply by conjugate:
cot x / (1 + csc x) × (1 − csc x) / (1 − csc x)
Distribute the denominator:
cot x (1 − csc x) / (1 − csc²x)
Use Pythagorean identity:
cot x (1 − csc x) / (-cot²x)
Divide:
(csc x − 1) / cot x
Answer:
Down below
Step-by-step explanation:
a. The range of y = sinθ is [-1,1]
b. The period of y = cosθ is 2π
c. The asymptotes of y = tanθ are -π2, π2, πn
d. The amplitude of y = sinθ is 1
e. The period of y = tanθ is π
f. The max value of y = cosθ is 1
Your answer is -0.51923077
