Answer:
Emperor tamarin, Mandrill, and Japanese macaque.
Step-by-step explanation:
Answer:
C. increases rapidly.
Step-by-step explanation:
tan(θ) = sin(θ)/cos(θ)
Now, when sin 90 = 1
and cos 90 = 0
so, tan(90) = 1/0 = not defined.
<em>(1/0 is infinity and its value is not defined)</em>
So, when angle θ increases to 90°, then the value of tan(θ) increases rapidly, as shown in the figure below.
Answer:
B) 25°
Step-by-step explanation:
Assuming the quadrilateral is the one shown in the picture attached, then a triangle ADC is form, where m∠ADC = 125° and m∠1 = 30°. The addition of the three angles of a triangle is equal to 180°, so:
m∠ADC + m∠1 + m∠2 = 180°
Replacing with the known values and isolating m∠2 we get:
125° + 30° + m∠2 = 180°
m∠2 = 180° - 125° - 30°
m∠2 = 25°
Answer:
364
Step-by-step explanation:
Answer:

Step-by-step explanation:
First, find the <em>rate of</em><em> </em><em>change</em><em> </em>[<em>slope</em>]:


Then plug these coordinates into the Slope-Intercept Formula instead of the <em>Point-Slope Formula</em> because you get it done much swiftly. It does not matter which ordered pair you choose:
1 = ⅖[6] + b
2⅖
−1⅖ = b
y = ⅖x - 1⅖ >> Line in <em>Slope-Intercept</em><em> </em><em>Form</em><em> </em>
If you need it written in <em>Standard Form</em>:
y = ⅖x - 1⅖
-⅖x -⅖x
_________
−⅖x + y = −1⅖ [We do not want fractions in our standard equation, so multiply by the denominator to get rid of it.]
−5[−⅖x + y = −1⅖]
2x - 5y = 7 >> Line in <em>Standard</em><em> </em><em>Form</em>
_______________________________________________
−3 = ⅖[−4] + b
−1⅗
−1⅖ = b
y = ⅖x - 1⅖ >> Line in <em>Slope-Intercept Form</em>
If you need it written in <em>Standard Form</em>:
y = ⅖x - 1⅖
-⅖x -⅖x
_________
−⅖x + y = −1⅖ [We do not want fractions in our standard equation, so multiply by the denominator to get rid of it.]
−5[−⅖x + y = −1⅖]
2x - 5y = 7 >> Line in <em>Standard</em><em> </em><em>Form</em>
** You see? I told you it did not matter which ordered pair you choose because you will ALWAYS get the exact same result.
I am joyous to assist you anytime.