Answer: -4
Step-by-step:
-x^2+5x-1 x _ = 4x^2-20x+4
4x^2-20x+4 / -x^2+5x-1
In the blank the answer would be -4 which is the monomial.
Answer:
Step-by-step explanation:
Use SOH CAH TOA to recall how the trig functions fit on a triangle
SOH: Sin(Ф)= Opp / Hyp
CAH: Cos(Ф)= Adj / Hyp
TOA: Tan(Ф) = Opp / Adj
5)
Adj = 14
Hyp = 26
∠X
so use
CAH
Cos(X) = 14/26
X = arcCos(14/26)
X = 57.421°
X = 57.4 ° ( rounded to nearest 10th )
6)
∠X
Hyp = 46
Opp = 12
use SOH
Sin(x) = 12/46
X = arcSin(12/46)
X = 15.121°
X = 15.1 ° ( rounded to nearest 10th )
7)
∠X
Adj = 29
Opp = 24
use TOA
Tan(x) = 29 / 24
X = arcTan( 29 /24)
X = 50.389
X = 50.4 ° ( rounded to nearest 10th )
8)
∠X
Adj = 22
Opp = 6
use TOA agian
Tan(x) = 6 / 22
X = arcTan(6/22)
X = 5.194
X = 5.2 ° ( rounded to the nearest 10th )
:)
So basycilly slope intercept form is y=mx+b where m=slope and b=y intercept
slope=-1
subsiute
y=-1x+b
y=-x+b
so we are given the point (-3,2)
x=-3 and y=2 is one solution
subsitue and solve for b
2=-(-3)+b
2=3+b
subtract 3 from both sides
-1=b
the slopt inercetp form is y=-x-1
you can only choose 4 unique ways Step-by-step explanation:
Answer:
10.50°C
Step-by-step explanation:
Given x = 2 + t , y = 1 + 1/2t where x and y are measured in centimeters. Also, the temperature function satisfies Tx(2, 2) = 9 and Ty(2, 2) = 3
The rate of change in temperature of the bug path can be expressed using the composite formula:
dT/dt = Tx(dx/dt) + Ty(dy/dt)
If x = 2+t; dx/dt = 1
If y = 1+12t; dy/dt = 1/2
Substituting the parameters gotten into dT/dt we will have;
dT/dt = 9(1)+3(1/2)
dT/dt = 9+1.5
dT/dt = 10.50°C/s
Hence the rate at which the temperature is rising along the bug's path is 10.50°C/s
