How can i use doubles when multiplying with 4 to find 4x8?

Solve this problem T=u+v/2 solve for u

AfilCa [17]

Answer:

$T = \frac{u + v}{2}$

Cross multiply

We have

2T = u + v

u + v = 2T

Move v to the right side of the equation to make u stand alone

We have the final answer as

Hope this helps you

5 0

3 years ago

Prove that sin3a-cos3a/sina+cosa=2sin2a-1

Sloan [31]

Answer:

$\frac{sin(3a)-cos(3a)}{sin(a)+cos(a)} =2sin(2a)-1$

Step-by-step explanation:

we are given

$\frac{sin(3a)-cos(3a)}{sin(a)+cos(a)} =2sin(2a)-1$

we can simplify left side and make it equal to right side

we can use trig identity

$sin(3a)=3sin(a)-4sin^3(a)$

$cos(3a)=4cos^3(a)-3cos(a)$

now, we can plug values

$\frac{(3sin(a)-4sin^3(a))-(4cos^3(a)-3cos(a))}{sin(a)+cos(a)}$

now, we can simplify

$\frac{3sin(a)-4sin^3(a)-4cos^3(a)+3cos(a)}{sin(a)+cos(a)}$

$\frac{3sin(a)+3cos(a)-4sin^3(a)-4cos^3(a)}{sin(a)+cos(a)}$

$\frac{3(sin(a)+cos(a))-4(sin^3(a)+cos^3(a))}{sin(a)+cos(a)}$

now, we can factor it

$\frac{3(sin(a)+cos(a))-4(sin(a)+cos(a))(sin^2(a)+cos^2(a)-sin(a)cos(a)}{sin(a)+cos(a)}$

$\frac{(sin(a)+cos(a))[3-4(sin^2(a)+cos^2(a)-sin(a)cos(a)]}{sin(a)+cos(a)}$

we can use trig identity

$sin^2(a)+cos^2(a)=1$

$\frac{(sin(a)+cos(a))[3-4(1-sin(a)cos(a)]}{sin(a)+cos(a)}$

we can cancel terms

$=3-4(1-sin(a)cos(a))$

now, we can simplify it further

$=3-4+4sin(a)cos(a))$

$=-1+4sin(a)cos(a))$

$=4sin(a)cos(a)-1$

$=2\times 2sin(a)cos(a)-1$

now, we can use trig identity

$2sin(a)cos(a)=sin(2a)$

we can replace it

$=2sin(2a)-1$

so,

$\frac{sin(3a)-cos(3a)}{sin(a)+cos(a)} =2sin(2a)-1$

7 0

3 years ago

Read 2 more answers

Rite the first ten terms of a sequence whose first term is -10 and whose common difference is -2.

Lelechka [254]

-10, -10-2·1= -12, -10-2·2=-14, -10-2·3=-16, -10-2·4= -18, -10-2·5= -20,
-10-2·6= -22, -10-2·7 =-24, -10-2·8=-26, -10-2·9=-28

a=-10
d=-2
xn=a+d(n-1)

the first ten terms are
-10, -12, -14, -16, -18, -20, -22, -24, -26, -28

8 0

3 years ago

Question Help

Sergeeva-Olga [200]

Answer:

if u divide it then slop it u will get your answer

3 0

3 years ago

Graph this rational equation. Identify the points of discontinuity, holes, vertical asymptotes, x-intercepts, and horizontal asy

irakobra [83]

Step-by-step explanation:

We have given,

A rational function : f(x) = $\frac{x-2}{x-4}$

W need to find :

Point of discontinuity : - At x = 4, f(x) tends to reach infinity, So we get discontinuity point at x =4.

For no values of x, we get indetermined form (i.e $\frac{0}{0}$ ), Hence there is no holes

Vertical Asymptotes:

Plug y=f(x) = ∞ in f(x) to get vertical asymptote {We can us writing ∞ = $\frac{1}{0}$ }

i.e ∞ = $\frac{x-2}{x-4}$

or $\frac{1}{0}=\frac{x-2}{x-4}$

or x-4 =0

or x=4, Hence at x = 4, f(x) has a vertical asymptote

X -intercept :

Plug f(x)=0 , to get x intercept.

i.e 0 = $\frac{x-2}{x-4}$

or x - 2 =0

or x = 2

Hence at x=2, f(x) has an x intercept

Horizontal asymptote:

Plug x = ∞ in f(x) to get horizontal asymptote.

i.e f(x) = $\frac{x-2}{x-4}$ = $\frac{x(1-\frac{2}{x} )}{x(1-\frac{4}{x} )}$

or f(x) = $\frac{(1-\frac{2}{∞} )}{(1-\frac{4}{∞} )}$

or f(x) = 1 = y

hence at y =f(x) = 1, we get horizontal asymptote

4 0

3 years ago