HEELLLPPP ME?!?!?!? ANYONE?

Solve for x. 25 - 3x = -5(1 - x) - 2x

grandymaker [24]

Answer:

$x = 5$

Step-by-step explanation:

25 - 3x = -5(1 - x) - 2x

We have to first get rid of the parenthesis:

$-5 * 1$ ==> -5

$-5 * -x$ ==> 5x

So now you should have:

25 - 3x = -5 + 5x - 2x

Combine like terms:

25 - 3x = -5 + 5x - 2x

25 - 3x = -5 + 3x

Add 3x to both sides:

25 - 3x = -5 + 3x

+ 3x = + 3x

And you should have:

25 = -5 + 6x

Add 5 to both sides:

25 = -5 + 6x

+5 = +5

30 = 6x

Divide 6 to both sides:

$\frac{30}{6}$ = $\frac{6x}{6}$

5 = x

3 0

3 years ago

Read 2 more answers

Given that sec x - tan x=1÷4

solniwko [45]

Answer:Let =tan2

t

=

tan

⁡

x

2

, then cos=1−21+2

cos

⁡

x

=

1

−

t

2

1

+

t

2

and tan=21−2

tan

⁡

x

=

2

t

1

−

t

2

.

Now

1+21−2+21−2(1+)21−21+1−tan2

Step-by-step explanation:

5 0

3 years ago

an assistant recieved a 3% raise bringing the salary to 43,989 what was the salary defore the raise? (round to the nearest dolla

lesya692 [45]

If her original salary is x, then her new salary is 103% of x.
To obtain x from 43989, you therefore divide 43989 by 1.03 to get:
42708 (to nearest dollar).
You can the check it by multiplying by 1.03.

4 0

3 years ago

Read 2 more answers

Write the given differential equation in the form L(y) = g(x), where L is a linear differential operator with constant coefficie

melamori03 [73]

Answer:

The complete solution is

$\therefore y= Ae^{3x}+Be^{-\frac13 x}-\frac43$

Step-by-step explanation:

Given differential equation is

3y"- 8y' - 3y =4

The trial solution is

$y = e^{mx}$

Differentiating with respect to x

$y'= me^{mx}$

Again differentiating with respect to x

$y''= m ^2 e^{mx}$

Putting the value of y, y' and y'' in left side of the differential equation

$3m^2e^{mx}-8m e^{mx}- 3e^{mx}=0$

$\Rightarrow 3m^2-8m-3=0$

The auxiliary equation is

$3m^2-8m-3=0$

$\Rightarrow 3m^2 -9m+m-3m=0$

$\Rightarrow 3m(m-3)+1(m-3)=0$

$\Rightarrow (3m+1)(m-3)=0$

$\Rightarrow m = 3, -\frac13$

The complementary function is

$y= Ae^{3x}+Be^{-\frac13 x}$

y''= D², y' = D

The given differential equation is

(3D²-8D-3D)y =4

⇒(3D+1)(D-3)y =4

Since the linear operation is

L(D) ≡ (3D+1)(D-3)

For particular integral

$y_p=\frac 1{(3D+1)(D-3)} .4$

$=4.\frac 1{(3D+1)(D-3)} .e^{0.x}$ [since $e^{0.x}=1$ ]

$=4\frac{1}{(3.0+1)(0-3)}$ [ replace D by 0 , since L(0)≠0]

$=-\frac43$

The complete solution is

y= C.F+P.I

$\therefore y= Ae^{3x}+Be^{-\frac13 x}-\frac43$

4 0

3 years ago

Two standard number cubes (i.e., dice) each with faces numbered 1-6 are rolled. What is the probability that the product of the

xeze [42]

16= 4*4
you need to roll two 4s out of presumably 36 possibilities so 2/36 since you need one 4 and a second 4. check the total possibilities tho

3 0

3 years ago