What are the soulutions of x^2+6x-6=10?

pls help this is due soon (sorry for the drawing on the side, my brain goes like a thousand mph and i tend to draw instead of fo

slava [35]

Answer:

I don't have an answer and I am sorry but I am only in 6th grade so maybe get a tutor and an app that is good

3 0

2 years ago

PLS HELP DUE SOON SHOW YOUR WORK!

Vladimir79 [104]

9514 1404 393

Answer:

615.8 square feet
9 truckloads

Step-by-step explanation:

The area of the circular ring is given by the formula ...

A = πr²

where r is the radius, or half the diameter.

A = π(14 ft)² = 196π ft² ≈ 615.752 ft²

The area of the training area is about 615.8 square feet.

__

The number of truckloads of dirt required for coverage will be ...

(615.8 ft²)/(75 ft²/truckload) ≈ 8.21 truckloads

Since the dirt comes only in whole truckloads, Hilary needs 9 truckloads.

8 0

3 years ago

10 points I really need help can you write it out

rodikova [14]

Answer: 125 inches

Step-by-step explanation:

8 3/4 x 2 6/7 = 25

25 x 5 = 125

6 0

4 years ago

Use the method of "undetermined coefficients" to find a particular solution of the differential equation. (The solution found ma

Naddika [18.5K]

Answer:

The particular solution of the differential equation

= $\frac{-1,36,656cos5x+1,89,800 sin5x}{-1204}$ + $\frac{1}{37}185e^{6x})$

Step-by-step explanation:

Given differential equation y''(x) − 10y'(x) + 61y(x) = −3796 cos(5x) + 185e6x

The differential operator form $(D^{2} -10D+61)y(x) = −3796 cos(5x) + 185e^{6x}$

Rules for finding particular integral in some special cases:-

let f(D)y = $e^{ax}$ then

the particular integral $\frac{1}{f(D)} (e^{ax} ) = \frac{1}{f(a)} (e^{ax} ) if f(a)$ ≠ 0

let f(D)y = cos (ax ) then

the particular integral $\frac{1}{f(D)} (cosax ) = \frac{1}{f(D^2)} (cosax ) =\frac{cosax}{f(-a^2)}$ f(-a^2) ≠ 0

Given problem

$(D^{2} -10D+61)y(x) = −3796 cos(5x) + 185e^{6x}$

Particular integral:-

$P.I = \frac{1}{f(D)}( −3796 cos(5x) + 185e^{6x})$

$P.I = \frac{1}{D^2-10D+61}( −3796 cos(5x) + 185e^{6x})$

$P.I = \frac{1}{D^2-10D+61}( −3796 cos(5x) + \frac{1}{D^2-10D+61}185e^{6x})$

P.I = $I_{1} +I_{2}$

we will apply above two conditions, we get

$I_{1}$ =

$\frac{1}{D^2-10D+61}( −3796 cos(5x) = \frac{1}{(-25)-10D+61}( −3796 cos(5x) ( since D^2 = - 5^2)$ $= \frac{1}{(36-10D}( −3796 cos(5x) \\= \frac{1}{(36-10D}X\frac{36+10D}{36+10D} ( −3796 cos(5x)$

on simplification we get

= $\frac{1}{(36^2-(10D)^2}36+10D( −3796 cos(5x)$

= $\frac{-1,36,656cos5x+1,89,800 sin5x}{1296-100(-25)}$

= $\frac{-1,36,656cos5x+1,89,800 sin5x}{-1204}$

$I_{2}$ =

$\frac{1}{D^2-10D+61}185e^{6x}) = \frac{1}{6^2-10(6)+61}185e^{6x})$

$\frac{1}{37}185e^{6x})$

Now particular solution

P.I = $I_{1} +I_{2}$

P.I = $\frac{-1,36,656cos5x+1,89,800 sin5x}{-1204}$ + $\frac{1}{37}185e^{6x})$

8 0

3 years ago

Solve 20 – 3.8x= -14 - 5.8x. The solution is x=

Darina [25.2K]

age Textfrom this Question

Step-by-step explanation:

8 0

3 years ago

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