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Burka [1]
3 years ago
14

Solve for x in the equation x^2-4x-9=29

Mathematics
1 answer:
Kipish [7]3 years ago
5 0
X = 2 + √42,2- √42
decimal form: 8.480741, -4.480741
sorry if my answer isn't right
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Factor 6p^2 +15p+9????
Ivanshal [37]
To factor the trinomial, you can use the ac method where you multiply the first and last terms together (6 and 9) and add up to 15.
In this case ac= 54
2 numbers that multiply to 54 and add up to 15 would be 6 and 9 so place these numbers where the 15 would be:
6p^{2} +9p + 6p+9
Then group terms to simplify:
(6p^{2} +9p)(6p+9)

Simplify any like terms:
3p(2p+3) 3(2p+3)
And group together again:
(3p+3)(2p+3)
You can simplify a 3 out of the first expression one more time to get:
3(p+1)(2p+3)
Which would be the factored form. I hope this helps :)
8 0
3 years ago
Read 2 more answers
write and equation in slope intercept form for the following scenario. andy’s rental car charges an initial fee of $20 plus an a
Elza [17]

Answer:

y = 30x + 20

Step-by-step explanation:

y = mx + b

find b

b=20

find m

m=30

8 0
3 years ago
Solve 5y'' + 3y' – 2y = 0, y(0) = 0, y'(0) = 2.8 y(t) = 0 Preview
mario62 [17]

Answer:  The required solution is

y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.

Step-by-step explanation:   We are given to solve the following differential equation :

5y^{\prime\prime}+3y^\prime-2y=0,~~~~~~~y(0)=0,~~y^\prime(0)=2.8~~~~~~~~~~~~~~~~~~~~~~~~(i)

Let us consider that

y=e^{mt} be an auxiliary solution of equation (i).

Then, we have

y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.

Substituting these values in equation (i), we get

5m^2e^{mt}+3me^{mt}-2e^{mt}=0\\\\\Rightarrow (5m^2+3y-2)e^{mt}=0\\\\\Rightarrow 5m^2+3m-2=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow 5m^2+5m-2m-2=0\\\\\Rightarrow 5m(m+1)-2(m+1)=0\\\\\Rightarrow (m+1)(5m-1)=0\\\\\Rightarrow m+1=0,~~~~~5m-1=0\\\\\Rightarrow m=-1,~\dfrac{1}{5}.

So, the general solution of the given equation is

y(t)=Ae^{-t}+Be^{\frac{1}{5}t}.

Differentiating with respect to t, we get

y^\prime(t)=-Ae^{-t}+\dfrac{B}{5}e^{\frac{1}{5}t}.

According to the given conditions, we have

y(0)=0\\\\\Rightarrow A+B=0\\\\\Rightarrow B=-A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)

and

y^\prime(0)=2.8\\\\\Rightarrow -A+\dfrac{B}{5}=2.8\\\\\Rightarrow -5A+B=14\\\\\Rightarrow -5A-A=14~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{Uisng equation (ii)}]\\\\\Rightarrow -6A=14\\\\\Rightarrow A=-\dfrac{14}{6}\\\\\Rightarrow A=-\dfrac{7}{3}.

From equation (ii), we get

B=\dfrac{7}{3}.

Thus, the required solution is

y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.

7 0
3 years ago
Simplify 3/5(10g-5k)-(-3g+2k)
dybincka [34]

Answer:

9g -5k

Step-by-step explanation:

6 0
4 years ago
The volume of a can of Pepsi is 0.33liter. There are 12 cans in 1 carton
marusya05 [52]

Answer:

19 liters 800 milliliters

Step-by-step explanation:

0.33 * 12 * 5 = 19.8 liters

6 0
2 years ago
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