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3 years ago

Solve the quadratic by factoring. x^2-5x-31=5

Mathematics

1 answer:

torisob [31]3 years ago

5 0

Answer:

$x = - 4\: or \: x = 9$

Step-by-step explanation:

make one side equal to zero:

${x}^{2} - 5x - 36 = 0$

find two factors that multiply to -36 and add to -5 (-9 & 4)

${x }^{2} - 9x + 4x - 36 = 0$

factorise

$x(x - 9) + 4(x - 9) = 0$

$(x + 4)(x - 9) = 0$

solve for x

$x = - 4 \: or \: x = 9$

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eimsori [14]

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3 years ago

Find slope A.1/2 B.-1/2 C.-2 D.2

fomenos

Answer:

Option B, -1/2

Step-by-step explanation:

The slope is change in y over change in x.

Since this line is facing downward, we already know that the slope is negative.

To find the ratio, find the change in y and divide the change in x.

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3 years ago

(15 pts) 4. Find the solution of the following initial value problem: y"-10y'+25y = 0 with y(0) = 3 and y'(0) = 13

jolli1 [7]

Answer:

$y(x)=3e^{5x}-2xe^{5x}$

Step-by-step explanation:

The given differential equation is $y''-10y'+25y=0$

The characteristics equation is given by

$r^2-10r+25=0$

Finding the values of r

$r^2-5r-5r+25=0\\\\r(r-5)-5(r-5)=0\\\\(r-5)(r-5)=0\\\\r_{1,2}=5$

We got a repeated roots. Hence, the solution of the differential equation is given by

$y(x)=c_1e^{5x}+c_2xe^{5x}...(i)$

On differentiating, we get

$y'(x)=5c_1e^{5x}+5c_2xe^{5x}+c_2e^{5x}...(ii)$

Apply the initial condition y (0)= 3 in equation (i)

$3=c_1e^{0}+0\\\\c_1=3$

Now, apply the initial condition y' (0)= 13 in equation (ii)

$13=5(3)e^{0}+0+c_2e^{0}\\\\13=15+c_2\\\\c_2=-2$

Therefore, the solution of the differential equation is

$y(x)=3e^{5x}-2xe^{5x}$

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3 years ago

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3 years ago

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frosja888 [35]

Answer:

Step-by-step explanation:

AB is parallel to CD. So, <ABC = <BCD.

Now, <BCD = <ABC = 180 - 111 - 31 = 38

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3 years ago