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Daniel [21]
3 years ago
14

Solving systems of equations using substitution p=q+2 4p+3q= -27

Mathematics
1 answer:
GREYUIT [131]3 years ago
6 0
Ok, so you are given the value of P=q+2

The substitution method tells us that we must insert the value we know, into the second equation, 4P+3q= -27


Doing so will give us 4(q+2)+3q= -27

For right now, lets just focus on the first part, 4(q+2)

We can simplify this by distributing(multiplying) the 4 to whats inside the variables.

This will give us 4q+8

now lets add this back to the rest of the equation >>>  4q+8+3q = -27

We can further simplify by adding like terms >>> 7q+8 = -27

subtract the 8 from both sides >>> 7q = -35

now divide both sides by 7 >>> q = -35/7

Therefor q = -5

EDIT* 

now that we know q = -5 we can put q into the equation for P !

we know that p=q+2

so lets put q in now >>> p=(-5)+2

and simplify>>> p = -3

I hope this helps:)
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Answer with Step-by-step explanation:

We are given that

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r'(t)=

\mid r'(t) \mid=\sqrt{(2t)^2+(2t^2)^2+1}=\sqrt{(2t^2+1)^2}=2t^2+1

T(t)=\frac{r'(t)}{\mid r'(t)\mid}=\frac{}{2t^2+1}

Now, substitute t=1

T(1)=\frac{}{2+1}=\frac{1}{3}

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T'(1)=-\frac{4}{9}+\frac{1}{3}

T'(1)=\frac{1}{9}=

\mid T'(1)\mid=\sqrt{(\frac{-2}{9})^2+(\frac{4}{9})^2+(\frac{-4}{9})^2}=\sqrt{\frac{36}{81}}=\frac{2}{3}

N(1)=\frac{T'(1)}{\mid T'(1)\mid}

N(1)=\frac{}{\frac{2}{3}}=

N(1)=

B(1)=T(1)\times N(1)

B(1)=\begin{vmatrix}i&j&k\\\frac{2}{3}&\frac{2}{3}&\frac{1}{3}\\\frac{-1}{3}&\frac{2}{3}&\frac{-2}{3}\end{vmatrix}

B(1)=i(\frac{-4}{9}-\frac{2}{9})-j(\frac{-4}{9}+\frac{1}{3})+k(\frac{4}{9}+\frac{2}{9})

B(1)=-\frac{2}{3}i+\frac{1}{3}j+\frac{2}{3}k

B(1)=\frac{1}{3}

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Step-by-step explanation:

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2 years ago
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kumpel [21]

Answer:

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Step-by-step explanation:

The equation of a circle is (x-h)^2+(y-k)^2=r^2 where (h,k) is the center of the circle and r is the radius of the circle.

Given that (h,k)\rightarrow(-4,3) and it passes (6,-4), their distance between each other must the radius of the circle, so we can use the distance formula to find the radius:

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Therefore, if the length of the radius is r=\sqrt{149} units, then r^2=149, making the final equation of the circle (x+4)^2+(y-3)^2=149

5 0
2 years ago
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