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Leni [432]
2 years ago
14

The equation of line A is y = 7x + 12. Line B is perpendicular to line A and passes through the point . What would be the soluti

on to the system of equations represented by line A and line B?
Mathematics
1 answer:
max2010maxim [7]2 years ago
3 0
Since B is perpendicular to A. We can say that the gradient of B will be -1/7 (product of the gradients of 2 perpendicular lines has to be -1).
Now we know that the equation for B is y=-(1/7)x + c with c being the y intercept.
Since the point isnt specified in the question, we could leave the equation like this.
But if there is a given point that B passes through, just plug in the x and y values into their respective places and solve to find c. That should give you the equation for b.
Now, to find the solution of x, we have 2 equations:
1) y=7x+12
2)y=-(1/7)x+c

In this simultaneous equation we see that y is equal to both the expressions. So,
7x+12=-(1/7)x+c

Now, since the value of c is not found, we cannot actually find the value of x, but if we would find c, we could also find x since it would only be a matter of rearranging the equation.
And there you go, that is your solution :)
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Answer: The area of a sector is a fraction of the area of the circle. The fraction is equal to the ratio of the measure of the sector’s central angle to one full rotation, or 360°.

Step-by-step explanation:

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2 years ago
PLS SOMEONE HELP ME ASAP
babymother [125]

Answer:

A. (2i)(8) = d. 16i

B. 16i³ = b. -16i

C. (2i)⁴ = a. 16

D. (2i)(8i) = c. -16

Step-by-step explanation:

A. Multiply 2i by 8 to get 16i, which corresponds to d.

B. The exponent is 3 more than a mulitple of 4 in 16i³, so the answer is negative. -16i corresponds to b.

C. (2i)⁴ has an exponent that is a multiple of 4, so the i isn't needed. 16 corresponds to a.

D. (2i)(8i) simplifies to 2(8) * i². The exponent is 2 more than a multiple of 4, so the answer is negative, without an i. -16 corresponds to c.

8 0
2 years ago
The eccentricity e of an ellipse is defined as the number c/a, where a is the distance of a vertex from the center and c is the
Anna71 [15]

Answer:

Check below, please.

Step-by-step explanation:

Hi, there!

Since we can describe eccentricity as e=\frac{c}{a}

a) Eccentricity close to 0

An ellipsis with eccentricity whose value is 0, is in fact, a degenerate one almost a circle. An ellipse whose value is close to zero is almost a degenerate circle. The closer the eccentricity comes to zero, the more rounded gets the ellipse just like a circle. (Check picture, please)

\frac{x^2}{a^2} +\frac{y^2}{b^2} =1 \:(Ellipse \:formula)\\a^2=b^2+c^2 \: (Pythagorean\: Theorem)\:a=longer \:axis.\:b=shorter \:axis)\\a^2=b^2+(0)^2 \:(c\:is \:the\: distance \: the\: Foci)\\\\a^2=b^2 \\a=b\: (the \:halves \:of \:each\:axes \:measure \:the \:same)

b) Eccentricity =5

5=\frac{c}{a} \:c=5a

An eccentricity equal to 5 implies that the distance between the Foci has to be five (5) times larger than the half of its longer axis! In this case, there can't be an ellipse since the eccentricity must be between 0 and 1 in other words:

If\:e=\frac{c}{a} \:then\:c>0 , and\: c>0 \:then \:1>e>0

c) Eccentricity close to 1

In this case, the eccentricity close or equal to 1 We must conceive an ellipse whose measure for the half of the longer axis a and the distance between the Foci 'c' they both have the same size.

a=c\\\\a^2=b^2+c^2\:(In \:the\:Pythagorean\:Theorem\: we \:should\:conceive \:b=0)

Then:\\\\a=c\\e=\frac{c}{a}\therefore e=1

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Murljashka [212]
<span>I got 0.10,Hopefully that helps!</span>
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127.5 is what percent of 51?
poizon [28]
Multiply 1.275 with 51 to get the answer (65.025)
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