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4 years ago

5. Solve using any method. y = -x + 8 y=x+4

Mathematics

1 answer:

vlada-n [284]4 years ago

7 0

Answer:

Step-by-step explanation:

$\bold{ELIMINATION\ METHOD}\\\\\left\{\begin{array}{ccc}y=-x+8\\y=x+4&\TEXT{change the signs}\end{array}\right\\\underline{+\left\{\begin{array}{ccc}y=-x+8\\-y=-x-4\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad0=-2x+4\qquad\text{add}\ 2x\ \text{to both sides}\\.\qquad2x=4\qquad\text{divide both sides by 2}\\.\qquad\boxed{x=2}\\\\\text{Put it to the second equation}\\y=2+4\\\boxed{y=6}$

$\bold{SUBSTITUTION\ METHOD}\\\\\left\{\begin{array}{ccc}y=-x+8&(1)\\y=x+4&(2)\end{array}\right\\\\\text{Substitute (1) to (2)}\\\\-x+8=x+4\qquad\text{subtract 8 from both sides}\\-x=x-4\qquad\text{subtract}\ x\ \text{from both sides}\\-x-x=-4\\-2x=-4\qquad\text{divide both sides by (-2)}\\\boxed{x=2}\\\\\text{Put it to the second equation}\\y=2+4\\\boxed{y=6}$

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In the figure to the right, if AC=19 and BC=16, what is the radius? <br><br> *For a circle*

ch4aika [34]

Answer:

The radius of the circle is 10.2 units

Step-by-step explanation:

<u><em>The complete question is</em></u>

In the figure to the right, if AC=19 and BC=16, what is the radius?

A circle has center A. Points B and D are on the circle, with B on the left and D near the bottom. Point C lies outside the circle such that the line segment A C passes through point D and the line segments A B and B C form a right angle.

The radius is approximately (Round to the nearest tenth as needed.)

The picture of the question in the attached figure

we know that

In the right triangle ABC

Applying the Pythagorean Theorem

$AC^2=AB^2+BC^2$

substitute the given values

$19^2=AB^2+16^2$

solve for AB

$AB^2=19^2-16^2$

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Remember that the radius is the same that the segment AB

therefore

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3 years ago

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4 years ago

I need help in partial fraction!! With simple explaination would be nice!

WITCHER [35]

$\dfrac{x^4-7x^2+17x-10}{x(x^2-3)}$

The degree of the numerator (4) is larger than the degree of the denominator (3), so first you need to divide. (Added screenshot of long division procedure.)

$\dfrac{x^4-7x^2+17x-10}{x(x^2-3)}=x-\dfrac{4x^2-17x+10}{x(x^2-3)}$

Now the second term can be decomposed into partial fractions.

$\dfrac{4x^2-17x+10}{x(x^2-3)}=\dfrac{r_1}x+\dfrac{r_2x+r_3}{x^2-3}$
$\dfrac{4x^2-17x+10}{x(x^2-3)}=\dfrac{r_1(x^2-3)+x(r_2x+r_3)}{x(x^2-3)}$
$4x^2-17x+10=r_1(x^2-3)+x(r_2x+r_3)$
$4x^2-17x+10=(r_1+r_2)x^2+r_3x-3r_1$
$\implies\begin{cases}r_1+r_2=4\\r_3=-17\\-3r_1=10\end{cases}\implies r_1=-\dfrac{10}3,r_2=\dfrac{22}3,r_3=-17=-\dfrac{51}3$
$\implies\dfrac{4x^2-17x+10}{x(x^2-3)}=-\dfrac{10}{3x}+\dfrac{22x-51}{x^2-3}$

So

$\dfrac{x^4-7x^2+17x-10}{x(x^2-3)}=x+\dfrac{10}{3x}-\dfrac{22x-51}{x^2-3}$

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