Solve the following system of equations using the substitution method

Mathematics

1 answer:

antiseptic1488 [7]2 years ago

4 0

Answer:

the answer is x5.5

Step-by-step explanation:

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I need help with #8. I need to use the quadratic formula. I have a calculator that is programmed for the quadratic formula so al

GuDViN [60]

Answer:

x = 14

Step-by-step explanation:

Extend line AB so that it intersects ray CE at point G. Then angles BGC and BAD are "alternate interior angles", hence congruent.

The angle at B is exterior to triangle BCG, and is equal to the sum of the interior angles at C and G:

138 = (376 -23x) +(x^2 -8x)

Subtracting 138 and collecting terms we have ...

x^2 -31x +238 = 0

For your calculator, a=1, b=-31, c=238.

<em>Additional comment</em>

You will find that the solutions to this are x = {14, 17}. You will also find that angle BCE will have corresponding values of 54° and -15°. That is, the solution x=17 is "extraneous." It is a solution to the equation, but not to the problem.

For x=14, the marked angles are A = 84°, C = 54°.

8 0

2 years ago

HELPPPP

krek1111 [17]

Answer:

12.60

Step-by-step explanation:

To find the tip, take the amount of the bill and multiply by 30%

42 * 30%

Change to decimal form

42 * .30

12.6

3 0

2 years ago

Read 2 more answers

Find the following integral

ololo11 [35]

There's nothing preventing us from computing one integral at a time:

$\displaystyle \int_0^{2-x} xyz \,\mathrm dz = \frac12xyz^2\bigg|_{z=0}^{z=2-x} \\\\ = \frac12xy(2-x)^2$

$\displaystyle \int_0^{1-x}\int_0^{2-x}xyz\,\mathrm dz\,\mathrm dy = \frac12\int_0^{1-x}xy(2-x)^2\,\mathrm dy \\\\ = \frac14xy^2(2-x)^2\bigg|_{y=0}^{y=1-x} \\\\= \frac14x(1-x)^2(2-x)^2$

$\displaystyle\int_0^1\int_0^{1-x}\int_0^{2-x}xyz\,\mathrm dz\,\mathrm dy\,\mathrm dx = \frac14\int_0^1x(1-x)^2(2-x)^2\,\mathrm dx$

Expand the integrand completely:

$x(1-x)^2(2-x)^2 = x^5-6x^4+13x^3-12x^2+4x$

Then

$\displaystyle\frac14\int_0^1x(1-x)^2(2-x)^2\,\mathrm dx = \left(\frac16x^6-\frac65x^5+\frac{13}4x^4-4x^3+2x^2\right)\bigg|_{x=0}^{x=1} \\\\ = \boxed{\frac{13}{240}}$