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lawyer [7]
3 years ago
9

How do you solve system of equations word problems?

Mathematics
2 answers:
PtichkaEL [24]3 years ago
7 0
Highlight important information
Define your Varibales
Write two equations
Use one of the methods for solving System of equations to solve
Check your answer by subsisting your order pair into the original Equation
kaheart [24]3 years ago
5 0
Well first you have to figure out what you need to solve. then you write out the two equations and solve the system s from there
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Find the measure of ∠ADB<br><br> ∠<br><br> A<br><br> D<br><br> B<br><br> if ∠ADC = 82°
sveta [45]
Assuming CDB is a straight line then angle ADB = 180-82=98
8 0
3 years ago
Solve dis attachment and show all work ( I got it all wrong and I want to know how to solve it )
DedPeter [7]
(a) First find the intersections of y=e^{2x-x^2} and y=2:

2=e^{2x-x^2}\implies \ln2=2x-x^2\implies x=1\pm\sqrt{1-\ln2}

So the area of R is given by

\displaystyle\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\left(e^{2x-x^2}-2\right)\,\mathrm dx

If you're not familiar with the error function \mathrm{erf}(x), then you will not be able to find an exact answer. Fortunately, I see this is a question on a calculator based exam, so you can use whatever built-in function you have on your calculator to evaluate the integral. You should get something around 0.5141.

(b) Find the intersections of the line y=1 with y=e^{2x-x^2}.

1=e^{2x-x^2}\implies 0=2x-x^2\implies x=0,x=2

So the area of S is given by

\displaystyle\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}(2-1)\,\mathrm dx+\int_{1+\sqrt{1-\ln2}}^2\left(e^{2x-x^2}-1\right)\,\mathrm dx
\displaystyle=2\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\mathrm dx

which is approximately 1.546.

(c) The easiest method for finding the volume of the solid of revolution is via the disk method. Each cross-section of the solid is a circle with radius perpendicular to the x-axis, determined by the vertical distance from the curve y=e^{2x-x^2} and the line y=1, or e^{2x-x^2}-1. The area of any such circle is \pi times the square of its radius. Since the curve intersects the axis of revolution at x=0 and x=2, the volume would be given by

\displaystyle\pi\int_0^2\left(e^{2x-x^2}-1\right)^2\,\mathrm dx
5 0
3 years ago
What is the solution to 89+ K= -42<br> A. K=131<br> B. K=47<br> C. K=-47<br> D. K=-131
lisabon 2012 [21]
89-131=-42 the answer is k=-131
5 0
3 years ago
PLS HELP ME ASAP FOR ALL 6!! (SHOW WORK!!) + LOTS AND LOTS OF POINTS!!! *GIVING THANKS!!*
aleksandr82 [10.1K]
This is the answer for #1

4 0
3 years ago
The point-slope form of the equation of the line that passes through (-9, -2) and (1, 3) is y-3 = {(x - 1). What is the slope-in
GrogVix [38]

Answer:

y=1/2x - 1/2

Step-by-step explanation:

Slope intercept form is y=mx+b

The m is the slope of the line and the b is the y-intercept.

You must first the slope. The formula of a slope is (y1-y2)/(x1-x2). In this example it is (-2-3)/(-9-1) which equals 1/2. By order of elimanation we can see that y=1/2x-1/2 is correct.

7 0
3 years ago
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