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3 years ago

How can one variable equations have more than one solution in some cases and no solutions in others?

2 answers:

7 0

Answer:A system of linear equations usually has a single solution, but sometimes it can have no solution (parallel lines) or infinite solutions (same line).

Step-by-step explanation:

andreev551 [17]3 years ago

5 0

Answer:

There can be more than one solution to a system of equations. A system of linear equations will have one point of intersection, or one solution. To graph a system of equations that are written in standard form, you must rewrite the equations in slope -intercept form.

Step-by-step explanation:

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Find the first three terms in the expansion , in ascending power of x , of (2+x)^6 and obtain the coefficient of x^2 in the expa

Nataly_w [17]

Answer:

The first 3 terms in the expansion of $(2 + x)^{6}$ , in ascending power of x are,

$64 , 192 \times x^{1} {\textrm{ and }}240 \times x^{2}$

coefficient of $x^{2}$ in the expansion of $(2+x - x^{2})^{6}$ = (240 - 192) = 48

Step-by-step explanation:

$(2+x)^{6}$

= $\sum_{k=0}^{6}(6_{C_{k}} \times x^{k} \times 2^{6 - k})$

= $6_{C_{0}} \times x^{0} \times 2^{6} + 6_{C_{1}} \times x^{1} \times 2^{5} + 6_{C_{2}} \times x^{2} \times 2^{4}$ + terms involving higher powers of x

= $64 + 192 \times x^{1} + 240 \times x^{2}$ + terms involving higher powers of x

so, the first 3 terms in the expansion of $(2 + x)^{6}$ , in ascending power of x are,

$64 , 192 \times x^{1} {\textrm{ and }}240 \times x^{2}$

Again,

$(2+x - x^{2})^{6}$

= $\sum_{k=0}^{6}(6_{C_{k}} \times (2 + x)^{k} \times (-x^{2})^{6 - k})$

Now, by inspection,

the term $x^{2}$ comes from k =5 and k = 6

for k = 5, the coefficient of $x^{2}$ is , $(-32) \times 6$ = -192

for k = 6 , the coefficient of $x^{2}$ is, $6_{C_{2}} \times 2^{4}$ = 240

so, coefficient of $x^{2}$ in the final expression = (240 - 192) = 48

3 0

3 years ago

Jasper and Corey have saved up a total of $78.00. Jasper has saved 6 dollars more than twice as much as Corey. How much has Core

Novay_Z [31]

Answer:

24 is the answer..for this concept

4 0

3 years ago

Read 2 more answers

Please help will give brainlist

RoseWind [281]

6(x²-4x+4-4)+1=0, 6(x-2)²-24+1=0, 6(x-2)²=23, x-2=±√(23/6), x=2±√(23/6)=2±1.95789, so x=3.95789 or 0.04211 approx. These are the zeroes.

6 0

3 years ago

Read 2 more answers

Make a table of values for the following equation:

loris [4]

*Not 100% sure what you want but here goes*

Step-by-step explanation:

So what you do is create a table for this example label the x values say -2 through 2. Then you plug in the x value to the equation so for 2 its <u>y=</u><u> </u><u>2</u><u>(</u><u>2</u><u>)</u><u>+</u><u>5</u> which is 9. Continue this for the rest of the graph and that should do it.

Hope this helps

3 0

2 years ago

Read 2 more answers

Which expression is equivalent to (16 x Superscript 8 Baseline y Superscript negative 12 Baseline) Superscript one-half?.

loris [4]

To solve the problem we must know the Basic Rules of Exponentiation.

<h2>Basic Rules of Exponentiation</h2>