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3 years ago

The solution of a system of two linear equations yields the equation 0=3. Describe what the graph of the system looks like.

Mathematics

1 answer:

Alenkasestr [34]3 years ago

4 0

Well TBH, does 0 really equal 3. no that means that they are parallel, there is no point of intersection.

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Can someone help? TvT please

zysi [14]

Answer:

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3 years ago

How do I solve this<br> 2x2+ 2x2 -2+2x2=

s2008m [1.1K]

Use BODMAS, this is the order in which you must
1. Brackets
2. Divide
3. Multiply
4. Add
5. Subtract
therefore 2x2+2x2-2+2x2=10

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3 years ago

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Find the value of x and y

Oksi-84 [34.3K]

180 - 120 = 60
x = y = 60/2 = 30
answer
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4 years ago

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Find the exact length of the curve. x = 1 + 6t2, y = 4 + 4t3, 0 ≤ t ≤ 3

vovangra [49]

$x=1+6t^2\implies\dfrac{\mathrm dx}{\mathrm dt}=12t$

$y=4+4t^3\implies\dfrac{\mathrm dy}{\mathrm dt}=12t^2$

The length of the curve $C$ is given by

$\displaystyle\int_C\mathrm ds=\int_0^3\sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\dfrac{\mathrm dy}{\mathrm dt}\right)^2}\,\mathrm dt$

$=\displaystyle\int_0^3\sqrt{144t^2+144t^4}\,\mathrm dt$

$=\displaystyle12\int_0^3t\sqrt{1+t^2}\,\mathrm dt$

$=\displaystyle6\int_0^3 2t\sqrt{1+t^2}\,\mathrm dt$

$=\displaystyle6\int_0^3\sqrt{1+t^2}\,\mathrm d(1+t^2)$

$=6\cdot\dfrac23(1+t^2)^{3/2}\bigg|_0^3$

$=4(10^{3/2}-1)$

$=40\sqrt{10}-4$

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3 years ago

Multiply (a-b)(a^2+ab+b^2)

AlexFokin [52]

Answer:

=(a-b)(a^2+ab+b^2)

= a^3+a^2b+ab+ab^2-ba^2-ab^2-b^3

SIMPLIFY

=a^3+ab-b^3

3 0

3 years ago