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Mashutka [201]
3 years ago
7

Write a system of linear inequalities so the points (1, 2) and (4, −3) are solutions of the system, but the point (−2, 8) is not

a solution of the system. PLEASE SOMEONE HELP AND SOLVE IT AS A SYSTEM OF LINEAR EQUATIONS
Mathematics
2 answers:
dalvyx [7]3 years ago
7 0
I can’t see it where is it it’s blurry
Travka [436]3 years ago
4 0

Answer:

<u>The sample system is given below:</u>

  • y < 5
  • y > x - 8

<em>see the attached for the graph</em>

The points (1, 2) and (4, -3) are in the intersected region but the point (-2, 8) is outside

  • <em>The easy way to get the inequalities to plot the points and add lines to divide the regions where desired point are inside and not required point is outside. </em>
  • <em>I put y = 5 and y = x - 8 lines and shaded the region below y = 5 line and above the other line.</em>

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4 0
4 years ago
Find the rational solution of x^3-3x^2+x+5=0. Show your use of the rational zero theorem to receive credit.
nasty-shy [4]
The rational root (or zero) theorem says a polynomial's rational zeros must have factors of the constant in the numerator and those of the leading coefficient in the denominator.

Here the leading coefficient is 1 and the constant is 5, so our only possibilities for rational roots are 1, -1, 5, -5

We try them each in term

1-3+1+5=6 nope

-1 -3 -1 +5 = 0 yes, x = -1 is a rational zero

5 and -5 don't work either so we're left with

Final answer: x = -1


5 0
3 years ago
Can someone explain me this please
AfilCa [17]

\purple{\maltese}\large\underline{\underline{\red{\sf\:\: Given :}}} \\

  • Mass (M) of space shuttle = 1.0 × 105 kg.
  • Altitude (r) = 200.0 km

\\\purple{\maltese}\large\underline{\underline{\red{\sf\:\: To \: \:   Find  :}}} \\

  • The force of gravity that the space shuttle experiences = ?

\\\purple{\maltese}\large\underline{\underline{\red{\sf\:\: Solution :}}} \\ \\

\purple{\maltese}\large\underline{\underline{\sf\:\: using \: formula :}} \\ \\

\bigstar \: \underline{ \boxed{\sf { \pink{g = \frac{G \: M}{{r }^{2} }}}}} \\  \\

\\ ❒\:  \: \underline{\textbf {Putting values \: in \: the \: \: given \: formula :}} \\

\sf \implies \: g = \frac{G \: M}{{r }^{2} }  \\

\sf \implies \: g = \frac{6.67×10^{-11} × 1.0×10^5}{(2×10^5)^{2} }  \\

\sf \implies \: g = \frac{6.67×10^{-6}}{4×10^{10}}\\

\sf \implies \: g = \frac{6.67×10^{-6}}{4×10^{10}} \\

\sf \implies \: g =1.6675×10^{-6-10}\\

\sf \implies \: g =1.6675×10^{-16}\: Newton\\\\

\:\:\:\:\:\:\:\qquad\rule{150pt}{2pt} \\\\

Henceforth, the force of gravity experienced by the shuttle is <u>1.6675×10^N</u>

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