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Lelechka [254]
3 years ago
5

Find a49a49 of the sequence 70,63,56,49,…70,63,56,49,…. A. -243 B. -273 C. -63 D. -266

Mathematics
1 answer:
ladessa [460]3 years ago
6 0
Well, from the sequence, notice <span>70,63,56,49,…

is simply dropping on the next term by 7 units, so,
70 - 7, 63
63 - 7, 56
and so on

thus, the "common difference", or the number you "add" to get the next term is -7..... as from the sequence itself, you can see the first term's value is 70.

</span>\bf n^{th}\textit{ term of an arithmetic sequence}\\\\&#10;a_n=a_1+(n-1)d\qquad &#10;\begin{cases}&#10;n=n^{th}\ term\\&#10;a_1=\textit{first term's value}\\&#10;d=\textit{common difference}\\&#10;----------\\&#10;a_1=70\\&#10;d=-7\\&#10;n=49&#10;\end{cases}&#10;\\\\\\&#10;a_{49}=70+(49-1)(-7)\implies a_{49}=70-336\implies a_{49}=-266<span>
</span>
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A quadratic function is a function of the form y=ax^2+bx+c where a, b, and c are constants. Given any 3 points in the plane, the
pochemuha

Answer:

The quadratic function whose graph contains these points is y=-x^{2}-2x-2

Step-by-step explanation:

We know that a quadratic function is a function of the form y=ax^{2}+bx+c. The first step is use the 3 points given to write 3 equations to find the values of the constants <em>a</em>,<em>b</em>, and <em>c</em>.  

Substitute the points (0,-2), (-5,-17), and (3,-17) into the general form of a quadratic function.

-2=a*0^{2}+b*0+c\\c=-2

-17=a*-5^{2}+b*-5+c\\c=-25a+5b-17

-17=a*3^{2}+b*3+c\\ c=-9a-3b-17

We can solve these system of equations by substitution

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-9a-3b-17=25a+5b-17\\-9a-3b-17=-2

  • Isolate a for the first equation

-9a-3b-17=-25a+5b-17\\a=\frac{b}{2}

  • Substitute a=\frac{b}{2} into the second equation

-9\left(-\frac{b}{2}\right)-3b-17=-2

  • Find the value of b

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  • Find the value of a

a=\frac{b}{2}\\  a=-1

The solutions to the system of equations are:

b=-2,a=-1,c=-2

So the quadratic function whose graph contains these points is

y=-x^{2}-2x-2

As you can corroborate with the graph of this function.

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