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bagirrra123 [75]
3 years ago
7

Axis of sym: x =

Mathematics
1 answer:
marta [7]3 years ago
4 0

Answer:

<h2>SEE BELOW</h2>

Step-by-step explanation:

<h3>to understand this</h3><h3>you need to know about:</h3>
  • quadratic function
  • PEMDAS
<h3>let's solve:</h3>

vertex:(h,k)

therefore

vertex:(-1,4)

axis of symmetry:x=h

therefore

axis of symmetry:x=-1

  • to find the quadratic equation we need to figure out the vertex form of quadratic equation and then simply it to standard form i.e ax²+bx+c=0

vertex form of quadratic equation:

  • y=a(x-h)²+k

therefore

  • y=a(x-(-1))²+4
  • y=a(x+1)²+4

it's to notice that we don't know what a is

therefore we have to figure it out

the graph crosses y-asix at (0,3) coordinates

so,

3=a(0+1)²+4

simplify parentheses:

3 = a(1 {)}^{2}  + 4

simplify exponent:

3 =  a + 4

therefore

a =  - 1

our vertex form of quadratic equation is

  • y=-(x+1)²+4

let's simplify it to standard form

simplify square:

y =  - ( {x}^{2}  + 2x + 1)  + 4

simplify parentheses:

y =  -  {x}^{2}  - 2x - 1 + 4

simplify addition:

y =  -  {x}^{2}  - 2x + 3

therefore our answer is D)y=-x²-2x+3

the domain of the function

x\in \mathbb{R}

and the range of the function is

y\leqslant 4

zeroes of the function:

-  {x}^{2}  - 2x + 3 = 0

\sf divide \: both \: sides \: by \:  - 1

{x}^{2}  + 2x - 3 = 0

\implies \:  {x}^{2} +   3x  - x  +  3 = 0

factor out x and -1 respectively:

\sf \implies \: x(x + 3)   - 1(x  + 3 )= 0

group:

\implies \: (x - 1)(x + 3) = 0

therefore

\begin{cases} x_{1} = 1 \\  x_{2}  =  - 3\end{cases}

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6x7.5=45

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An observation deck extends 120 feet out above a valley. The deck sits 70 feet above the valley floor. If an object is dropped f
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Answer:

2

Step-by-step explanation:

The height formula given is:

h = -16t^2 + 70

That means the object will be initially (t=0) at the height 70 feet, from where it will be dropped.

If we want to know the time when the object will be at height 6 feet, we just need to use h=6 in the equation, and then calculate the value of t:

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Helppp me plsssssssss<br><br>​
Oliga [24]

Answer:

The class 35 - 40 has maximum frequency. So, it is the modal class.

From the given data,

  • \sf \:\:\:\:\:\:\:\:\:\:x_{k}=35
  • \sf \:\:\:\:\:\:\:\:\:\:f_{k}=50
  • \sf \:\:\:\:\:\:\:\:\:\:f_{k-1}=34
  • \sf \:\:\:\:\:\:\:\:\:\:f_{k+1}=42
  • \sf \:\:\:\:\:\:\:\:\:\:h=5

{\bf \:\: {By\:using\:the\: formula}} \\ \\

\:\dag\:{\small{\underline{\boxed{\sf {Mode,\:M_{o} =\sf\red{x_k + {\bigg(h \times \: \dfrac{ ( f_k - f_{k-1})}{ (2f_k - f_{k - 1} - f_{k +1})}\bigg)}}}}}}} \\ \\

\sf \:\:\:\:\:\:\:\:\:= 35+ {\bigg(5 \times \dfrac{(50 - 34)}{ ( 2 \times 50 - 34 - 42)}\bigg)} \\ \\

\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= 35 +{\bigg(5 \times \dfrac{16}{24}\bigg)} \\ \\

\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= {\bigg(35+\dfrac{10}{3}\bigg)} \\ \\

\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:(35 + 3.33) =.38.33 \\ \\

\:\:\sf {Hence,}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\ \large{\underline{\mathcal{\gray{ mode\:=\:38.33}}}} \\ \\

{\large{\frak{\pmb{\underline{Additional\: information }}}}}

MODE

  • Most precisely, mode is that value of the variable at which the concentration of the data is maximum.

MODAL CLASS

  • In a frequency distribution the class having maximum frequency is called the modal class.

{\bf{\underline{Formula\:for\: calculating\:mode:}}} \\

{\underline{\boxed{\sf {Mode,\:M_{o} =\sf\red{x_k + {\bigg(h \times \: \dfrac{ ( f_k - f_{k-1})}{ (2f_k - f_{k - 1} - f_{k +1})}\bigg)}}}}}} \\ \\

Where,

\sf \small\pink{ \bigstar} \: x_{k}= lower\:limit\:of\:the\:modal\:class\:interval.

\small \blue{ \bigstar}\sf \: f_{k}=frequency\:of\:the\:modal\:class

\sf \small\orange{ \bigstar}\: f_{k-1}=frequency\:of\:the\:class\: preceding\:the\;modal\:class

\sf \small\green{ \bigstar}\: f_{k+1}=frequency\:of\:the\:class\: succeeding\:the\;modal\:class

\small \purple{ \bigstar}\sf \: h= width \:of\:the\:class\:interval

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Answer:

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Step-by-step explanation:

The volume of a sphere is given by the formula ...

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  V = (π/6)d^3

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