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3 years ago

Hi someone can help me with, 5 problems of polinomial functions with degree three or more, that apply to real life

Mathematics

1 answer:

dusya [7]3 years ago

4 0

Answer:8

Step-by-step explanation:

you just add the numbers up and then that si how u get your answer for your problem u are trying to work out.

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4. Gloria the grasshopper is working on her hops.

aivan3 [116]

The path that Gloria follows when she jumped is a path of parabola.

The equation of the parabola that describes the path of her jump is $\mathbf{y = -\frac{5}{49}(x - 14)^2 + 20}$

The given parameters are:

$\mathbf{Height = 20}$

$\mathbf{Length = 28}$

<em>Assume she starts from the origin (0,0)</em>

The midpoint would be:

$\mathbf{Mid = \frac 12 \times Length}$

$\mathbf{Mid = \frac 12 \times 28}$

$\mathbf{Mid = 14}$

So, the vertex of the parabola is:

$\mathbf{Vertex = (Mid,Height)}$

Express properly as:

$\mathbf{(h,k) = (14,20)}$

A point on the graph would be:

$\mathbf{(x,y) = (28,0)}$

The equation of a parabola is calculated using:

$\mathbf{y = a(x - h)^2 + k}$

Substitute $\mathbf{(h,k) = (14,20)}$ in $\mathbf{y = a(x - h)^2 + k}$

$\mathbf{y = a(x - 14)^2 + 20}$

Substitute $\mathbf{(x,y) = (28,0)}$ in $\mathbf{y = a(x - 14)^2 + 20}$

$\mathbf{0 = a(28 - 14)^2 + 20}$

$\mathbf{0 = a(14)^2 + 20}$

Collect like terms

$\mathbf{a(14)^2 =- 20}$

Solve for a

$\mathbf{a =- \frac{20}{14^2}}$

$\mathbf{a =- \frac{20}{196}}$

Simplify

$\mathbf{a =- \frac{5}{49}}$

Substitute $\mathbf{a =- \frac{5}{49}}$ in $\mathbf{y = a(x - 14)^2 + 20}$

$\mathbf{y = -\frac{5}{49}(x - 14)^2 + 20}$

Hence, the equation of the parabola that describes the path of her jump is $\mathbf{y = -\frac{5}{49}(x - 14)^2 + 20}$

See attachment for the graph

Read more about equations of parabola at:

brainly.com/question/4074088

7 0

3 years ago

Two terms of a geometric sequence are given. Find the first five terms. Please help asap

Zepler [3.9K]

Answer:

4, 8, 16, 32, 64

Step-by-step explanation:

The nth term of a geometric sequence is

$a_{n}$ = a₁ $(r)^{n-1}$

Given

a₇ = 256 and a₁₀ = 2048 , then

a₁ $r^{6}$ = 256 → (1)

a₁ $r^{9}$ = 2048 → (2)

Divide (2) by (1)

$\frac{a_{1}r^{9} }{a_{1}r^{6} }$ = $\frac{2048}{256}$

r³ = 8 ( take the cube root of both sides )

r = $\sqrt[3]{8}$ = 2

Substitute r = 2 into (1)

a₁ × $2^{6}$ = 256

a₁ × 64 = 256 ( divide both sides by 64 )

a₁ = 4

Then

a₁ = 4

a₂ = 2a₁ = 2 × 4 = 8

a₃ = 2a₂ = 2 × 8 = 16

a₄ = 2a₃ = 2 × 16 = 32

a₅ = 2a₄ = 2 × 32 = 64

7 0

3 years ago

given the points a (-1,2) and (7,14), find the coordinates of the point p on directed line segment ab that partitions ab in the

Vinvika [58]

Check the picture below.

$\bf \left. \qquad \right.\textit{internal division of a line segment}
\\\\\\
a(-1,2)\qquad b(7,14)\qquad
\qquad 1:3
\\\\\\
\cfrac{aP}{Pb} = \cfrac{1}{3}\implies \cfrac{a}{b} = \cfrac{1}{3}\implies 3a=1b\implies 3(-1,2)=1(7,14)
\\\\
-------------------------------\\\\
{ P=\left(\cfrac{\textit{sum of "x" values}}{r1+r2}\quad ,\quad \cfrac{\textit{sum of "y" values}}{r1+r2}\right)}$

$\bf -------------------------------\\\\
P=\left(\cfrac{(3\cdot -1)+(1\cdot 7)}{1+3}\quad ,\quad \cfrac{(3\cdot 2)+(1\cdot 14)}{1+3}\right)
\\\\\\
P=\left( \cfrac{-3+7}{4}~,~\cfrac{6+14}{4} \right)$

3 0

3 years ago

Pleasssssssss help 5<br> points

love history [14]

What is the situation from the problem

7 0

3 years ago

The mathematical language we use when introducing base-ten words is important to the development of the ideas. Identify the stat

barxatty [35]

The answer is D

brainliest?

4 0

3 years ago