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Allushta [10]
3 years ago
13

PLEASE HELP!!!!how to solve 9x^3-16x^2=0 by factoring WILL GIVE BRAINLIEST IF CORRECT!!!!

Mathematics
1 answer:
andreev551 [17]3 years ago
7 0

Answer:

x=5.3

Step-by-step explanation:

Graph each side of the equation. The solution is the x-value of the point of intersection.

x

≈

5.3

Hope this helps, can you now give me brainliest??

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How do you write a quadratic function in vertex and standard form
Marta_Voda [28]

Answer:

  • vertex form: f(x) = a(x -h)^2 +k
  • standard form: f(x) = ax^2 +bx +c

Step-by-step explanation:

Often, you're given one form and asked to write the equation in the other form. Here, we can show the relationship between the two forms.

In <u>vertex form</u>, the vertex of the function (h, k) is obvious in the way the function expression is written:

  f(x) = a(x -h)^2 +k . . . . . . . for vertex (h, k) and vertical scale factor "a"

If we "simplify" this form, we get ...

  f(x) = a(x^2 -2hx +h^2) +k

  f(x) = ax^2 -2ah + (ah^2 +k) . . . . . "standard form" from vertex form

Comparing this to standard form, we can see the relations between the coefficients are ...

  • a = a
  • b = -2ah
  • c = ah^2 +k

__

In <u>standard form</u>, terms are written in descending order of the exponent of the variable.

  f(x) = ax^2 +bx +c

Generally, coefficients are named in alphabetical order, starting with "a" for the leading coefficient (the coefficient of the highest-degree term).

We can use the relations shown above to find the vertex from from these coefficients.

  b = -2ah

  h = -b/(2a) . . . . . divide by the coefficient of h

And the other coefficient of the vertex is ...

  k = c - ah^2 . . . . subtract ah^2 from the equation for c

  k = c - b^2/(4a)

Then ...

  f(x) = a(x +b/(2a))^2 +(c -b^2/(4a)) . . . . . "vertex form" from standard form

_____

You may notice that the key relationship is that between "b" and "h". It is useful to remember it:

  h = -b/(2a)

3 0
4 years ago
Tina is playing a computer game. She starts with 100 points, and she loses points based on the following rules: • Each time a pl
Natasha_Volkova [10]

Answer: she must complete 4 levels to have a point fewer than 20

Step-by-step explanation:

Given the following :

Starting point = 100 points

Passing a level = - 8 points

Catching a flower = - 3 points

Suppose Tina catches 6 flowers per level

Number of levels she must complete to have fewer Than 20 points

Total number of points lost per level:

Catching 6 flowers = -(6 × 3)

Passing the level = - 8

= - 18 + - 8 = - 26 points

Number of levels she must complete to have < 20

Let number of levels = y

Starting points - (26 × number of levels) < 20

100 - (26y) < 20

100 - 26y < 20

-26y < 20 - 100

-26y < - 80

y > 80/26

y > 3.07

Hence she must complete 4 levels to have a point fewer than 20

8 0
4 years ago
Read 2 more answers
Anyone know how to solve this
lisabon 2012 [21]

Hey!

----------------------------

Solution:

5 1/5% = 0.052

325 x 0.052 = 16.9

----------------------------

Answer:

D. 16.899...

----------------------------

Hope This Helped! Good Luck!

8 0
4 years ago
Read 2 more answers
What is the degree of vertex F
VikaD [51]

Answer: 1

Step-by-step explanation:

7 0
3 years ago
Set up but do not solve for the appropriate particular solution yp for the differential equation y′′+4y=5xcos(2x) using the Meth
taurus [48]

Answer:

So, solution of  the differential equation is

y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos  2x+c_1e^{-2it}+c_2e^{2it}\\

Step-by-step explanation:

We have the given differential equation: y′′+4y=5xcos(2x)

We use the Method of Undetermined Coefficients.

We first solve the homogeneous differential equation y′′+4y=0.

y''+4y=0\\\\r^2+4=0\\\\r=\pm2i\\\\

It is a homogeneous solution:

y_h(t)=c_1e^{-2i t}+c_2e^{2i t}

Now, we finding a particular solution.

y_p(t)=A5x\cos 2x\\\\y'_p(t)=A5\cos 2x-A10x\sin 2x\\\\y''_p(t)=-A20\sin 2x-A20x\cos 2x\\\\\\\implies y''+4y=5x\cos 2x\\\\-A20\sin 2x-A20x\cos 2x+4\cdot A5x\cos 2x=5x\cos 2x\\\\-A20\sin 2x=5x\cos 2x\\\\A=-\frac{x}{4} \cot 2x\\

we get

y_p(t)=A5\cos 2x\\\\y_p(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos  2x\\\\\\y(t)=y_p(t)+y_h(t)\\\\y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos  2x+c_1e^{-2it}+c_2e^{2it}\\

So, solution of  the differential equation is

y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos  2x+c_1e^{-2it}+c_2e^{2it}\\

7 0
3 years ago
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