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3 years ago

Please help meeeeeee pt5

Mathematics

2 answers:

FinnZ [79.3K]3 years ago

8 0

Answer:

roots and zeroes.

Explanation:

Zeroes are the solution of any polynomial equation, but when it comes to quadratic equation, we use the term, roots.

So therefore, we use the term zero, for it is a polynomial, and root, for it's a quadratic equation.

Hope you understand

Thank You

Please mark as brainliest

Marta_Voda [28]3 years ago

7 0

Answer:

see explanation

Step-by-step explanation:

The solutions of a quadratic equation are referred to as

x- intercepts, roots , zeros

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Solve this simultaneous equation by using substitution method 3x-2y=12 and x+3y=-7

ratelena [41]

Answer:

Solving this simultaneous equation by using substitution method 3x-2y=12 and x+3y=-7 we get x=2 and y=-3

Solution set = {2,-3}

Step-by-step explanation:

We need to solve this simultaneous equation by using substitution method 3x-2y=12 and x+3y=-7

let:

$3x-2y=12--eq(1)\\x+3y=-7 --eq(2)$

Finding value of x from equation 2 and putting in eq(1)

$From \ eq(2)\\x+3y=-7\\x=-7-3y$

Putting value of x in eq(1)

$3x-2y=12\\Put \ x=-7-3y\\3(-7-3y)-2y=12\\Solving\\-21-9y-2y=12\\-11y=12+21\\-11y=33\\y=\frac{33}{-11}\\y=-3$

So, we get y=-3

Now put value of y in equation 2 to find value of x

$x+3y=-7\\x+3(-3)=-7\\x-9=-7\\x=-7+9\\x=2$

So, we get value of x =2

So, solving this simultaneous equation by using substitution method 3x-2y=12 and x+3y=-7 we get x=2 and y=-3

Solution set = {2,-3}

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2 years ago

Given a cylinder with a volume of 350 cubic inches and height of 7 inches. Find B, the area of the base. Use 3.14 for pie

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UzusuxhzjaisuzAnswer:

Step-by-step explanation:

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3 years ago

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For the function y=3x2: (a) Find the average rate of change of y with respect to x over the interval [3,6]. (b) Find the instant

nirvana33 [79]

Answer:

The instantaneous rate of change of $y$ with respect to $x$ at the value $x = 3$ is 18.

Step-by-step explanation:

a) Geometrically speaking, the average rate of change of $y$ with respect to $x$ over the interval by definition of secant line:

$r = \frac{y(b) -y(a)}{b-a}$ (1)

Where:

$a$ , $b$ - Lower and upper bounds of the interval.

$y(a)$ , $y(b)$ - Function exaluated at lower and upper bounds of the interval.

If we know that $y = 3\cdot x^{2}$ , $a = 3$ and $b = 6$ , then the average rate of change of $y$ with respect to $x$ over the interval is:

$r = \frac{3\cdot (6)^{2}-3\cdot (3)^{2}}{6-3}$

$r = 27$

The average rate of change of $y$ with respect to $x$ over the interval $[3,6]$ is 27.

b) The instantaneous rate of change can be determined by the following definition:

$y' = \lim_{h \to 0}\frac{y(x+h)-y(x)}{h}$ (2)

Where:

$h$ - Change rate.

$y(x)$ , $y(x+h)$ - Function evaluated at $x$ and $x+h$ .

If we know that $x = 3$ and $y = 3\cdot x^{2}$ , then the instantaneous rate of change of $y$ with respect to $x$ is:

$y' = \lim_{h \to 0} \frac{3\cdot (x+h)^{2}-3\cdot x^{2}}{h}$

$y' = 3\cdot \lim_{h \to 0} \frac{(x+h)^{2}-x^{2}}{h}$

$y' = 3\cdot \lim_{h \to 0} \frac{2\cdot h\cdot x +h^{2}}{h}$

$y' = 6\cdot \lim_{h \to 0} x +3\cdot \lim_{h \to 0} h$

$y' = 6\cdot x$

$y' = 6\cdot (3)$

$y' = 18$

The instantaneous rate of change of $y$ with respect to $x$ at the value $x = 3$ is 18.

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