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3 years ago

What are the zeros of the function f(x) = x2 – 10x + 21? help bro

Mathematics

1 answer:

Nady [450]3 years ago

5 0

Answer:

3, 7

Step-by-step explanation:

"Zeros" means solutions to the equation f(x) = 0. In other words, what are the values of x that make the function's value equal to zero?

This one can be solved by factoring.

$x^2-10x+21=0\\(x-7)(x-3)=0\\x-7=0 \text{ or } x-3=0\\x=7 \text{ or } x=3$

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Millie has a bank account balance of $1200. Each week, her balance changes by

Hatshy [7]

Answer:

at least 8

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Use the graph of the function to find f(-2)

lesya692 [45]

Answer:

Answer D: f(-2) = 2

Step-by-step explanation:

To evaluate f(-2) we need to find what is the y-value that the graph shows for the x-value "-2".

To find such, we look for the value x = -2 on the horizontal (x) axis (located two units to the left of the origin of coordinates (0,0)), and from there, we investigate at what y-value (value on the vertical y-axis) a line that follows the vertical grid passing through x = -2, intercepts the graph of the function.

That is pictured with a red dot in the attached image. Notice that the y-value at which such intersection occurs is y = 2.

Therefore f(-2) = 2

4 0

3 years ago

Can someone answer this question please?

Mekhanik [1.2K]

Steps:
1. calculate the values of y at x=0,1,2. using y=5-x^2
2. calculate the areas of trapezoids (Bottom+Top)/2*height
3. add the areas.

1.
x=0, y=5-0^2=5
x=1, y=5-1^2=4
x=2, y=5-2^2=1
2.
Area of trapezoid 1 = (5+4)/2*1=4.5
Area of trapezoid 2 = (4+1)/2*1=2.5

Total area of both trapezoids = (4.5+2.5) = 7

Exact area by integration:
integral of (5-x^2)dx from 0 to 2
=[5x-x^3/3] from 0 to 2
=[5(2-0)-(2^3-0^3)/3]
=10-8/3
=22/3
=7 1/3, slight greater than the estimation by trapezoids.

8 0

3 years ago

four families go to the cinema. the reid family take 2 adults and 2 children for £18 ; the mghee family take 1 senior, 2 adults

RoseWind [281]

Answer:

$\pounds 26.50$

Step-by-step explanation:

Let

x = cost for one adult

y = cost for one child

z = cost for one senior.

The Reid family:

take 2 adults and 2 children and paid $2x+2y$ that is $\pounds 18$ , so

$2x+2y=18$

The Mghee family:

take 1 senior, 2 adults and 1 child and paid $x+2x+y$ that is $\pounds 18.50$ , so

$x+2x+y=18.50$

The Griffiths family:

take 1 senior and 3 adults and paid $z+3x$ that is $\pounds 19.50$ , so

$z+3x=19.50$

You get the system of three equations:

$\left\{\begin{array}{l}2x+2y=18\\ \\z+2x+y=18.50\\ \\z+3x=19.50\end{array}\right.$

From the first equation:

$2y=18-2x\\ \\y=9-x$

From the third equation:

$z=19.50-3x$

Substitute them into the second equation:

$19.50-3x+2x+9-x=18.50\\ \\-3x+2x-x=18.50-19.50-9\\ \\-2x=-10\\ \\2x=10\\ \\x=5$

Then

$y=9-5=4\\ \\z=19.50-3\cdot 5=19.50-15=4.50$

Hence,

the cost for one adult is $\pounds 5$

the cost for one child is $\pounds 4$

the cost for one senior is $\pounds 4.50$

The Linton family takes 1 senior, 2 adults and 3 children and paid

$\pounds 4.50+2\cdot \pounds 5+3\cdot \pounds 4=\pounds 26.50$

3 0

3 years ago

manuel deposits $10000 for 12 yr in an account paying 4% compounded annually.He then puts this total amount on deposit in anothe

Naddik [55]

$\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$10000\\
r=rate\to 4\%\to \frac{4}{100}\to &0.04\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &12
\end{cases}
\\\\\\
A=10000\left(1+\frac{0.04}{1}\right)^{1\cdot 12}\implies A=1000(1.04)^{12}\\\\\\ A\approx 16010.32$

he then turns around and grabs that money and sticks it for another 9 years,

$\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
~~
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$16010.32\\
r=rate\to 5\%\to \frac{5}{100}\to &0.05\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{semi-annually, thus twice}
\end{array}\to &2\\
t=years\to &9
\end{cases}
\\\\\\
A=16010.32\left(1+\frac{0.05}{2}\right)^{2\cdot 9}\implies A=16010.32(1.025)^{18}
\\\\\\
A\approx 24970.64$

add both amounts, and that's how much is for the whole 21 years.

6 0

3 years ago