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3 years ago

Been looking for answers all day. Please.

Mathematics

1 answer:

Papessa [141]3 years ago

8 0

Answer:88

Step-by-step explanation:

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A 9.85 m ladder is placed against a wall. The height to the top of the ladder is 5 m more than the distance between the wall and

Dafna1 [17]

Given:

Length of the ladder = 9.85 m

The height to the top of the ladder is 5 m more than the distance between the wall and the foot of the ladder.

To find:

The height to the top and the distance between the wall and the foot of the ladder.

Solution:

let x be the distance between the wall and the foot of the ladder. Then the height to the top of the ladder is (x+5).

Pythagoras theorem: In a right angle triangle,

$Hypotenuse^2=Base^2+Perpendicular^2$

In the given situation, hypotenuse is the length of ladder, i.e., 9.85 m. The base is x m and the height is (x+5) m.

Using the Pythagoras theorem, we get

$(9.85)^2=x^2+(x+5)^2$

$97.0225=x^2+x^2+10x+25$

$0=2x^2+10x+25-97.0225$

$0=2x^2+10x-72.0225$

Here, $a=2, b=10,c=-72.0225$ . Using the quadratic formula, we get

$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\dfrac{-10\pm \sqrt{(10)^2-4(2)(-72.0225)}}{2(2)}$

$x=\dfrac{-10\pm \sqrt{676.18}}{4}$

Approximating the value, we get

$x=\dfrac{-10\pm 26}{4}$

$x=\dfrac{-10+26}{4},\dfrac{-10-26}{4}$

$x=\dfrac{16}{4},\dfrac{-36}{4}$

$x=4,-9$

Distance cannot be negative so $x\neq -9.$

Now we have $x=4$

$x+5=4+5$

$x+5=9$

Therefore, the base is 4 m and the height is 9 m.

5 0

3 years ago

You are saving money to buy an electric guitar. You deposit $1000 in an account that earns interest compounded annually. The exp

Katena32 [7]

Let's move like a crab, backwards some.

after 2 years?

$\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$1000\\
r=rate\to 3\%\to \frac{3}{100}\to &0.03\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &2
\end{cases}
\\\\\\
A=1000\left(1+\frac{0.03}{1}\right)^{1\cdot 2}\implies A=1000(1.03)^2$

after 3 years?

$\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$1000\\
r=rate\to 3\%\to \frac{3}{100}\to &0.03\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &3
\end{cases}
\\\\\\
A=1000\left(1+\frac{0.03}{1}\right)^{1\cdot 3}\implies A=1000(1.03)^3$

is that enough to pay the $1100?

now, let's write 1000(1+r)² in standard form

1000( 1² + 2r + r²)

1000(1 + 2r + r²)

1000 + 2000r + 1000r²

1000r² + 2000r + 1000 <---- standard form.

8 0

3 years ago

The table represents an exponential function.

kkurt [141]

Answer:

b=2/3

Step-by-step explanation:

The table is given as:

$\left|\begin{array}{c|c}x&y\\--&--\\1&9\\2&6\\3&4\\4&\dfrac83\\\\5&\dfrac{16}{9}\end{array}\right|$