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Diano4ka-milaya [45]

3 years ago

10

elisa and joey work for their uncle mike, repairs skateboards and bicycles. he asks them to order 20 new wheels for the 8 skateb

oards and bicycles in his repair shop. how many bicycles and how many skateboards are in uncle mikes shop?

1 answer:

Juli2301 [7.4K]3 years ago

6 0

8 I guess? The question isn’t very well written

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The slope f '(x) at each point (x, y) on a curve. y = f (x) is given along with a particular point (a, b) on the curve. Use this

Advocard [28]

Answer:

Step-by-step explanation:

Integrating each term with respect to x, we get:

x^3 x^2

f(x) = 9--------- + 4------- - 4x + C

3 2

We are told that if x = 0, f(x) = -7, and so C must equal - 7.

The solution is

x^3 x^2

f(x) = 9--------- + 4------- - 4x - 7, or f(x) = 3x^3 + 2x^2 - 4x - 7

3 2

5 0

2 years ago

The gradiant y=2x³-5x+1 at point (2,7)

olga2289 [7]

Given:

The equation of the curve is:

$y=2x^3-5x+1$

To find:

The gradient (slope) of the given curve at point (2,7).

Solution:

We have,

$y=2x^3-5x+1$

Differentiate the given equation with respect to x.

$y=2(3x^2)-5(1)+(0)$

$y'=6x^2-5$

Now we need to find the value of this derivative at (2,7).

$y'_{(2,7)}=6(2)^2-5$

$y'_{(2,7)}=6(4)-5$

$y'_{(2,7)}=24-5$

$y'_{(2,7)}=19$

Therefore, the gradient (slope) of the given curve at point (2,7) is 19.

8 0

2 years ago

What is i^30?<br>0A. -i<br>0B. -1<br>0C. 1<br>0D. i

kozerog [31]

Answer: B

Step-by-step explanation:

-1 will be your answer. (B)

6 0

2 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

2 years ago

What is 34 divided by 45

Mrrafil [7]

34 ÷ 45 $\frac{34}{45}$ = 0.7555555555556 ≈ 0.76

7 0

3 years ago