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2 years ago

Cody rented a bike from John's Bikes. It

Mathematics

1 answer:

adelina 88 [10]2 years ago

4 0

It would be two hours. This is because 24-16=8 and 8 divided by 4 is 2. Hope this helps

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Which of the following statements best describes the value of the expression 6x + 7 when x = 5?

Tanzania [10]

Answer:

? is there a option?

Step-by-step explanation:

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3 years ago

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Josie took a road trip. She drove for 45 minutes at 70 miles per hour. Then she drove for 15 minutes at 20 miles per hour. How f

fomenos

C. I'm pretty sure that's the answer I hope it helps

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2 years ago

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What is the y- intercept of the line?

Nadusha1986 [10]

The y-intercept of the line is 40

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2 years ago

Finding Derivatives Implicity In Exercise, find dy/dx implicity.<br> 4x3 + ln y2 + 2y = 2x

lisov135 [29]

Answer:

$\frac{dy}{dx}=\frac{y(1-6x^2)}{(1+y)}$

Step-by-step explanation:

Given function : 4x³ + ln(y²) + 2y = 2x

on differentiating both sides with respect to 'x', we get

$\frac{d(4x^3 + ln(y^2) + 2y)}{dx}= \frac{d(2x)}{dx}$

$12x^2+\frac{2}{y}(\frac{dy}{dx})+2(\frac{dy}{dx})=2$

$12x^2+(\frac{2}{y}+2)\frac{dy}{dx}=2$

$(\frac{2+2y}{y})\frac{dy}{dx}=2-12x^2$

$\frac{dy}{dx}=\frac{y(2-12x^2)}{2+2y}$

$\frac{dy}{dx}=\frac{2y(1-6x^2)}{2(1+y)}$

$\frac{dy}{dx}=\frac{y(1-6x^2)}{(1+y)}$

3 0

3 years ago

Plz help<br>urgent !!!!<br>will give the brainliest !!

Mkey [24]

Answer:

$X = \begin{bmatrix}1&3\\ 2&4\end{bmatrix}$

Step-by-step explanation:

The question we have at hand is, in other words,

$\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}\left(X\right)=\begin{bmatrix}8&20\\ \:3&7\end{bmatrix}$ - where we have to solve for the value of $X$

If we have to isolate $X$ here, then we would have to take the inverse of the following matrix ...

$\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}$ ... so that it should be as follows ... $\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}^{-1}$

Therefore, we can conclude that the equation as to solve for " $X$ " will be the following,

$X=\begin{bmatrix}4&2\\ 1&1\end{bmatrix}^{-1}\begin{bmatrix}8&20\\ 3&7\end{bmatrix}$ - First find the 2 x 2 matrix inverse of the first portion,

$\begin{bmatrix}4&2\\ 1&1\end{bmatrix}^{-1}$ = $\frac{1}{\det \begin{pmatrix}4&2\\ 1&1\end{pmatrix}}\begin{pmatrix}1&-2\\ -1&4\end{pmatrix}$ = $\frac{1}{2}\begin{bmatrix}1&-2\\ -1&4\end{bmatrix}$ = $\begin{bmatrix}\frac{1}{2}&-1\\ -\frac{1}{2}&2\end{bmatrix}$

At this point we have to multiply the rows of the first matrix by the rows of the second matrix,

$X = \begin{bmatrix}\frac{1}{2}&-1\\ -\frac{1}{2}&2\end{bmatrix}\begin{bmatrix}8&20\\ 3&7\end{bmatrix}$ ,

$X = \begin{pmatrix}\frac{1}{2}\cdot \:8+\left(-1\right)\cdot \:3&\frac{1}{2}\cdot \:20+\left(-1\right)\cdot \:7\\ \left(-\frac{1}{2}\right)\cdot \:8+2\cdot \:3&\left(-\frac{1}{2}\right)\cdot \:20+2\cdot \:7\end{pmatrix}$ - Simplifying this, we should get ...

$\begin{bmatrix}1&3\\ 2&4\end{bmatrix}$ ... which is our solution.

7 0

3 years ago