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2 years ago

9x=10 please help…….

Mathematics

1 answer:

Natasha_Volkova [10]2 years ago

8 0

Let's solve your equation step-by-step.

9x=10

Step 1: Divide both sides by 9.

9x / 9 = 10 /9

x = 10 / 9

Answer:

x = 10 / 9

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-5 and -3 are your numbers.

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2 years ago

During which time period does Landon's elevation change the fastest? Explain how you know?

bogdanovich [222]

<h3>1. How many inches per minute does London's elevation change between 4 minutes and 8 minutes. </h3>

The question actually asks for the slope of the line that stands for the points $(4,3) \ and \ (8,6)$ why? because the questions tells us that London's elevation changes between 4 minutes and 8 minutes here. Hence, to find the slope of this line we have to use the following formula:

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \\ \\ But: \\ \\ P(x_{1},y_{1})=P(4,3) \\ P(x_{2},y_{2})=P(8,6)$

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \\ \\ But: \\ \\ P(x_{1},y_{1})=P(4,3) \\ P(x_{2},y_{2})=P(8,6) \\ \\ So: \\ \\ m=\frac{6-3}{8-4}=0.75in/min$

<em>So London's elevation changes 0.75 inches per minute</em>

<h3>2. During which time period does London's elevation change the fastest?</h3>

The greater the absolute value of the slope of the line the faster London's elevation changes. Since this is a Piecewise function, we must analyze each period.

FIRST:

→ Between 0 minutes and 4 minutes the function is constant, so there is no any change here.

→ Between 10 minutes and 14 minutes the function is constant, so there is no any change here.

→ Between 18 minutes and 22 minutes the function is constant, so there is no any change here.

So the solution is not in these parts of the function.

SECOND:

→ Between 4 minutes and 10 minutes the function has a positive slope, so there is change here.

In the previous item we calculated the slope between 4 and 8 minutes that is the same slope between 4 and 8 minutes and equals 0.75.

→ Between 14 minutes and 18 minutes the function has a positive slope, so there is change here.

Let's take two points here, say, $(16,5) \ and \ (18,3)$

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \\ \\ But: \\ \\ P(x_{1},y_{1})=P(16,5) \\ P(x_{2},y_{2})=P(18,3) \\ \\ So: \\ \\ m=\frac{3-5}{18-16}=-1 in/min$

As you can see, the absolute value here is 1 that is greater than 0.75.

<em>In conclusion, London's elevation changes the fastest between 14 and 18 minutes</em>

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2 years ago

In a class full of men and women, 5/9 of the class are women. What is the ratio of men to women in its simplest form?

andrew-mc [135]

Answer:

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Step-by-step explanation:

5x are women

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2 years ago

What is 33 1/3% profit if $4800

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Answer:

The left side 33.3 does not equal to the right side 4800, which means that the given statement is false Hope this helps

Step-by-step explanation:

I don't know if u meant converting to a percentage But i guess thats your Answer

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Read 2 more answers

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Answer:

Since,

$\frac{d}{dx}x^n = nx^{n-1}$

$\frac{d}{dx}(f(x).g(x)) = f(x).\frac{d}{dx}(g(x)) + g(x).\frac{d}{dx}(f(x))$

$\frac{d}{dx}(\frac{f(x)}{g(x)})=\frac{g(x).f'(x) - f(x) g'(x)}{(g(x))^2}$

1) $y = 8x e^x$

Differentiating with respect to x,

$\frac{dy}{dx}=8( x \times e^x + e^x) = 8(xe^x + e^x) = 8e^x(x+1)$

2) $y = 5x e^{x^4}$

Differentiating w. r. t x,

$\frac{dy}{dx}=5(x\times 4x^3 e^{x^4}+e^{x^4})=5e^{x^4}(4x^4+1)$

3) $y = x^8 + \frac{5}{x^4}$

Differentiating w. r. t. x,

$\frac{dy}{dx}=8x^7 - \frac{5}{x^5}\times 4 = 8x^7 - \frac{20}{x^5}=\frac{8x^{12}-20}{x^5}$

4) $f(t) = te^{11}-6t^5$

Differentiating w. r. t. t,

$f'(t) = e^{11} - 30t^4$

5) $g(p) = p\ln(2p+3)$

Differentiating w. r. t. p,

$g'(p) = p\frac{1}{2p+3}(2) + \ln(2p+3) = \frac{2p}{2p+3}+\ln(2p+3)$

6) $z = (te^{6t}+e^{5t})^7$

Differentiating w. r. t. t,

$\frac{dz}{dt}=7(te^{6t}+e^{5t})^6 ( 6te^{6t}+e^{6t} + 5e^{5t})$

7) $w =\frac{2y + y^2}{7+y}$

Differentiating w. r. t. y,

$\frac{dw}{dy} = \frac{(7+y)(2+2y)-(2y+y^2)}{(7+y)^2} = \frac{14 + 2y + 14y +2y^2 - 2y - y^2}{(7+y)^2}=\frac{14+14y+y^2}{(7+y)^2}$

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