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3 years ago

Solve -2a-5>3. which graph shows the solutions?

Mathematics

1 answer:

Flura [38]3 years ago

8 0

Answer:

see explanation

Step-by-step explanation:

Given

- 2a - 5 > 3 ( add 5 to both sides )

- 2a > 8

Divide both sides by - 2, reversing the direction of the inequality sign as a consequence of dividing by a negative quantity

a < - 4

Since a is less than we require an open circle at - 4 on the number line.

Since less than the arrow head is pointing to the left, that is

Graph A, the top graph illustrates the solution

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The function c(x) = 18x -10 represents the cost in dollars. How many tickets can you buy?

Mekhanik [1.2K]

Note: It seems you may have unintentionally missed writing the complete question. As total cost is missing.

So, I am assuming how many tickets Mr. XYZ can buy if he/she pays 530 dollars.

The solution would still clear your concept though.

Answer:

Please check the explanation.

Step-by-step explanation:

Given the function

c(x) = 18x -10

The slope-intercept form of the line equation

We know that the slope-intercept form of the line equation

$y = mx+b$

where m is the slope and b is the y-intercept

c(x) = 18x - 10

comparing with the slope-intercept form of the line equation

rate of change = 18

c(x) or y represents the cost
'x' represents the number of tickets

Assuming the total cost i.e. c(x) = $530

In order to determine the value of x, set c(x) = 530

i.e.

530 = 18x -10

switch sides

18x - 10 = 530

add 10 to both sides

18x - 10 + 10 = 530 + 10

18x = 540

Divide 18 to both sides

18x/18 = 540/18

x = 30

Therefore, if you can buy x = 30 tickets if you pay 530 dollars.

5 0

3 years ago

What is the value of x?

satela [25.4K]

Answer:

x = 4

Step-by-step explanation:

By property of intersecting secants outside of circle.

$PC \times PA = PB \times PD \\ 10(2x + 10) = 9(2x + 3 + 9) \\ 20x + 100 = 18x + 108 \\ 20x - 18x = 108 - 100 \\ 2x = 8 \\ x = \frac{8}{2} \\ \huge \red{ \boxed{ x = 4}}$

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3 years ago

True / False: ANOVA, Part I. Determine if the following statements are true or false in ANOVA, and explain your reasoning for st

Juliette [100K]

The answer is B hope that helps

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3 years ago

Find the exact area of the surface obtained by rotating the curve about the x-axis. y = 1 + ex , 0 ≤ x ≤ 9

tekilochka [14]

The surface area is given by

$\displaystyle2\pi\int_0^9(1+e^x)\sqrt{1+e^{2x}}\,\mathrm dx$

since $y=1+e^x\implies y'=e^x$ . To compute the integral, first let

$u=e^x\implies x=\ln u$

so that $\mathrm dx=\frac{\mathrm du}u$ , and the integral becomes

$\displaystyle2\pi\int_1^{e^9}\frac{(1+u)\sqrt{1+u^2}}u\,\mathrm du$

$=\displaystyle2\pi\int_1^{e^9}\left(\frac{\sqrt{1+u^2}}u+\sqrt{1+u^2}\right)\,\mathrm du$

Next, let

$u=\tan t\implies t=\tan^{-1}u$

so that $\mathrm du=\sec^2t\,\mathrm dt$ . Then

$1+u^2=1+\tan^2t=\sec^2t\implies\sqrt{1+u^2}=\sec t$

so the integral becomes

$\displaystyle2\pi\int_{\pi/4}^{\tan^{-1}(e^9)}\left(\frac{\sec t}{\tan t}+\sec t\right)\sec^2t\,\mathrm dt$

$=\displaystyle2\pi\int_{\pi/4}^{\tan^{-1}(e^9)}\left(\frac{\sec^3t}{\tan t}+\sec^3 t\right)\,\mathrm dt$

Rewrite the integrand with

$\dfrac{\sec^3t}{\tan t}=\dfrac{\sec t\tan t\sec^2t}{\sec^2t-1}$

so that integrating the first term boils down to

$\displaystyle2\pi\int_{\pi/4}^{\tan^{-1}(e^9)}\frac{\sec t\tan t\sec^2t}{\sec^2t-1}\,\mathrm dt=2\pi\int_{\sqrt2}^{\sqrt{1+e^{18}}}\frac{s^2}{s^2-1}\,\mathrm ds$

where we substitute $s=\sec t\implies\mathrm ds=\sec t\tan t\,\mathrm dt$ . Since

$\dfrac{s^2}{s^2-1}=1+\dfrac12\left(\dfrac1{s-1}-\dfrac1{s+1}\right)$

the first term in this integral contributes

$\displaystyle2\pi\int_{\sqrt2}^{\sqrt{1+e^{18}}}\left(1+\frac12\left(\frac1{s-1}-\frac1{s+1}\right)\right)\,\mathrm ds=2\pi\left(s+\frac12\ln\left|\frac{s-1}{s+1}\right|\right)\bigg|_{\sqrt2}^{\sqrt{1+e^{18}}}$

$=2\pi\sqrt{1+e^{18}}+\pi\ln\dfrac{\sqrt{1+e^{18}}-1}{1+\sqrt{1+e^{18}}}$

The second term of the integral contributes

$\displaystyle2\pi\int_{\pi/4}^{\tan^{-1}(e^9)}\sec^3t\,\mathrm dt$

The antiderivative of $\sec^3t$ is well-known (enough that I won't derive it here myself):

$\displaystyle\int\sec^3t\,\mathrm dt=\frac12\sec t\tan t+\frac12\ln|\sec t+\tan t|+C$

so this latter integral's contribution is

$\pi\left(\sec t\tan t+\ln|\sec t+\tan t|\right)\bigg|_{\pi/4}^{\tan^{-1}(e^9)}=\pi\left(e^9\sqrt{1+e^{18}}+\ln(e^9+\sqrt{1+e^{18}})-\sqrt2-\ln(1+\sqrt2)\right)$

Then the surface area is

$2\pi\sqrt{1+e^{18}}+\pi\ln\dfrac{\sqrt{1+e^{18}}-1}{1+\sqrt{1+e^{18}}}+\pi\left(e^9\sqrt{1+e^{18}}+\ln(e^9+\sqrt{1+e^{18}})-\sqrt2-\ln(1+\sqrt2)\right)$