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2 years ago

What value of x makes this equation true. X/3 - 3 = X/9 + 3

Mathematics

1 answer:

OLga [1]2 years ago

6 0

27, you need to simplify both sides of the equation to get x on one side.

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At a particular restaurant, each mini hotdog has 80 calories and each pizza roll has 50 calories. A combination meal with mini h

cestrela7 [59]

There are 15 mini hotdogs and 5 mini pizza rolls in the combination meal.

Step-by-step explanation:

Given,

Calories in each hotdog = 80 calories

Calories in each mini pizza = 50 calories

Combined meal = 20

Combined calories = 1450 calories

Let,

x represent the number of hot dogs.

y represent the number of mini pizzas.

According to given statement;

x+y=20 Eqn 1

80x+50y=1450 Eqn 2

Multiplying Eqn 1 by 80

$80(x+y=20)\\80x+80y=1600\ \ \ Eqn\ 3$

Subtracting Eqn 2 from Eqn 3

$(80x+80y)-(80x+50y)=1600-1450\\80x+80y-80x-50y=150\\30y=150$

Dividing both sides by 30

$\frac{30y}{30}=\frac{150}{30}\\y=5$

Putting y=5 in Eqn 1

$x+5=20\\x=20-5\\x=15$

There are 15 mini hotdogs and 5 mini pizza rolls in the combination meal.

Keywords: linear equation, elimination method

Learn more about elimination method at:

#LearnwithBrainly

7 0

2 years ago

The area enclosed by the graphs of y = 1/x, y = 1, and x = 3 is rotated about the line y = -1. Find the volume and show steps.

Natalija [7]

Answer:

$A=\pi\displaystyle\biggr[\frac{16}{3}-2\ln(|3|)\biggr]\approx9.8524$

Step-by-step explanation:

Use the Washer Method $A=\pi\displaystyle \int\limits_{a}^{b}{\bigr[R(x)^2-r(x)^2\bigr] \, dx$ where $R(x)$ is the outer radius and $r(x)$ is the inner radius.

If we sketch out the graph, we see that $y=1$ intersects points $(1,1)$ and $(3,1)$ , which will be our bounds of integration.

Here, our outer radius will be $R(x)=(-1-1)=-2$ and our inner radius will be $r(x)=-1-\frac{1}{x}$ .

Thus, we can compute the integral and find the volume:

$A=\pi\displaystyle\int\limits^{3}_{1} {(-2)^2-\biggr(-1-\frac{1}{x}\biggr)^2 } \, dx\\ \\A=\pi\displaystyle\int\limits^{3}_{1} {4-\biggr(1+\frac{2}{x}+\frac{1}{x^2} \biggr) } \, dx\\\\A=\pi\displaystyle\int\limits^{3}_{1} {4-1-\frac{2}{x}-\frac{1}{x^2}} \, dx\\\\A=\pi\displaystyle\int\limits^{3}_{1} {3-\frac{2}{x}-\frac{1}{x^2}} \, dx\\\\A=\pi\displaystyle\biggr[3x-2\ln(|x|)+\frac{1}{x}\biggr]\Biggr|_{1}^{3}\\$

$A=\pi\displaystyle\biggr[\biggr(3(3)-2\ln(|3|)+\frac{1}{3}\biggr)-\biggr(3(1)-2\ln(|1|)+\frac{1}{1}\biggr)\biggr]\\\\A=\pi\displaystyle\biggr[\biggr(9-2\ln(|3|)+\frac{1}{3}\biggr)-\biggr(3+1\biggr)\biggr]\\\\A=\pi\displaystyle\biggr[\biggr(\frac{28}{3}-2\ln(|3|)\biggr)-\biggr(4\biggr)\biggr]\\A=\pi\displaystyle\biggr[\frac{16}{3}-2\ln(|3|)\biggr]\\A\approx9.8524$