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3 years ago

I need help with this other one oof lol

Mathematics

1 answer:

natima [27]3 years ago

3 0

Answer:

Step-by-step explanation:

2.2/3 × 2.3/4

6/3×11/4

24×33_ ÷ 12 = 66 I hope u understand

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Which of the following integrals represents the area of the region bounded by x = e and the functions f(x) = ln(x) and g(x) = lo

Neko [114]

Notice that

$\log_{1/e}x=\dfrac{\ln x}{\ln\frac1e}=\dfrac{\ln x}{-\ln e}=-\ln x$

$f(x)=\ln x$ and $g(x)=-\ln x$ intersect when $x=1$ . For all $x>1$ , we have $\ln x>0$ and $-\ln x$ , so $f(x)>g(x)$ . Then the area we want is given by the integral,

$\displaystyle\int_1^e\ln x-(-\ln x)\,\mathrm dx=2\int_1^e\ln x\,\mathrm dx$

or in terms of $\log_{1/e}x$ ,

$\displaystyle\int_1^e\ln x-\log_{1/e}x\,\mathrm dx$

5 0

3 years ago

The friendly sausage factory (fsf) can produce hot dogs at a rate of 5,000 per day. fsf supplies hot dogs to local restaurants a

juin [17]

Answer:

a. 21 327 hot dogs/run

b. 70 runs/yr

c. 4 da/run

Step-by-step explanation:

Data:

Production rate (p) = 5000/da

Usage rate (u) = 260/da

Setup cost (S) = $66

Annual carrying cost (H) = $0.45/hot dog

Production days (d) = 294 da

Calculations:

a. Optimal run size

(i) Annual demand (D) = pd = (5000 hot dogs/1 day) × (294 days/1 yr)

= 1 470 000 hot dogs/yr

(ii) Economic run size

$Q_{0}= \sqrt{\frac{2DS }{ h}\times\frac{ p}{p-u }}$

$= \sqrt{\frac{2\times1470000\times66 }{ 0.45}\times\frac{ 5000}{5000-260 }}$

$= \sqrt{431200000\times\frac{ 5000}{4740 }}$

$= \sqrt{454852321}$

= 21 327 hot dogs/run

b. Number of runs per year

Runs = D/Q₀ = (1 470 000 hot dogs/1yr) × (1 run/21 327 hotdogs)

= 70 runs/yr

c. Length of a run

Length = Q₀/p = (21 327 hot dogs/1 run) × (1 da/5000 hot dogs)

= 4 da/run

8 0

3 years ago

Please help!!!!!!!!!!!!!

Afina-wow [57]

Answer:

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Solve for (a+1)/2=1/a.

LenKa [72]

Hello!

To solve this, you are trying to convert one side to a quadratic equation and the other to just 0. Then, solve the equation.

----

First, move the /2 to the other side of the equation.

(a+1)/2 = 1/a

a+1 = 2/a

Next, multiply both sides by a.

a+1 = 2/a

$a^{2} +a=2$

And then subtract both sides by 2. This will get you to a quadratic equation.

$a^{2} +a=2$

$a^{2} +a-2=0$

Now, factor the equation. When we're factoring, we are looking for the form (a + b)(a + c), where b + c must equal 1 (the coefficient of a) and b*c must equal -2. In this case, b = 2, and c = -1.

Therefore:

$a^{2} +a-2=0$

(a + 2)(a - 1) = 0

To get the answer, recall that anything multiplied by 0 is equal to 0. Therefore, to get the left side to equal 0, either (a + 2) or (a - 1) must equal 0. To do this, a must either be -2 or 1.

Therefore, a = -2, or a = 1.

----

Check your work:

(-2 + 1)/2 = 1/-2

-1/2 = -1/2

(1 + 1)/2 = 1/1

2/2 = 1/1

1 = 1

----

Hope this helps!

6 0

3 years ago

How do you solve this?

Jet001 [13]

Answer:

There are 8 boys in the chorus and 16 girls in the chorus

The graph in the attached figure

Step-by-step explanation:

Let

x----> the number of boys

y----> the number of girls

we know that

$y=2x$ -----> equation A

$x=y-8$ ----> equation B

Solve the system of equations by graphing

Remember that the solution of the system of equations is the intersection point both graphs

The solution is the point (8,16)

see the attached figure

therefore

There are 8 boys in the chorus

There are 16 girls in the chorus

6 0

4 years ago