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3 years ago

I need help with my math homework x+23=9/10

Mathematics

1 answer:

BartSMP [9]3 years ago

8 0

Answer:

-221/10

Step-by-step explanation:

In the attached file

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What side length would you specify if you were required to create a regular hexagonal plate that was composed of 33 cm2 of sheet

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3 years ago

A cable that weighs 8 lb/ft is used to lift 750 lb of coal up a mine shaft 500 ft deep. Find the work done. Show how to approxim

Annette [7]

Answer:

13.8 × 10⁵ lb/ft

Step-by-step explanation:

Suppose the distance(ft) below the top of the shaft is represented by x

The weight of the cable = 8 lb/ft

Weight of the coal to be lifted from mine = 750 lb

Recall that:

The work done by a force f to move an object through a distance x can be expressed as:

W = force (f) × displacement (x)

So, the force implies the total weight which should be lifted at any height x

f(x) = 750 + 8x

Using Riemann sum

where; the coal is lifted from x = 0 to x = 500 i.e. [a,b] = [0,500]

Dividing the interval into n subintervals

$\Deltax = \dfrac{b -a }{n}$

$\Deltax = \dfrac{500- 0 }{n}$

Suppose $[x_{i-1},x_i ]$ to represent the $i^{th}$ subinterval, then the work done can be estimated as:

$W_i = f(x_i) \Delta x$

$W_i =(750 + 8x_i) \Delta x$

Therefore; the total work done in between all the n subintervals is:

$W = \sum \limits ^n_{i=1} W_1$

$W = \sum \limits ^n_{i=1} f(x_i) \Delta x$

$W = \sum \limits ^n_{i=1}(750 +8x_I)\Delta x$

Therefore;

dW = f(x) dx

$\int \ dW = \int \ f(x) \ dx$ where x ranges from 0 to 500

$W = \int ^{500}_{0} \ 750 + 8x \ dx$

$W =\bigg [750x + 4x^2 \bigg ]^{500}_{0}$

$W =\bigg [750(500) + 4(500)^2-0 \bigg ]$

W = 375000 + 1000000

W = 1375000

Thus; the total work done W = 13.8 × 10⁵ lb/ft

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3 years ago

At the start of the day there was £129.75 in the till

ella [17]

They made £15.75 in one hour

6 0

3 years ago

Suppose that $p$ and $q$ are positive numbers for which \[\log_9 p = \log_{12} q = \log_{16} (p + q).\] what is the value of $q/

SVEN [57.7K]

Given $p,q>0$ and $\log_9p=\log_{12}q=\log_{16}(p+q)=x$ (say).

Then,

$p=9^x\\ q=12^x\\ p+q=16^x$

From the above 3 equations,

$\frac{q}{p} =(\frac{12}{9} )^x\\ \frac{q}{p} =(\frac{4}{3} )^x\\ \frac{p+q}{p} =(\frac{16}{9} )^x\\ \frac{p+q}{p} =(\frac{4}{3} )^{2x}\\$

From the equations, we get

$\frac{p+q}{p}=(\frac{q}{p})^2\\ 1+\frac{q}{p}=(\frac{q}{p})^2\\ (\frac{q}{p})^2-\frac{q}{p}-1=0\\ \frac{q}{p}=\frac{1 \pm \sqrt{5}}{2}$

Since $p,q>0$ , the negative value is rejected.

$\frac{q}{p}=\frac{1 + \sqrt{5}}{2}$

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3 years ago

HELP I WILL GIVE BRAINLIEST!

kvv77 [185]

Answer:

1: 38

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5: 6

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3 years ago