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GaryK [48]
2 years ago
15

To arrive to his appointment on time, Mr. Jones had to drive all the way from his home with the average speed of 60 mph. Due to

heavy traffic, he was driving 15 mph slower than he planned and arrived to the appointment 20 minutes later. How many miles from Mr. Jones' home was his appointment?
Mathematics
1 answer:
FinnZ [79.3K]2 years ago
8 0
35,35 35353553535353553535353535535353535535353553
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The data reflects the number of hours an employee spends working a checkout line (x), paired with the number of customers served
Whitepunk [10]

The slope of the line can be interpreted that with each additional hour, nearly 18 customers are served; option B

<h3>What is the slope of a line?</h3>

The slope of a line is the ratio of climb to that of run

Slope = change in y/change in x

The given graph is a graph of the number of hours an employee spends working a checkout line (x), against the number of customers served (y).

The slope of the graph can be written as follows:

Slope = change in the number of customers served/change in the number of hours an employee spends working

Therefore, the slope of the line can be interpreted that with each additional hour, nearly 18 customers are served.

In conclusion, slope of a line is ratio of change in y to change in x.

Learn more about slope at: brainly.com/question/16949303

#SPJ1

7 0
1 year ago
Sara has 24 sweets
Tresset [83]
<h2>Hello!</h2>

The answer is:

The expression for the number of sweets that Sara and Tim have now, are:

Sara=(17-x)Sweets\\\\Tim=\frac{(24+x)Sweets}{2}

<h2>Why?</h2>

To write the expressions for the number of sweets that Sara and Tim have now, we need to follow the next steps:

Sara starts with 24 sweets and Tim starts with 24 sweets

Sara=24 Sweets\\Tim=24 Sweets

Then, Sara gives Tim x Sweets

Sara=(24-x)Sweets\\Tim=(24+x)Sweets

Then. Sara eats 7 of her Sweets

Sara=(24-x-7) Sweets\\Sara=(17-x )Sweets\\Tim=(24+x) Sweets

Then, Tim eats half of his sweets

Sara=(17-x)Sweets\\\\Tim=\frac{(24+x)Sweets}{2}

So, the expression for the number of sweets that Sara and Tim have now, are:

Sara=(17-x) Sweets\\\\Tim=\frac{(24+x)Sweets}{2}

Have a nice day!

8 0
3 years ago
What is the value of x if x/ 3 + 1 = -2 ?​
Serga [27]
<h2>Answer:</h2><h3>x = -9</h3>

<h2>Step-by-step explanation:</h2>

<h3><u>Step 1</u>: Simplify both sides of the equation.</h3><h3 /><h3>1/3x+1=−2</h3><h3 /><h3><u>Step 2</u>: Subtract 1 from both sides.</h3><h3 /><h3>1/3x+1−1=−2−1</h3><h3 /><h3>1/3x=−3</h3><h3 /><h3><u>Step 3</u>: Multiply both sides by 3.</h3><h3 /><h3>3*(1/3x)=(3)*(−3)</h3><h3 /><h3>x=−9</h3>
8 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Csf%5Clim_%7Bx%20%5Cto%200%20%7D%20%5Cfrac%7B1%20-%20%5Cprod%20%5Climits_%
xxTIMURxx [149]

To demonstrate a method for computing the limit itself, let's pick a small value of n. If n = 3, then our limit is

\displaystyle \lim_{x \to 0 } \frac{1 - \prod \limits_{k = 2}^{3} \sqrt[k]{\cos(kx)} }{ {x}^{2} }

Let a = 1 and b the cosine product, and write them as

\dfrac{a - b}{x^2}

with

b = \sqrt{\cos(2x)} \sqrt[3]{\cos(3x)} = \sqrt[6]{\cos^3(2x)} \sqrt[6]{\cos^2(3x)} = \left(\cos^3(2x) \cos^2(3x)\right)^{\frac16}

Now we use the identity

a^n-b^n = (a-b)\left(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots a^2b^{n-3}+ab^{n-2}+b^{n-1}\right)

to rationalize the numerator. This gives

\displaystyle \frac{a^6-b^6}{x^2 \left(a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5\right)}

As x approaches 0, both a and b approach 1, so the polynomial in a and b in the denominator approaches 6, and our original limit reduces to

\displaystyle \frac16 \lim_{x\to0} \frac{1-\cos^3(2x)\cos^2(3x)}{x^2}

For the remaining limit, use the Taylor expansion for cos(x) :

\cos(x) = 1 - \dfrac{x^2}2 + \mathcal{O}(x^4)

where \mathcal{O}(x^4) essentially means that all the other terms in the expansion grow as quickly as or faster than x⁴; in other words, the expansion behaves asymptotically like x⁴. As x approaches 0, all these terms go to 0 as well.

Then

\displaystyle \cos^3(2x) \cos^2(3x) = \left(1 - 2x^2\right)^3 \left(1 - \frac{9x^2}2\right)^2

\displaystyle \cos^3(2x) \cos^2(3x) = \left(1 - 6x^2 + 12x^4 - 8x^6\right) \left(1 - 9x^2 + \frac{81x^4}4\right)

\displaystyle \cos^3(2x) \cos^2(3x) = 1 - 15x^2 + \mathcal{O}(x^4)

so in our limit, the constant terms cancel, and the asymptotic terms go to 0, and we end up with

\displaystyle \frac16 \lim_{x\to0} \frac{15x^2}{x^2} = \frac{15}6 = \frac52

Unfortunately, this doesn't agree with the limit we want, so n ≠ 3. But you can try applying this method for larger n, or computing a more general result.

Edit: some scratch work suggests the limit is 10 for n = 6.

6 0
2 years ago
What s is y= x+ 5 I need to knoww for math
IrinaK [193]
If you’re looking for the y-intercept: y=5
If you’re looking for the x-intercept: x=-5
4 0
2 years ago
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