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Oksi-84 [34.3K]
2 years ago
5

Sandy takes a day off once every 4 days, and Morgan takes a day off once every 10 days. Today, both of them have taken the day o

ff. How many days will it be until they have their next day off together?
Mathematics
2 answers:
Alenkasestr [34]2 years ago
4 0
Lets say that the day they both take off is day 0.
The next day Sandy takes off will be day 4, and Morgans will be day 10.
Sandy will take off 4, 8, 12, 16, and 20.
Morgan will take off 10, 20, 30, 40, and 50.

Both Morgan and Sandy take off day 20, so the next time they have a day off together will be in 20 days.
Lubov Fominskaja [6]2 years ago
4 0
The interval for each time they take the day off would be every 20 days.
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Determine which equations have the same solution set as StartFraction 2 Over 3 EndFraction minus x plus StartFraction 1 Over 6 E
alex41 [277]

Answer:

rfrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
Match each set of points with the slope of the line formed by the ordered pairs.
Fynjy0 [20]

1) The slope of the point (0,0) and (2,1) is 1/2.

2) The slope of the point (0,-1) and (2,-5) is -2.

3) The slope of the point (0,7) and (2,4) is -3/2.

4) The points (-1,3) and (-1,-3) has No slope.

5) The slope of the point (1,-3) and (-1,-3) is zero.

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

There are five set of points given in the question.

You need to find out the slope of the line formed by the points.

The formula used to find the slope of a line is given by,

Slope = \frac{y2-y1}{x2-x1}

1) The given points are (0,0) and (2,1)

(x1,y1) ⇒ (0,0)

(x2,y2) ⇒ (2,1)

Slope = \frac{1-0}{2-0}

⇒ 1/2 is the slope.

2) The given points are (0,-1) and (2,-5)

(x1,y1) ⇒ (0,-1)

(x2,y2) ⇒ (2,-5)

Slope = \frac{-5+1}{2-0}

⇒ \frac{-4}{2}

⇒ -2 is the slope.

3) The given points are (0,7) and (2,4)

(x1,y1) ⇒ (0,7)

(x2,y2) ⇒ (2,4)

Slope = \frac{4-7}{2-0}

⇒ -3/2 is the slope.

4) The given points are (-1,3) and (-1,-3)

(x1,y1) ⇒ (-1,3)

(x2,y2) ⇒ (-1,-3)

Slope = \frac{-3-3}{-1+1}

⇒ -6/0 (No slope)

If the denominator of a fraction is zero, the expression is not a legal fraction because it's overall value is undefined.

∴ There is no slope.

5) The given points are (1,-3) and (-1,-3)

(x1,y1) ⇒ (1,-3)

(x2,y2) ⇒ (-1,-3)

Slope = \frac{-3+3}{-1-1}

⇒ 0/-2

∴ The slope is zero.

5 0
2 years ago
Read 2 more answers
Please answer these questions thanks
xz_007 [3.2K]
Well...
$6.50/64= about $0.10
assuming that the cost of the 48oz bottle is $4.20,
$4.20/48= about $0.08

the 48oz is better buy, as it is $0.08/oz
(/ means per)

7 0
2 years ago
John's commute to work is 20kmhr while Sheri's commute is 500mmin. Who has the fastest commute to work in mihrif 1.61km=1mi? A S
Andreyy89

Answer:

A) Sheri has the faster commute by 6.2 miles/hr.

Step-by-step explanation:

Given

John's commute to work =20\ km/hr

Sheri's commute to work  =500\ m/min

1.61\ km = 1\ mile

John's commute to work in miles per hour = \frac{20\ km}{1 hr}\times \frac{1\ mile}{1.61\ km}= 12.42\ miles/hr

Sheri's commute to work in miles per hour =\frac{500\ m}{1\ min}\times \frac{1\ km}{1000\ m}\times \frac{1\ mile}{1.61\ km}\times \frac{60\ min}{1\ hr}= 18.63\ miles/ hr

We can see that Sheri has a faster commute.

Difference between the rates =18.63\ miles/ hr-12.42\ miles/hr=6.21\ miles/hr\approx 6.2\ miles/hr

∴ Sheri has the faster commute by 6.2 miles/hr.

3 0
2 years ago
If p is a polynomial show that lim x→ap(x)=p(a
Lostsunrise [7]

Let p(x) be a polynomial, and suppose that a is any real number. Prove that

lim x→a p(x) = p(a) .

 

Solution. Notice that

 

2(−1)4 − 3(−1)3 − 4(−1)2 − (−1) − 1 = 1 .

 

So x − (−1) must divide 2x^4 − 3x^3 − 4x^2 − x − 2. Do polynomial long division to get 2x^4 − 3x^3 − 4x^2 – x – 2 / (x − (−1)) = 2x^3 − 5x^2 + x – 2.

 

Let ε > 0. Set δ = min{ ε/40 , 1}. Let x be a real number such that 0 < |x−(−1)| < δ. Then |x + 1| < ε/40 . Also, |x + 1| < 1, so −2 < x < 0. In particular |x| < 2. So

 

|2x^3 − 5x^2 + x − 2| ≤ |2x^3 | + | − 5x^2 | + |x| + | − 2|

= 2|x|^3 + 5|x|^2 + |x| + 2

< 2(2)^3 + 5(2)^2 + (2) + 2

= 40

 

Thus, |2x^4 − 3x^3 − 4x^2 − x − 2| = |x + 1| · |2x^3 − 5x^2 + x − 2| < ε/40 · 40 = ε.

3 0
2 years ago
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