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Nina [5.8K]
2 years ago
5

Samuel bought 148 pearl-colored beads and 36 strings from a shop. He wants to make a certain number of necklaces using a combina

tion of the beads and the strings, with no pieces left over. What is the greatest number of necklaces he can make?
Mathematics
1 answer:
Lina20 [59]2 years ago
3 0

Answer:

4 necklaces

Step-by-step explanation:

This questions test our knowledge of highest common factor.

-We find the value that goes into 148 and 36 at the same time:

148=2\times 2\times37\\\\36=2\times2\times 3\times 3\\\\\therefore HCF=2\times 2=2^2=4

Since the HCF is 4, the greatest number of necklaces he can make is 4

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Evren used 24 centimeters of tape to wrap 3 presents. How many presents would Evren be able to wrap with 112 cm of tape?
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14

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Since we're given the conversion 24 cm for every 3 presents wrapped, we can divide our total amount of tape by this conversion, assuming a constant proportion.

Evren has 112 cm of tape. We can write the following proportion:

\frac{112}{x}=\frac{24}{3}, where x is the number of presents he can wrap with 112 cm.

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2 years ago
Tia takes $50 from her bank account to pay for bus fare for the month. She does this each month for 7 months in a row. Which equ
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  • B. - $50 * 7 = - $350

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Read 2 more answers
answer A radio station located 120 miles due east of Collinsville has a listening radius of 100 miles. A straight road joins Col
melamori03 [73]

Answer:

168.7602 miles

Step-by-step explanation:

One way to solve this problem is by using an equation that describes the listening radius of the station, and another for the road, then the points where this two-equation intersect each other will represent when the driver starts and stops listening to the station, and the distance between the points is the miles that the driver will receive the signal.

The equation for the listening radius (the radio station is at (0,0)):

x^2+y^2=100^2

The equation for the road that past through the points (-120,0) and (80,100) (Collinsville and Harmony respectively):

m=\frac{y_2-y_1}{x_2-x_1} =\frac{100-0}{80-(-120)}=\frac{100}{200}=\frac{1}{2}

y-y_1=m(x-x_1)\\y-0=\frac{1}{2}(x-(-120))\\ y=\frac{1}{2}x+60

Substitutes the value of y in the equation of the circle:

x^2+(\frac{1}{2}x+60)^2=100^2\\x^2+\frac{1}{4} x^2+60x+3600=10000\\\frac{5}{4} x^2+60x+3600=10000\\\frac{5}{4} x^2+60x+3600-10000=0\\\frac{5}{4} x^2+60x-6400=0\\5 x^2+240x-25600=0\\x^2+48x-5120=0\\

The formula to solve second-degree equations:

x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac} }{2a} \\x_{1,2}=\frac{-48\pm\sqrt{48^2-4(1)(-5120)} }{2(1)}\\x_{1,2}=\frac{-48\pm\sqrt{2304+20480} }{2}\\x_{1,2}=\frac{-48\pm\sqrt{22784} }{2}\\x_{1,2}=\frac{-48\pm16\sqrt{89} }{2}\\x_{1,2}=-24\pm8\sqrt{89} \\x_1=-24+8\sqrt{89}\approx51.4718\\x_2=-24-8\sqrt{89}\approx-99.4718\\

Using the values in x to find the values in y:

y_1=\frac{1}{2}x_1+60\\y_1=\frac{1}{2}(-24+8\sqrt{89} )+60\\y_1=-12+4\sqrt{89}+60\\ y_1=48+4\sqrt{89}\approx85.7359

y_2=\frac{1}{2}x_2+60\\y_2=\frac{1}{2}(-24-8\sqrt{89} )+60\\y_1=-12-4\sqrt{89}+60\\ y_1=48-4\sqrt{89}\approx10.2641

The distance between the points (51.4718,85.7359) and (-99.4718,10.2641) :

d=\sqrt{(x_1 -x_2 )^2+(y_1 -y_2)^2} \\d=\sqrt{(-24+8\sqrt{89} -(-24-8\sqrt{89}) )^2+(48+4\sqrt{89} -(48-4\sqrt{89}) )^2}\\d=\sqrt{(-24+8\sqrt{89} +24+8\sqrt{89} )^2+(48+4\sqrt{89} -48+4\sqrt{89} )^2}\\d=\sqrt{(16\sqrt{89} )^2+(8\sqrt{89} )^2}\\d=\sqrt{22784+5696}\\d=\sqrt{28480}\\d=8\sqrt{445}\approx168.7602miles

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