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2 years ago

Question 1 of 10

Mathematics

1 answer:

sergiy2304 [10]2 years ago

6 0

Answer: C. 2 and 3 only

Step-by-step explanation: i did the same quiz

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If Isaac uses a coupon entitling him to a 25% discount off the purchase price before tax, how much will his bill be? Assume that

lions [1.4K]

Answer:

b) 25/100 x 79

= 19.75

= 79 - 19.75

= $59.25 (before tax)

7/100 x 59.25

= 414.75/100

= 4.15

= 59.25 + 4.15

= $63.4 (after adding tax)

First we have to take out the percentage of discount and minus it from the total cost . To find the discounted price with tax, we have to take out the percentage of the tax from the discounted amount and add it to the discounted amount to get the total cost.

Step-by-step explanation:

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2 years ago

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Can all edges have the same length? Yes or No

DedPeter [7]

Yes they can all have the same

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2 years ago

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HELPPP DOES ANYONE KNOW THE ANSWER???

IRISSAK [1]

Answer:

Step-by-step explanation:

solve equation.

5 0

1 year ago

PLZ HELP ME ☻ <img src="https://tex.z-dn.net/?f=%5C%5B%5Cfrac%7Bxy%7D%7Bx%20%2B%20y%7D%20%3D%201%2C%20%5Cquad%20%5Cfrac%7Bxz%7D%

Yanka [14]

Answer:

$x=\frac{12}{7} \\y=\frac{12}{5} \\z=-12$

Step-by-step explanation:

Let's re-write the equations in order to get the variables as separated in independent terms as possible \:

First equation:

$\frac{xy}{x+y} =1\\xy=x+y\\1=\frac{x+y}{xy} \\1=\frac{1}{y} +\frac{1}{x}$

Second equation:

$\frac{xz}{x+z} =2\\xz=2\,(x+z)\\\frac{1}{2} =\frac{x+z}{xz} \\\frac{1}{2} =\frac{1}{z} +\frac{1}{x}$

Third equation:

$\frac{yz}{y+z} =3\\yz=3\,(y+z)\\\frac{1}{3} =\frac{y+z}{yz} \\\frac{1}{3}=\frac{1}{z} +\frac{1}{y}$

Now let's subtract term by term the reduced equation 3 from the reduced equation 1 in order to eliminate the term that contains "y":

$1=\frac{1}{y} +\frac{1}{x} \\-\\\frac{1}{3} =\frac{1}{z} +\frac{1}{y}\\\frac{2}{3} =\frac{1}{x} -\frac{1}{z}$

Combine this last expression term by term with the reduced equation 2, and solve for "x" :

$\frac{2}{3} =\frac{1}{x} -\frac{1}{z} \\+\\\frac{1}{2} =\frac{1}{z} +\frac{1}{x} \\ \\\frac{7}{6} =\frac{2}{x}\\ \\x=\frac{12}{7}$

Now we use this value for "x" back in equation 1 to solve for "y":

$1=\frac{1}{y} +\frac{1}{x} \\1=\frac{1}{y} +\frac{7}{12}\\1-\frac{7}{12}=\frac{1}{y} \\ \\\frac{1}{y} =\frac{5}{12} \\y=\frac{12}{5}$

And finally we solve for the third unknown "z":

$\frac{1}{2} =\frac{1}{z} +\frac{1}{x} \\\\\frac{1}{2} =\frac{1}{z} +\frac{7}{12} \\\\\frac{1}{z} =\frac{1}{2}-\frac{7}{12} \\\\\frac{1}{z} =-\frac{1}{12}\\z=-12$

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2 years ago

Write 10.895.73 in word form and 11.273.<br> 26 in word form

bija089 [108]

I don't think a decimal number could have 2 decimal points.

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3 years ago

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