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3 years ago

Solve the system with substitution: y=3x-1 and -9x+3y=-3

Mathematics

1 answer:

kati45 [8]3 years ago

7 0

I hope this helps. the answer is infinite

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Mike paid $11.71 total

yan [13]

Answer:

Bill received 13.68 change back.

Step-by-step explanation:

20-6.32=13.68

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3 years ago

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Write a word problem for 8x - 5 = 40

masya89 [10]

Answer:

Bob only has 40 dollars but he wants to buy a membership at the gym. It cost 8 dollars per month to receive a gym card. Bob went to the Gym in person so they gave him a 5 dollar off coupon. How many months can Bob buy?

Step-by-step explanation:

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2 years ago

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How many shapes would I have on term 104?

pishuonlain [190]

Answer:

313 shapes

Step-by-step explanation:

The nth term is 3n+1

so 3x104=312+1= 313 shapes

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2 years ago

Which is the equation for the line that has a slope of 12 and passes through the point (3, 20)?

sasho [114]

Answer:

y=12x-16

Step-by-step explanation:

y-y1=m(x-x1)

y-20=12(x-3)

y=12x-36+20

y=12x-16

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2 years ago

Six sophomores and 14 freshmen are competing for two alternate positions on the debate team. Which expression represents the pro

Elanso [62]

Answer:

<em>Choose the first alternative</em>

$\displaystyle P=\frac{_{1}^{6}\textrm{C}\ _{1}^{5}\textrm{C}}{_{2}^{20}\textrm{C}}$

Step-by-step explanation:

<u>Probabilities</u>

The requested probability can be computed as the ratio between the number of ways to choose two sophomores in alternate positions $(N_s)$ and the total number of possible choices $(N_t)$ , i.e.

$\displaystyle P=\frac{N_s}{N_t}$

There are 6 sophomores and 14 freshmen to choose from each separate set. There are 20 students in total

We'll assume the positions of the selections are NOT significative, i.e. student A/student B is the same as student B/student A.

To choose 2 sophomores out of the 6 available, the first position has 6 elements to choose from, the second has now only 5

$_{1}^{6}\textrm{C}\ _{1}^{5}\textrm{C} \text{ ways to do it}$

The total number of possible choices is

$_{2}^{20}\textrm{C} \text{ ways to do it}$

The probability is then

$\boxed{\displaystyle P=\frac{_{1}^{6}\textrm{C}\ _{1}^{5}\textrm{C}}{_{2}^{20}\textrm{C}}}$

Choose the first alternative

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2 years ago

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