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2 years ago

Dont solve just find the system of equations

Mathematics

1 answer:

Black_prince [1.1K]2 years ago

4 0

Answer:

3y + x = 6

4y + 1.50x = 8.50

Step-by-step explanation:

Cost of walnut = $1.50

Cost of cashew = $1.00

Cost of dates = $1.50

Let

Dates = x

Cashew = y

Walnuts = 2y

Pounds of total mixture = 6 lb

Total cost of the mixture = $8.50

First equation (from Quantity)

Walnuts + cashew + dates = Total mixture

2y + y + x = 6

3y + x = 6

Second equation (from price× Quantity)

2y(1.50) + y(1.00) + x(1.50) = 8.50

3y + y + 1.50x = 8.50

4y + 1.50x = 8.50

Therefore, the system of equation to solve the question is

3y + x = 6

4y + 1.50x = 8.50

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MAX is equiangular If m

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What’s the question?

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2 years ago

Let C(n, k) = the number of k-membered subsets of an n-membered set. Find (a) C(6, k) for k = 0,1,2,...,6 (b) C(7, k) for k = 0,

vladimir1956 [14]

Answer:

(a) $C(6,0) = 1$ , $C(6,1) = 6$ , $C(6,2) = 15$ , $C(6,3) = 20$ , $C(6,4) = 15$ , $C(6,5) = 6$ , $C(6,6) = 1$ .

(b) $C(7,0) = 1$ , $C(7,1) = 7$ , $C(7,2) = 21$ , $C(7,3) = 35$ , $C(7,4) = 35$ , $C(7,5) = 21$ , $C(7,6) = 7$ , $C(7,7)=1$ .

Step-by-step explanation:

In this exercise we only need to recall the formula for C(n,k):

$C(n,k) = \frac{n!}{k!(n-k)!}$

where the symbol $n!$ is the factorial and means

$n! = 1\cdot 2\cdot 3\cdot 4\cdtos (n-1)\cdot n$ .

By convention 0!=1. The most important property of the factorial is $n!=(n-1)!\cdot n$ , for example 3!=1*2*3=6.

(a) The explanations to the solutions is just the calculations.

$C(6,0) = \frac{6!}{0!(6-0)!} = \frac{6!}{6!} = 1$

$C(6,1) = \frac{6!}{1!(6-1)!} = \frac{6!}{5!} = \frac{5!\cdot 6}{5!} = 6$

$C(6,2) = \frac{6!}{2!(6-2)!} = \frac{6!}{2\cdot 4!} = \frac{5!\cdot 6}{2\cdot 4!} = \frac{4!\cdot 5\cdot 6}{2\cdot 4!} = \frac{5\cdot 6}{2} = 15$

$C(6,3) = \frac{6!}{3!(6-3)!} = \frac{6!}{3!\cdot 3!} = \frac{5!\cdot 6}{6\cdot 6} = \frac{5!}{6} = \frac{120}{6} = 20$

$C(6,4) = \frac{6!}{4!(6-4)!} = \frac{6!}{4!\cdot 2!} = frac{5!\cdot 6}{2\cdot 4!} = \frac{4!\cdot 5\cdot 6}{2\cdot 4!} = \frac{5\cdot 6}{2} = 15$

$C(6,5) = \frac{6!}{5!(6-5)!} = \frac{6!}{5!} = \frac{5!\cdot 6}{5!} = 6$

$C(6,6) = \frac{6!}{6!(6-6)!} = \frac{6!}{6!} = 1$ .

(b) The explanations to the solutions is just the calculations.

$C(7,0) = \frac{7!}{0!(7-0)!} = \frac{7!}{7!} = 1$

$C(7,1) = \frac{7!}{1!(7-1)!} = \frac{7!}{6!} = \frac{6!\cdot 7}{6!} = 7$

$C(7,2) = \frac{7!}{2!(7-2)!} = \frac{7!}{2\cdot 5!} = \frac{6!\cdot 7}{2\cdot 5!} = \frac{5!\cdot 6\cdot 7}{2\cdot 5!} = \frac{6\cdot 7}{2} = 21$

$C(7,3) = \frac{7!}{3!(7-3)!} = \frac{7!}{3!\cdot 4!} = \frac{6!\cdot 7}{6\cdot 4!} = \frac{5!\cdot 6\cdot 7}{6\cdot 4!} = \frac{120\cdot 7}{24} = 35$

$C(7,4) = \frac{7!}{4!(7-4)!} = \frac{6!\cdot 7}{4!\cdot 3!} = frac{5!\cdot 6\cdot 7}{4!\cdot 6} = \frac{120\cdot 7}{24} = 35$

$C(7,5) = \frac{7!}{5!(7-2)!} = \frac{7!}{5!\cdot 2!} = 21$

$C(7,6) = \frac{7!}{6!(7-6)!} = \frac{7!}{6!} = \frac{6!\cdot 7}{6!} = 7$

$C(7,7) = \frac{7!}{7!(7-7)!} = \frac{7!}{7!} = 1$

For all the calculations just recall that 4! =24 and 5!=120.

6 0

2 years ago

A car is traveling at a rate of

LUCKY_DIMON [66]

50 meters per second

3600 seconds per hour

1000 meters per km

50/1000 = 0.05

1/3600 = .0002777

0.05/0.0002777 = 180km per hour

180*4 = 720km in 4 hours

8 0

3 years ago

Read 2 more answers

Can someone please show me how 19169000*e^(0.15)= 222712201 When e= 2.7182818284?

Ierofanga [76]

Answer:

It doesn't. x = 22 271 201

Step-by-step explanation:

You are going to need a calculator no matter how you do it.

$x =19 169 000 \times e^{0.15}$

(a) The direct method

$x = 19 169 000\times 2.718 281 828^{0.15} = 19 169 000 \times 1.161 834 243 = 22 271 201$

(b) The indirect method

$\ln \left (19 169 000\times 2.718 281 828^{0.15} \right ) = \ln(19 169 000) + 0.15 = 16.768 805 + 0.15 = 16.918 804\\\\e^{16.918 804} = 22 271 201$

8 0

2 years ago

which linear system has been correctly solved for one of the variables from the following system? 2x - y = -1 2x y = -7 2x - y =

mr Goodwill [35]

If you would like to solve the system of linear equations, you can do this using the following steps:

2x - y = -1 ... y = 2x + 1
2x + y = -7 ... y = -7 - 2x

The correct result would be: y = 2x + 1, 2x + y = -7.

4 0

3 years ago