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PSYCHO15rus [73]
4 years ago
8

Solve:

Mathematics
1 answer:
s2008m [1.1K]4 years ago
8 0
I hope this helps you


y+z=4-x


5x+4-x=16


4x=12


x=3


x+y+z+x-y+4z=4+27


2x+5z=31


2.3+5z=31


5z=25


z=5


x+y+z=4


3+y+5=4


y=-4
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Find the solution to the system of equations.
goldenfox [79]

Substitute

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8 0
3 years ago
Dons signature coffee blend is 60% dark roast and 40% light roast. He has 10 kg of blend A, which is 80% dark roast and 20% ligh
astraxan [27]

Answer

Find out the how many kilograms of blend A will don need to use to make 10 kg of his signature blend .

To proof

As given

Dons signature coffee blend is 60% dark roast and 40% light roast.

He has 10 kg

60 % is written in the decimal form

= \frac{60}{100}

= 0.6

40 % is written in the decimal form

= \frac{40}{100}

= 0.4

Now 60% dark roast of  10 kg = 0.6 ×10

                                                 = 6 kg

Now 40% of  10 kg = 0.4 × 10

                               = 4 kg

As given

He has 10 kg of blend A, which is 80% dark roast and 20% light roast  i.e 80% of the 6kg is dark roast and 20% of the  4kg is light roast .

80% is written in the decimal form

= \frac{80}{100}

= 0.8

80% of the 6kg is dark roast = 0.8 × 6

                                                = 4.8 kg

20% is written in the decimal form

= \frac{20}{100}

= 0.2

20% of the 4kg is light roast = 0.2 × 4

                                               = .8kg

Total kilograms of blend A will don need to use to make 10 kg of his signature blend =  4.8 kg + .8 kg

                          = 5.6 kg

Hence proved




 


8 0
3 years ago
For the function given​ below, find a formula for the riemann sum obtained by dividing the interval​ [a,b] into n equal subinter
Nata [24]

We split [2, 4] into n subintervals of length \dfrac{4-2}n=\dfrac2n,

[2,4]=\left[2,2+\dfrac2n\right]\cup\left[2+\dfrac2n,2+\dfrac4n\right]\cup\left[2+\dfrac4n,2+\dfrac6n\right]\cup\cdots\cup\left[2+\dfrac{2(n-1)}n,4\right]

so that the right endpoints are given by the sequence

x_i=2+\dfrac{2i}n=\dfrac{2(n+i)}n

for 1\le i\le n. Then the Riemann sum approximating

\displaystyle\int_2^42x\,\mathrm dx

is

\displaystyle\sum_{i=1}^nf(x_i)\dfrac{4-2}n=\frac8{n^2}\sum_{i=1}^n(n+i)=\frac8{n^2}\left(n^2+\frac{n(n+1)}2\right)=\frac{12n+4}n

The integral is given exactly as n\to\infty, for which we get

\displaystyle\int_2^42x\,\mathrm dx=\lim_{n\to\infty}\frac{12n+4}n=12

To check: we have

\displaystyle\int_2^42x\,\mathrm dx=x^2\bigg|_2^4=4^2-2^2=16-4=12

7 0
3 years ago
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