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Katena32 [7]
3 years ago
14

Help asap do all these show work posting more after this one

Mathematics
1 answer:
ivanzaharov [21]3 years ago
4 0

Answer:

1) (2,10) is solution of y=4x+2

2) (7,5) is not solution of 2x+y=5

3) (-3,3) is not solution of y=6-x

4) (10,-1) is solution of x+8y=2

5) (2,21) is not solution of y=6x+7

6) (6,-8) is solution of 3x-y=26

Step-by-step explanation:

1) y=4x+2 (2,10)

Put x=2 and y=10 and check if both sides are equal or not

10=4(2)+2

10=8+2

10=10

So, (2,10) is solution of y=4x+2

2) 2x+y=5 (7,5)

Put x=7 and y=5 and check if both sides are equal or not

2x+y=5

2(7)+5=5

14+5=5

19≠5

So, (7,5) is not solution of 2x+y=5

3) y=6-x (-3,3)

Put x=-3 and y=3 and check if both sides are equal or not

y=6-x

3=6-(-3)

3=6+3

3≠9

So, (-3,3) is not solution of y=6-x

4) x+8y=2 (10,-1)

Put x=10 and y=-1 and check if both sides are equal or not

x+8y=2

10+8(-1)=2

10-8=2

2=2

So, (10,-1) is solution of x+8y=2

5) y=6x+7 (2,21)

Put x=2 and y=21 and check if both sides are equal or not

y=6x+7

21=6(2)+7

21=12+7

21≠19

So, (2,21) is not solution of y=6x+7

6) 3x-y=26 (6,-8)

Put x=6 and y=-8 and check if both sides are equal or not

3x-y=26

3(6)-(-8)=26

18+8=26

26=26

(6,-8) is solution of 3x-y=26

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What is the focus of the parabola? y=−1/4x2−x+3
alex41 [277]

Answer:  Focus = (-2, 3)

<u>Step-by-step explanation:</u>

y=-\dfrac{1}{4}x^2-x+3\\\\\rightarrow a=-\dfrac{1}{4},\ b=-1

First let's find the vertex. We do that by finding the Axis-Of-Symmetry:

AOS: x=\dfrac{-b}{2a}\quad =\dfrac{-(-1)}{2(\frac{-1}{4})}=\dfrac{1}{-\frac{1}{2}}=-2

Then finding the maximum by inputting x = -2 into the given equation:

y=-\dfrac{1}{4}(-2)^2-(-2)+3\\\\y=-1+2+3\\\\y=4

The vertex is: (-2, 4)

Now let's find p, which is the distance from the vertex to the focus:

a=\dfrac{1}{4p}\\\\\\-\dfrac{1}{4}=\dfrac{1}{4p}\\\\\\p=-1

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3 years ago
For each month of a given year except December, a worker earned the same monthly salary and donated one-tenth of that salary to
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Answer:

D) 11/3

Step-by-step explanation:

Let x represent monthly salary.

We have been given that for each month of a given year except December, a worker earned the same monthly salary and donated one-tenth of that salary to charity.        

Salary earned in 11 months would be 11x.

Money donated in 11 months would be \frac{11x}{10}.

Further we are told that in December, the worker earned N times his usual monthly salary and donated one-fifth of his earnings to charity.

Salary for December would be Nx.

Money donated in December would be \frac{Nx}{5}.  

The worker's charitable contributions totaled one-eighth of his earnings for the entire year that is \frac{1}{8}\cdot (11x+Nx).

\frac{11x}{10}+\frac{Nx}{5}=\frac{1}{8}\cdot (11x+Nx)

Dividing whole equation by x, we will get:

\frac{11x}{10x}+\frac{Nx}{5x}=\frac{1}{8x}\cdot x(11+N)                

\frac{11}{10}+\frac{N}{5}=\frac{1}{8}\cdot (11+N)

\frac{11}{10}+\frac{N}{5}=\frac{11}{8}+\frac{N}{8}

Combine like terms:

\frac{N}{5}-\frac{N}{8}=\frac{11}{8}-\frac{11}{10}

\frac{N}{5}*40-\frac{N}{8}*40=\frac{11}{8}*40-\frac{11}{10}*40

8N-5N=11*5-11*4

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\frac{3N}{3}=\frac{11}{3}\\\\N=\frac{11}{3}

Therefore, the value of N is 11/3 and option D is the correct choice.

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Answer:

the first, second and fifth ones

Step-by-step explanation:

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