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3 years ago

Subtract. (6x - 3) - (2x -2) What is the answer?

Mathematics

1 answer:

kirill115 [55]3 years ago

6 0

4x-1 all u got to do is remove the parentheses and collect the like terms

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Sedaia [141]

The answer is B.
Explanation: A doesn’t make sense because he only has 3 tries not 4. C doesn’t make sense because he tried 3 times not 2 times

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3 years ago

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a local park is shaped like a square with sides 6 meters long. A circular splash fountain is being placed in the middle of the p

DENIUS [597]

Answer: 624 Square meters

Step-by-step explanation:

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3 years ago

Solve the system using elimination. 3x + 3y = –9 3x – 3y = 21

scoundrel [369]

Answer:

y=-5,x=2

Step-by-step explanation:

3x+3y=-9________________eqn 1

3x-3y=21________________eqn 2

using elimination method

3x+3y=-9

3x-3y=21

subtract equation 2 from equation 1

0+6y=-30

6y=-30

y=-30/6

y=-5

substitute the value of y in equation 1

3x+3y=-9

3x+3(-5)=-9

3x-15=-9

3x=-9+15

3x=6

x=6/3

x=2

6 0

3 years ago

Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y

irina [24]

Answer:

Therefore the value of y(1)= 0.9152.

Step-by-step explanation:

According to the Euler's method

y(x+h)≈ y(x) + hy'(x) ....(1)

Given that y(0) =3 and step size (h) = 0.2.

$y'(x)= x^2y(x)-\frac12y^2(x)$

Putting the value of y'(x) in equation (1)

$y(x+h)\approx y(x) +h(x^2y(x)-\frac12y^2(x))$

Substituting x =0 and h= 0.2

$y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]$

$\Rightarrow y(0.2)\approx 3+0.2[-\frac12 \times3]$ [∵ y(0) =3 ]

$\Rightarrow y(0.2)\approx 2.7$

Substituting x =0.2 and h= 0.2

$y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2]$

$\Rightarrow y(0.4)\approx 2.7+0.2[(0.2)^2\times 2.7- \frac12(2.7)^2]$

$\Rightarrow y(0.4)\approx 1.9926$

Substituting x =0.4 and h= 0.2

$y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2]$

$\Rightarrow y(0.6)\approx 1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2]$

$\Rightarrow y(0.6)\approx 1.6593$

Substituting x =0.6 and h= 0.2

$y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]$

$\Rightarrow y(0.8)\approx 1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]$

$\Rightarrow y(0.6)\approx 0.8800$

Substituting x =0.8 and h= 0.2

$y(0.8+0.2)\approx y(0.8)+0.2[(0.8)^2\times y(0.8)-\frac12 (y(0.8))^2]$

$\Rightarrow y(1.0)\approx 0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]$

$\Rightarrow y(1.0)\approx 0.9152$

Therefore the value of y(1)= 0.9152.

4 0

3 years ago

PLEASE HELP

ycow [4]

Answer:

$11$ bushels of apples were sold

Step-by-step explanation:

Let

x----> bushels of peaches

y----> bushels of apples

we know that

$7x+6y=346$ -----> equation A

$x=y+29$ -----> equation B

substitute equation B in equation A and solve for y

$7(y+29)+6y=346$

$7y+203+6y=346$

$13y=346-203$

$13y=143$

$y=143/13=11$

5 0

3 years ago