All categories

3 years ago

Solve this system of equations: y = 7; y = -2x+1 *

Mathematics

2 answers:

k0ka [10]3 years ago

8 0

Answer:

x = -3

Step-by-step explanation:

y = 7

y = -2x+1

substitute for y

7 = -2x+1

subtract 1 from each side

6 = -2x

divide both sides by -2

6/-2 = x

x = -3

Gnoma [55]3 years ago

6 0

Answer:

x = -3

y = 7

Step-by-step explanation:

y = -2x + 1

7 = -2x + 1

-1 -1

6 = -2x

divide by -2

x = -3

You might be interested in

You choose two cards at random. find p(the cards are both hearts)

Tcecarenko [31]

The probability is (13 2)/(52 2)

7 0

3 years ago

Please explain, 2 (4m-5) - (6m-11) =33

zlopas [31]

Answer:

m = 16

Step-by-step explanation:

2(4m - 5) - (6m - 11) = 33 (Given)

8m - 10 - 6m + 11 = 33 (Distribute)

2m + 1 = 33 (Combine like terms)

2m = 32 (Subtract 1 on both sides)

m = 16 (Divide 2 on both sides)

You can check by substituting your answer into the equation:

2(4(16) - 5) - (6(16) - 11) = 33

2(64 - 5) - (96 - 11) = 33

2(59) - (85) = 33

118 - 85 = 33

33 = 33

7 0

3 years ago

Which process in the water cycle that is not driven by the force of gravity? *

Deffense [45]

Answer : The answer is option c evaporation

8 0

3 years ago

Read 2 more answers

Prove the identity: (a-b)(a+b)(a^2+b^2)(a^4+b^4) Please help quickly for brainliest

Verdich [7]

Answer: a^8-b^8

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Y=x.arctan(x)^1/2. find dy/dx. pls show steps

Whitepunk [10]

$y=x(\arctan x)^{1/2}$

Use the product rule first:

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dx}{\mathrm dx}(\arctan x)^{1/2}+x\dfrac{\mathrm d(\arctan x)^{1/2}}{\mathrm dx}$

$\dfrac{\mathrm dy}{\mathrm dx}=(\arctan x)^{1/2}+x\dfrac{\mathrm d(\arctan x)^{1/2}}{\mathrm dx}$

Use the chain rule to compute the derivative of $(\arctan x)^{1/2}$ . Let $z=(\arctan x)^{1/2}$ and take $u=\arctan x$ , so that by the chain rule

$\dfrac{\mathrm dz}{\mathrm dx}=\dfrac{\mathrm dz}{\mathrm du}\dfrac{\mathrm du}{\mathrm dx}$

$\dfrac{\mathrm dz}{\mathrm du}=\dfrac{\mathrm du^{1/2}}{\mathrm du}=\dfrac12u^{-1/2}$

$\dfrac{\mathrm du}{\mathrm dx}=\dfrac{\mathrm d\arctan x}{\mathrm dx}=\dfrac1{1+x^2}$

$\implies\dfrac{\mathrm d(\arctan x)^{1/2}}{\mathrm dx}=\dfrac1{2(\arctan x)^{1/2}(1+x^2)}$

So we have

$\dfrac{\mathrm dy}{\mathrm dx}=(\arctan x)^{1/2}+\dfrac x{2(\arctan x)^{1/2}(1+x^2)}$

You can rewrite this a bit by factoring $(\arctan x)^{-1/2}$ , just to make it look neater:

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac1{2(\arctan x)^{1/2}}\left(2\arctan x+\dfrac x{1+x^2}\right)$

3 0

3 years ago