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3 years ago

Maths question PLEASE ASAP HELP PLEASE

Mathematics

2 answers:

beks73 [17]3 years ago

8 0

The <em><u>correct answers</u></em> are:

#1) no solution; #2) (3, 11), because both lines pass through this point ; #3) There is no graph to use to answer the question; #4) There is no graph to use to answer the question; #5) Line y = -2x + 3 intersects the line y = -x + 1; #6) There is no graph to use to answer the question; #7) There is no graph to use to answer the question; #8) 4h + 1 = 3h + 5 ; #9) No solution; #10) (-6, -4)

Explanation:

#1) For the equations x+y=5 and x+y=4, if we were to eliminate a variable, we would wind up eliminating both variables and have unequal constants. Thus there is no solution.

#2) To solve the system y=x+8 and y=3x+2, we can substitute for y in the first equation:

3x+2 = x+8

Subtract x from each side:

3x+2-x = x+8-x

2x+2 = 8

Subtract 2 from each side:

2x+2-2 = 8-2

2x = 6

Divide both sides by 2:

2x/2 = 6/2

x = 3

Substitute this into the first equation:

y=x+8

y=3+8

y=11

This means that both lines pass through (3, 11) and this is the solution to the system.

#3) No graph to use to answer the question.

#4) No graph to use to answer the question.

#5) The slopes of the lines y = -2x+3 and y = -x+1 are different (-2 and -1). This means that the lines will intersect.

#6) No graph to use to answer the question.

#7) No graph to use to answer the question.

#8) To solve this, substitute the value of g in the first equation into g in the second equation:

4h+1 = 3h+5

#9) The slopes of these equations are the same, and the y-intercepts are different; this means that the lines are parallel and never intersect, so there is no solution.

#10) To solve this system, substitute the value of a from the second equation into the first equation:

a - 4b = 10

a = b-2

b - 2 - 4b = 10

Combine like terms:

-3b - 2 = 10

Add 2 to each side:

-3b - 2 + 2 = 10 + 2

-3b = 12

Divide both sides by -3:

-3b/-3 = 12/-3

b = -4

Substitute this value into the second equation:

a = b - 2

a = -4 - 2

a = -6

This gives us the solution (-6, -4)

Sphinxa [80]3 years ago

8 0

1.) No solution.

x + y = 5

x + y = 4

Solve by elimination. Make one equation negative to solve for x.

-x - y = 5

x + y = 4

*Can't have both x's and y's cancel at the same time*

2.) (3,11) - both pass through point.

y = x + 8

y = 3x + 2

Solve by elimination. Solve for y.

-3y = -3x - 24

y = 3x + 2

-2y = -22

-y = -11

y = 11

Plug in y to one of the original equations.

11 = x + 8

3 = x

Solution: (3,11)

4.) Not enough information provided for me to solve problem 4. Feel free to send it to me!

5.) Line y = -2x + 3 intersects line y = -x + 1

6.) Not enough information provided for me to solve.

7.) Not enough information provided for me to solve.

8.) 4h+1 = 3h+5

9.) No solution. Same setup as #1

10.) (-6,-4)

a - 4b = 10

a = b - 2

Solve by substituting.

b - 2 - 4b = 10

-3b - 2 = 10

-3b = 12

b = -4

Plug in b to solve for a.

a = -4 - 2

a = -6

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Svet_ta [14]

Answer:

$\dfrac{dh}{dt}=21 \text{m/min}$

the rate of change of height when the water is 1 meter deep is 21 m/min

Step-by-step explanation:

First we need to find the volume of the trough given its dimensions and shape: (it has a prism shape so we can directly use that formula OR we can multiply the area of its triangular face with the length of the trough)

$V = \dfrac{1}{2}(bh)\times L$

here L is a constant since that won't change as the water is being filled in the trough, however 'b' and 'h' will be changing. The equation has two independent variables and we need to convert this equation so it is only dependent on 'h' (the height of the water).

As its an isosceles triangle we can find a relationship between b and h. the ratio between the b and h will be always be the same:

$\dfrac{b}{h} = \dfrac{5}{7}$

$b=\dfrac{5}{7}h$ this can be substituted back in the volume equation

$V = \dfrac{5}{14}h^2L$

the rate of the water flowing in is:

$\dfrac{dV}{dt} = 6$

The question is asking for the rate of change of height (m/min) hence that can be denoted as: $\frac{dh}{dt}$

Using the chainrule:

$\dfrac{dh}{dt}=\dfrac{dh}{dV}\times \dfrac{dV}{dt}$

the only thing missing in this equation is dh/dV which can be easily obtained by differentiating the volume equation with respect to h

$V = \dfrac{5}{14}h^2L$

$\dfrac{dV}{dh} = \dfrac{5}{7}hL$

reciprocating

$\dfrac{dh}{dV} = \dfrac{7}{5hL}$

plugging everything in the chain rule equation:

$\dfrac{dh}{dt}=\dfrac{dh}{dV}\times \dfrac{dV}{dt}$

$\dfrac{dh}{dt}=\dfrac{7}{5hL}\times 6$

$\dfrac{dh}{dt}=\dfrac{42}{5hL}$

L = 12, and h = 1 (when the water is 1m deep)

$\dfrac{dh}{dt}=\dfrac{42}{5(1)(12)}$

$\dfrac{dh}{dt}=21 \text{m/min}$

the rate of change of height when the water is 1 meter deep is 21 m/min

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Read 2 more answers

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sergeinik [125]

Answer:

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Step-by-step explanation:

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Vesnalui [34]

Answer:

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Anvisha [2.4K]

Answer:

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Step-by-step explanation:

Given

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Required

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This goes thus;

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