Find the equation of the line with a slope of -3, passing through the point (2,5).

Whitepunk [10]

Answer:

y = 8

Step-by-step explanation:

$-2(3y - 6) + 4(5y - 8) = 92\\\\\mathrm{Expand\:}-2\left(3y-6\right)+4\left(5y-8\right):\quad 14y-20\\14y-20=92\\\\\mathrm{Add\:}20\mathrm{\:to\:both\:sides}\\14y-20+20=92+20\\\\Simplify\\14y=112\\\\\mathrm{Divide\:both\:sides\:by\:}14\\\frac{14y}{14}=\frac{112}{14}\\\\Simplify\\y= 8$

3 0

2 years ago

Read 2 more answers

9 t-shirts and a hat costs £93 4 t-shirts and a hat costs £43 how much does the hat and t-shirt cost

alexandr402 [8]

Answer:

T-shirt costs £10

Hat costs £3

Step-by-step explanation:

£93-£43=£50

9-4=5

£50/5=10

1 t-shirt=£10

£10x9=£90

£93-£90=£3

T-shirt = £10

Hat= £3

4 0

3 years ago

Read 2 more answers

Identify the interval on which the quadratic function is positive.

Alenkasestr [34]

Answer:

$\textsf{1. \quad Solution: $1 < x < 4$,\quad Interval notation: $(1, 4)$}$

$\textsf{2. \quad Solution: $-2 < x < 4$,\quad Interval notation: $(-2, 4)$}$

Step-by-step explanation:

<h3>Question 1</h3>

The intervals on which a quadratic function is positive are those intervals where the function is above the x-axis, i.e. where y > 0.

The zeros of the quadratic function are the points at which the parabola crosses the x-axis.

As the given quadratic function has a negative leading coefficient, the parabola opens downwards. Therefore, the interval on which y > 0 is between the zeros.

To find the zeros of the given quadratic function, substitute y = 0 and factor:

$\begin{aligned}y&= 0\\\implies -7x^2+35x-28& = 0\\-7(x^2-5x+4)& = 0\\x^2-5x+4& = 0\\x^2-x-4x+4& = 0\\x(x-1)-4(x-1)&= 0\\(x-1)(x-4)& = 0\end{aligned}$

Apply the zero-product property:

$\implies x-1=0 \implies x=1$

$\implies x-4=0 \implies x=4$

Therefore, the interval on which the function is positive is:

Solution: 1 < x < 4
Interval notation: (1, 4)

<h3>Question 2</h3>

The intervals on which a quadratic function is negative are those intervals where the function is below the x-axis, i.e. where y < 0.

The zeros of the quadratic function are the points at which the parabola crosses the x-axis.

As the given quadratic function has a positive leading coefficient, the parabola opens upwards. Therefore, the interval on which y < 0 is between the zeros.

To find the zeros of the given quadratic function, substitute y = 0 and factor:

$\begin{aligned}y&= 0\\\implies 2x^2-4x-16& = 0\\2(x^2-2x-8)& = 0\\x^2-2x-8& = 0\\x^2-4x+2-8& = 0\\x(x-4)+2(x-4)&= 0\\(x+2)(x-4)& = 0\end{aligned}$

Apply the zero-product property:

$\implies x+2=0 \implies x=-2$

$\implies x-4=0 \implies x=4$

Therefore, the interval on which the function is negative is:

Solution: -2 < x < 4
Interval notation: (-2, 4)

3 0

1 year ago

I need help with all of them and if you can't help with all of them just answer one of them please. Look at the picture

likoan [24]

All you have to do is substitute the y value from the 1st equation into the second equation and solve...

a) y= 2-x
5x + 4y = 5

Substitute (2-x) into the second equation anywhere there is a y...

5x + 4y = 5
5x + 4(2-x) = 5

Now solve

5x + 8 - 4x = 5
5x - 4x + 8 = 5
x + 8 = 5
x = -3

Now that you have a solution for x, substitute -3 into either of the original equations anywhere there is an x then solve for y...

y = 2 - x
y = 2 - (-3)
y = 2+3 = 5

You solved for x and got -3 and solved for y and got 5, so your solution set is
(-3, 5).

Now check it by substituting both numbers into one of the original equations and you should have a true statement if it is correct...

y = 2 - x
5 = 2 - (-3)
5 = 2+3
5 = 5

True statement... it checks!

note* during the check, if the equation would have worked out to something like 2 = 5, then that is a false statement therefore the solution set would be wrong and you'd have to go back and find the mistake.

5 0

3 years ago

Lamar has a summer job. He gets paid $7.25 an hour and a $150 bonus at the end of summer. The total Lamar earns can be modeled b

Alex17521 [72]

B.

Slope is the first number and intercept is the second

4 0

3 years ago