I need help on my algebra test. I’m reaching out because I’m really stuck and have no teachers available. Can you help me solve for y?

Answer:

A

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

y = [tex]\frac{1}{4}[/tex] x - [tex]\frac{2}{3}[/tex] ← is in slope- intercept form

with y- intercept c = - [tex]\frac{2}{3}[/tex] → A

Answer:

-3 - (2w)

Step-by-step explanation:

hope this helps!

Answer:

Hope It's Help

Answer: He sold 1,800 merchandise

Step-by-step explanation: 426 - 300 = 126 ÷0.07 = 1,800

I dont know if im right though

Answer:

Step-by-step explanation:

6 times 2 is 12

42 multiplied by 2 is 84

therefore the answer is 12 days.

Answer:

  c.  x + y = 0

Step-by-step explanation:

The terminal ray of the angle θ = 3π/4 extends to the upper left through the 2nd quadrant with a slope of -1. It lies on the line ...

  y = -x

  x + y = 0 . . . . . . add x to get this standard form equation

Answer:

yes...

Step-by-step explanation:

k

Answer:

[tex]\frac{1}{4}[/tex]

Step-by-step explanation:

The problem [tex]2x - 3x = 3[/tex] is incorrect. The correct answer for that one is [tex]-x \:\sf{or\:-1[/tex]

[tex]\sf{Simplify\: the\: expression.[/tex]

[tex]\sf{\frac{16x}{64x} = \frac{1}{4}[/tex]

[tex]x=\frac{1}{4}[/tex]

Therefore, the answer is [tex]\frac{1}{4}[/tex]

Answer:

(x, y)

(-1, -4)

Step-by-step explanation:

[tex]-20x=-16-9y\\-14x=-22-9y[/tex]

Solve by elimination, subtract the 2nd row from the 1st row:

[tex](-20x-(-14x))=(-16-(-22))+(-9y-(-9y))\\(-20x+14x)=(-16+22)+(-9y+9y)\\-6x=6[/tex]

Solve for x in the equation above

[tex]-6x=6\\x=-1[/tex]

Now we have the value of X, substitute it into any of the 2 original equations:

[tex]-20x=-16-9y\\-20(-1)=-16-9y[/tex]

Solve for y:

[tex]-20(-1)=-16-9y\\20=-16-9y\\36=-9y\\-4=y\\y=-4[/tex]

Please mark brainliest if this answer helps you. I can provide more information if needed.

Answer:

.923

Step-by-step explanation:

1.3 X 0.71 = .923

:))

Answer:

:)

Step-by-step explanation:

:)

Answer:

Neither.

Step-by-step explanation:

Manipulate the formulas into slope-intercept form which is [tex]y=mx+b[/tex].

In order for them to be parallel, m needs to be the same. Perpendicular, m for one needs to be the negative reciprocal of the other [tex]m=-\frac{1}{m}[/tex].

Let's see what we get.

(1)

[tex]3x+4y=2\\4y=-3x+2\\y=-\frac{3}{4}x+1/2[/tex]

(2)

[tex]4x+3y=2\\3y=-4x+2\\y=-\frac{4}{3}x+2/3[/tex]

They are reciprocals, but they are both negative. So, they are not the same line, perpendicular, nor parallel.

Also, you can graph these equations on desmos and see easily that they have none of these relationships.

9514 1404 393

Answer:

  (c) 5y - x = -20

Step-by-step explanation:

When the x- and y-terms are on the same side of the equation, the slope of the line will be ...

  m = -(coefficient of x)/(coefficient of y)

So, the slopes of the offered choices are ...

  a) -5/1 = -5

  b) -1/5

 c) -(-1)/5 = 1/5 . . . . . parallel to given line

  d) -5/-1 = 5

The lines will be parallel when their slopes are the same. The slope of the given line is the coefficient of x, 1/5. Choice C has the same slope.

-2 1/2 - (-1 3/4)

When subtracting a negative vile the equation becomes addition:

-2 1/2 + 1 3/4 = -3/4

Answer: -3/4

Step-by-step explanation:

sorry for my bad handwriting, hope it helps

Answer:

$20

Step-by-step explanation:

correct me if im wrong(┬┬﹏┬┬)

Answer:

The average rate of change between x = 1 to x = 2 is 2, from x = 2 to x = 3 is 4 and from x = 3 to x = 4 is 8.

Step-by-step explanation:

9514 1404 393

Answer:

  (x, y) = (1.25, 8.5)

Step-by-step explanation:

I choose to put these equations into reduced general form, then use the "cross multiplication method" to solve.

  2x -y +6 = 0

  10x -y -4 = 0

d1 = 2(-1) -10(-1) = 8

d2 = -1(-4) -(-1)(6) = 10

d3 = 6(10) -(-4)(2) = 68

Solutions are ...

  x = d2/d1 = 10/8 = 1.25

  y = d3/d1 = 68/8 = 8.5

The solution is (x, y) = (1.25, 8.5).

_____

Additional comment

I like a graphing calculator for a quick and easy solution. When the values are not integers, I like an algebraic solution to see what the exact values are.

__

The "cross multiplication method" is similar to Cramer's rule in that determinants of pairs of coefficients are computed. If you look up this method, videos will show you a variation that is slightly different. Here, we determined d1, d2, and d3 as the 2×2 determinant of the coefficients in adjacent columns of the array you get from the general form equations with the first column repeated:

  [tex]\begin{array}{cccc}2&-1&6&2\\10&-1&-4&10\end{array}[/tex]

In each pair, the determinant is the difference between the product of numbers on the down-diagonal and the product of numbers on the up-diagonal.

The determinants relate to the x- and y-values by ...

  1/d1 = x/d2 = y/d3   ⇒   x = d2/d1, y = d3/d1

There is a Vedic solution method for linear equations that is similar to this, but starts from equations in standard form. As a consequence, the cross-differences are done in a different order.

__

Standard form: ax +by = c

General form: ax +by +c = 0

Answer:

y = 5/3x - 3 and y = 1/2x - 10

Step-by-step explanation:

A. -5x + 3y = -9

In order to go from standard to slope-intercept form, we want to isolate y on the left side.  We start by adding 5x to both sides:

3y = -9 + 5x

Next, we divide both sides by 3:

y = (-9 + 5x)/3

And then distribute:

y = -9/3 + 5x/3

And finally simplify:

[tex]y = -3 + \frac{5}{3}x[/tex]

And move the numbers on the right to fit the form y = mx + b:

[tex]y = \frac{5}{3}x-3[/tex]

B. -x + 2y = -20

In order to go from standard to slope-intercept form, we want to isolate y on the left side.  We start by adding x to both sides:

2y = -20 + x

Next, we divide both sides by 2:

y = (-20 + x)/2

And then distribute:

y = -20/2 + x/2

And finally simplify:

[tex]y = -10 + \frac{1}{2}x[/tex]

And move the numbers on the right to fit the form y = mx + b:

[tex]y = \frac{1}{2}x - 10[/tex]

a. By the chain rule,

[tex]\dfrac{dy}{dx} = \dfrac{dy}{dt}\cdot\dfrac{dt}{dx} = \dfrac{dy}{dt}\cdot\dfrac{1}{\frac{dx}{dt}}[/tex]

Given that [tex]y=e^t+e^{-t}[/tex] and [tex]x=e^{-t}[/tex], we have the derivatives

[tex]\dfrac{dy}{dt} = e^t -e^{-t}[/tex]

[tex]\dfrac{dx}{dt} = -e^{-t}[/tex]

and so

[tex]\dfrac{dy}{dx} = \dfrac{e^t - e^{-t}}{-e^{-t}} = 1-e^{2t}[/tex]

Set this equal to zero and solve for t :

[tex]1-e^{2t} = 0[/tex]

[tex]1 = e^{2t}[/tex]

[tex]\ln(1) = \ln\left(e^{2t}\right)[/tex]

[tex]0 = 2t \ln(e)[/tex]

[tex]0 = 2t[/tex]

[tex]t=0[/tex]

This value of t corresponds to x = e⁰ = 1 and y = e⁰ - 1/e⁰ = 1 - 1 = 0. So the only point on the curve where the derivative dy/dx is zero is (1, 0).

b. Compute the second derivative. Since dy/dx is a function of t, we'll momentarily replace it with f(t). By the chain rule,

[tex]\dfrac{d^2}{dx^2} = \dfrac d{dx} \dfrac{dy}{dx} = \dfrac{df}{dx} = \dfrac{df}{dt}\cdot\dfrac{dt}{dx} = \dfrac{df}{dt}\cdot\dfrac{1}{\frac{dx}{dt}}[/tex]

We have

[tex]\dfrac{df}{dt} = -2e^{2t}[/tex]

and we've already committed dx/dt. So

[tex]\dfrac{d^2}{dx^2} = \dfrac{-2e^{2t}}{-e^{-t}} = 2e^{3t} [/tex]

Substitute the first and second derivative into the differential equation:

[tex]\left(\dfrac{d^2y}{dx^2}\right)^2 + \dfrac{dy}{dx} - 1 = 0[/tex]

[tex]\left(2e^{3t}\right)^2 + (1-e^{2t}) - 1 = 0[/tex]

[tex]4e^{6t} -e^{2t}= 0[/tex]

[tex]e^{2t} (4e^{4t} - 1) = 0[/tex]

[tex]4e^{4t} - 1 = 0[/tex]

[tex]4e^{4t} = 1[/tex]

[tex]e^{4t} = \dfrac14[/tex]

[tex]\ln\left(e^{4t}\right) = \ln\left( \dfrac14\right)[/tex]

[tex]4t \ln(e) = -\ln(4)[/tex]

[tex]4t = -\ln(4)[/tex]

[tex]t = -\dfrac{\ln(4)}4[/tex]

Step-by-step explanation:

All angles of the ∆ are same. So it's a equilateral ∆ . So all sides will also be equal .

16 + y = 46

y = 46 -16 = 30

And ,

60 + x + (180-80) = 180

-20 +x = 0

x = 0 +20

x = 20°

Answer: 12 feet up

Step-by-step explanation:

This problem models a right triangle, in which the ladder is the hypotenuse and the distance away from the wall is one of the legs.  Thus, you may use the Pythagorean Theorem (a^2 + b^2 = c^2)

[tex]a^2+b^2=c^2\\5^2+b^2=13^2\\25+b^2=169\\b^2=144\\b=12[/tex]

N.B. There are some very useful sets of numbers that can help you with this unit.

They are called Pythagorean triplets, and they are the sets of whole numbers that make right triangles.  The important ones are 3, 4, 5, and 5, 12, 13 (And their multiples, like 6, 8, 10(3,4,5 *2))

Hope it helps :) and let me know if you want me to elaborate further.

Answer:

[tex]-\frac{x(15x^4-2x^3+3)}{3}[/tex]

Step-by-step explanation:

[tex]\frac{30x^8-4x^7+6x^4}{-6x^3}[/tex]

The first thing I would do is to factor the numerator. All coefficients there are divisible by 2, and all variables are divisible by x⁴. I'll also move the negative sign to the left.

[tex]-\frac{2x^4(15x^4-2x^3+3)}{6x^3}[/tex]

Next, you can simplify 2/6 to 1/3:

[tex]-\frac{x^4(15x^4-2x^3+3)}{3x^3}[/tex]

Finally, the last thing you can do is simplify x⁴/x³ to x/0:

[tex]-\frac{x(15x^4-2x^3+3)}{3}[/tex]

That's as simplified as it gets.

Answer:

100º

Step-by-step explanation:

Cooresponding angles are congruent

11x + 1 = 10x + 10

Subtract 10x from both sides

x + 1 = 10

subtract 1 from both sides

x = 9

θ = 11x + 1

θ = 11(9) + 1

θ = 99 + 1

θ = 100º

Answer:

Ms. April will not have enough students.

Step-by-step explanation:

This is because 180 days of the year divided by 3 students per day = 60 students per year. The class only has 20 students, therefore, there is not enough students needed to collect milk.

Answer:

Ms. April will not have enough students.

Step-by-step explanation:

This is because 180 days of the year divided by 3 students per day = 60 students per year. The class only has 20 students, therefore, there is not enough students needed to collect milk.

As the hint suggests, convert to spherical coordinates using

x = p cos(u) sin(v)

y = p sin(u) sin(v)

z = p cos(v)

dV = dx dy dz = p² sin(v) dp du dv

Then U is the set

[tex]U = \left\{ (p,u,v) \mid 0\le p\le2 \text{ and } 0\le u\le \dfrac{\pi}2 \text{ and } 0\le v\le\dfrac{\pi}2\right\}[/tex]

and the integral of y over U is

[tex]\displaystyle \iiint_U y \, dV = \iiint_U p\sin(u)\sin(v) \cdot p^2 \sin(v) \, dV[/tex]

[tex]\displaystyle \iiint_U y \, dV = \int_0^{\frac\pi2} \int_0^{\frac\pi2} \int_0^2 p^3 \sin(u) \sin^2(v) \, dp \, du \, dv [/tex]

[tex] \displaystyle \iiint_U y \, dV = \frac{2^4-0^4}4 \int_0^{\frac\pi2} \int_0^{\frac\pi2} \sin(u) \sin^2(v) \, du \, dv [/tex]

[tex]\displaystyle \iiint_U y \, dV = 4 \cdot \left(-\cos\left(\frac\pi2\right) + \cos(0)\right) \int_0^{\frac\pi2} \sin^2(v) \, dv [/tex]

[tex]\displaystyle \iiint_U y \, dV = 4 \cdot \frac12 \int_0^{\frac\pi2} (1-\cos(2v)) \, dv [/tex]

[tex]\displaystyle \iiint_U y \, dV = 2 \left(\left(\frac\pi2 - \frac12 \sin\left(2\cdot\frac\pi2\right)\right) - \left(0 - \frac12 \sin\left(2\cdot0\right)\right) \right)[/tex]

[tex]\displaystyle \iiint_U y \, dV = \pi - \sin(\pi) = \boxed{\pi}[/tex]