There are 14 girls and 15 boys in a sixth grade math class. What is the ratio of girls to total students?​

Answer:

yes...

Step-by-step explanation:

k

Answer:

i think it is :

5×5×5=125

2×2×2=8

Answer:

y = 5 x + 7

Step-by-step explanation:

Answer:479001600

Step-by-step explanation:

12 factorial is 12 * 11 * 10 ..... all the way to one

A way to do it by hand is

12 * 1

11 * 2

10 * 3

9 * 4

8 * 5

7 * 6

stop here because 6 * 7 and 5 *8 are already listed

This gives you 6 products then multiply those 6 products by pairs which gives you three pairs which gives you three products then compute a pair then multiply by last product.

Easy math but a lot of written work

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:

[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:

.923

Step-by-step explanation:

1.3 X 0.71 = .923

:))

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Answer:

  (c)  ΔTRS

Step-by-step explanation:

Perhaps the easiest way to find corresponding vertices is to look at the angle markings:

  1 mark: angles J and T

  2 marks: angles K and R

  3 marks: angles I and S

Then, in order of corresponding angles, we have ...

  ΔJKI ≅ ΔTRS

Step-by-step explanation:

sorry for my bad handwriting, hope it helps

Answer:

10.4

Step-by-step explanation:

x=3⇒f(3)= 4×3+3=15

x=3⇒g(3)=8/3+2≈4.6

f(3)-g(3)=15-4.6=10.4

Answer: Yes

Step-by-step explanation: They are similar because it the same kind of triangle just ones smaller and the other is bigger.

Answer:

$20

Step-by-step explanation:

correct me if im wrong(┬┬﹏┬┬)

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]

Answer:

-3 - (2w)

Step-by-step explanation:

hope this helps!

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:

[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:

  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

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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.

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.

Answer:

A. 20

Step-by-step explanation:

1. To solve this problem, instead of doing 80/100 times 16, we can do 100/80 times 16 (inverse of percentage, basically).

Therefore, the answer is A. 20.

-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

Answer:

3:4 itself.

3:4 itself.

Step-by-step explanation:

3:4 itself.

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]

Answer:

:)

Step-by-step explanation:

:)