pls halllllllpppppppp

Answer:

It's option a. 36 + 6x^2 = 28x

Step-by-step explanation:

Answer:

0.85714285714286 x 100 = 85.7143%.

Step-by-step explanation:

Answer:

240

Step-by-step explanation:

! means factorial or to multiply by the previous number down to one so 5!=5*4*3*2*1=120 120*2=240.

Answer:

Step-by-step explanation:

Divide by y

kl/y = f/(p + n) + r/u              Subtract r/u from both sides

kl/y - r/u = f/(p + n)               Multiply both sides by (p + n)

(kl/y - r/u ) (p + n) = f            Divide by (kl/y - r/u)

p + n = f / (kl/y - r/u)             Subtract n from both sides

p  =  f / (kl/y - r/u)  - n

I think I'd leave this as the answer. I don't think you are expected to make it a 2 tier fraction.

Answer:

6+2+9+5+2+3+6+3 = 36

Step-by-step explanation:

If you like my answer than please mark me brainliest

2 digits number = 10-99

Lets find the multiples of 2 and 7 both

Which are = 14, 28,42,56,70,84,98

Therefore there are 7 two digit multiples of both 2 and 7

Must click thanks and mark brainliest

9514 1404 393

Answer:

  7

Step-by-step explanation:

Those numbers are 14, 28, 42, 56, 70, 84, 98. There are 7 numbers that are multiples of 2 and 7 (multiples of 14).

Answer:

x=29

Does the answer help you?

Answer:

x=29

Step-by-step explanation:

Answer:

5 a) Total = 20.83 hrs = 20 hrs and 50 mins  (1250mins total)

5 b) Total = 96 meters. = 0.096km in 8 hrs.

Step-by-step explanation:

1km = 1000 meters

5 mins = 4 meters

1000/4 = 250 multiplier

250 x 5mins = 1250 minutes

1250/60 = 20hrs + 50 minutes

50 / 60 =  0.83 = 20.83hrs

b)  8 hrs = 8 x 60 = 480 minutes

480/5 = 24 multiplier of 4 meters

24 x 4 = 96 meters

Answer: Look it up on internet

Step-by-step explanation: INTERNET

Answer:

Step-by-step explanation:

You've typed the equation incorrectly. 3^5x-1 = 3⁵x-1.

It needs to be typed as 3^(5x-1) = 7.

Answer:

B

Step-by-step explanation:

Since LR=LA, RU=RA. 3m+21=6m, 3m=21, m=7

The table that represents a function in the options given is: Table A. (see image attached below).

Recall:

In the tables given, only the first table has exactly one y-value assigned or related to every x-value given.

Therefore, the table that represents a function in the options given is: Table A. (see image attached below).

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Answer:

Graph A

Step-by-step explanation:

Answer:

y(x)=6^(x)-3

Step-by-step explanation:

Let the exponential function be y(x) = ab^(x) but since the graph is translated 3 units down, y(x) = ab^(x)-3. Now, y(0)=-2=a*b^(0)-3. a=1. The equation is nearly complete but we need b, we can find it by using the point y(1)=3. y(x)=b^(x) - 3. y(1)=3=b-3, b=6. The equation of the function is y(x)=6^(x)-3

Answer:

I agree with the first one

Answer:

Let r = amount of the 7% solution

y represent the amount of the 15 percent solution

r =7.5 L

y = 12.5 L

Step-by-step explanation:

Let r = amount of the 7% solution

y represent the amount of the 15 percent solution

.07r + .15 y = (r+y) .12

r+y = 20

y = 20-r

.07r + .15 (20-r) = (20) .12

0.07r+0.15(20-r)=2.4

.07r+ 3 - .15r = 2.4

-.08r = 2.4-3

-.08r = -.6

Divide by-.08

r =7.5

y = 20-7.5

y = 12.5

Answer:

Slope passes through (-6, 5) and (-2, 7)

[tex]m_1=\frac{y_2-y_1}{x_2-x_1} =m_1=\frac{7-5}{-2+6}[/tex]

[tex]m_1=\frac{2}{4} =\frac{1}{2}[/tex]

Slope passes through (4, 2) and (6, 6)

[tex]m_2=\frac{6-2}{6-4}[/tex]

[tex]\frac{4}{2} =2[/tex]

[tex]m_1\times m_2 \neq -1[/tex] [tex]m_1\neq m_2[/tex]

Answer:- A) neither perpendicular nor parallel.

OAmalOHopeO

Answer:

[tex]y = \sqrt{x - 5} + 7[/tex]

9514 1404 393

Answer:

Step-by-step explanation:

The derivative of P with respect to t is ...

  P' = (x', y') = (4t, 4)

so the slope of the curve for some value of t is ...

  dy/dx = y'/x' = 4/(4t) = 1/t

The line we want to be a tangent is y = x +2, which has a slope of 1. The curve will have a slope of 1 where ...

  1/t = 1   ⇒   t = 1

At t=1, the point P is (2·1², 4·1) = (2, 4). For x=2, the point on the desired tangent is y = x +2 = 2 +2 = 4, or (x, y) = (2, 4).

The curve and the given line both have a slope of 1 at the point (2, 4), so that is the tangent point.

__

The normal to the curve at (2, 4) will have a slope that is the opposite reciprocal of the slope of the tangent: -1/1 = -1. Then point-slope form of the equation of the normal line is ...

  y -k = m(x -h) . . . . . . line with slope m through point (h, k)

  y -4 = -1(x -2) . . . . . . line with slope -1 through point (2, 4)

  y = -x +6 . . . . . . . . equation of the normal line in slope-intercept form

Answer:

Please do it by your shelf because if we measure it and send you may not be able to do by online .So, please do it by by yourself using your scale .

Answer:

Step-by-step explanation:

1 km = 0.621 mi

1 hr = 60 min

(5 km)/(4.5 hr) × (0.621 mi)/km × (1 hr)/(60 min) = (0.0115 mi)/min

9514 1404 393

Answer:

Step-by-step explanation:

1. The negative leading coefficient (-2) tells you the parabola opens downward.

__

2. The fact that the parabola opens downward tells you the vertex is a maximum.

__

3. For quadratic ax^2 +bx +c, the axis of symmetry is x = -b/(2a). For this parabola, that is x = -4/(2(-1)) = 2. The y-value of the vertex is f(2) = -2^2+4(2)-3 = -4+8-3 = 1. The y-intercept is the constant, c = -3.

Answer:

(x-5) so translated 5 units to the right

Multiplied with 2, p vertically compressed

+4 means translated 4 units up

Step-by-step explanation:

28000 ÷ 100

=280

280 × 5

=1400

9514 1404 393

Answer:

  (x/y)^(m+n)

Step-by-step explanation:

  [tex]\displaystyle\frac{\left(x+\frac{1}{y}\right)^m\left(x-\frac{1}{y}\right)^n}{\left(y+\frac{1}{x}\right)^m\left(y-\frac{1}{x}\right)^n}=\left(\frac{x}{y}\right)^m\left(\frac{x}{y}\right)^n\\\\=\boxed{\left(\frac{x}{y}\right)^{m+n}}[/tex]

Answer:

[tex]x=4[/tex]

Step-by-step explanation:

Solving for [tex]x[/tex], we get:

[tex]16=2(3x-6)+x[/tex]

[tex]16=2*3x+2(-6)+x[/tex] (Distributive Property of Multiplication)

[tex]16=6x-12+x[/tex] (Multiply distributed terms)

[tex]16=7x-12[/tex] (Combine like terms)

[tex]16+12=7x-12+12[/tex] (Add [tex]12[/tex] to both sides of the equation to isolate [tex]x[/tex])

[tex]28=7x[/tex] (Combine like terms / Simplify)

[tex]\frac{28}{7}=\frac{7x}{7}[/tex] (Divide both sides of the equation by [tex]7[/tex] to get rid of [tex]x[/tex]'s coefficient)

[tex]\bf x=4[/tex] (Simplify / Symmetric Property of Equality)

Hope this helps!

Step-by-step explanation:

[tex]16 = 2(3x - 6) + x \\ 16 = 6x - 12 + x \\ - 6x - x = - 12 - 16 \\ - 7x = - 12 - 16 \\ 7x = 12 + 16 \\ 7x = 28 \\ x = \frac{28}{7} \\ x = 4[/tex]

Answer:

-7/8

Step-by-step explanation:

-5/8+x=-3/2

x= -3/2+5/8=-12/8+5/8= -7/8

Answer:

D

Step-by-step explanation:

The correct answer is D, try graphing them on desmos.

Answer: Choice B) 50 square units

Work Shown:

A = h*(b1+b2)/2

A = 5*(12+8)/2

A = 5*(20)/2

A = 100/2

A = 50 square units (choice B)

Notice how the b1 and b2 are the parallel bases, while the height h = 5 is perpendicular to both of these bases.

The area of Trapezium ABCD is 50 square unit.

The formula to calculate the area of trapezium is:

Area = ½ x Sum of parallel sides x Distance between the parallel sides

Given:

base 1= 8 , base 2 = 12 unit

height= 5 unit

so, the area of Trapezium

= 1/2 ( base 1 + base 2) /2

= 1/2 (8 + 12) x5

= 1/2 x 20 x 5

= 10 x 5

= 50 square unit.

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Answer:

[tex]\displaystyle \left(\frac{-(m^{2}-1)\, x + 2\, m\, y - 2\, m \, c}{m^{2} + 1},\, \frac{(m^{2} - 1)\, y + 2\, m \, x + 2\, c}{m^{2} + 1}\right)[/tex].

Step-by-step explanation:

Consider the line that is perpendicular to [tex]y = m\, x + c[/tex] and goes through [tex](x,\, y)[/tex].

Both [tex](x,\, y)[/tex] and the reflection would be on this new line. Besides, the two points would be equidistant from the intersection of this new line and line [tex]y = m\, x + c[/tex].

Hence, if the vector between [tex](x,\, y)[/tex] and that intersection could be found, adding twice that vector to [tex](x,\, y)\![/tex] would yield the coordinates of the reflection.

Since this new line is perpendicular to line [tex]y = m\, x + c[/tex], the slope of this new line would be [tex](-1/m)[/tex].

Hence, [tex]\langle 1,\, -1/m\rangle[/tex] would be a direction vector of this new line.

[tex]\langle m,\, -1\rangle[/tex] (a constant multiple of [tex]\langle 1,\, -1/m\rangle[/tex] would also be a direction vector of this new line.)

Both [tex](x,\, y)[/tex] and the aforementioned intersection are on this new line. Hence, their position vectors would differ only by a constant multiple of a direction vector of this new line.

In other words, for some constant [tex]\lambda[/tex], [tex]\langle x,\, y \rangle + \lambda\, \langle m,\, -1 \rangle = \langle x + \lambda \, m,\, y - \lambda \rangle[/tex] would be the position vector of the reflection of [tex](x,\, y)[/tex] (the position vector of [tex](x,\, y)\![/tex] is [tex]\langle x,\, y \rangle[/tex].)

[tex]( x + \lambda \, m,\, y - \lambda )[/tex] would be the coordinates of the intersection between the new line and [tex]y = m\, x + c[/tex]. [tex]\lambda\, \langle m,\, -1 \rangle[/tex] would be the vector between [tex](x,\, y)[/tex] and that intersection.

Since that intersection is on the line [tex]y = m\, x + c[/tex], its coordinates should satisfy:

[tex]y - \lambda = m\, (x + \lambda \, m) + c[/tex].

Solve for [tex]\lambda[/tex]:

[tex]y - \lambda = m\, x + m^{2}\, \lambda + c[/tex].

[tex]\displaystyle \lambda = \frac{y - m\, x - c}{m^{2} + 1}[/tex].

Hence, the vector between the position of [tex](x,\, y)[/tex] and that of the intersection would be:

[tex]\begin{aligned} & \lambda\, \langle m,\, -1 \rangle \\= \; & \left\langle \frac{m\, (y - m\, x - c)}{m^{2} + 1},\, \frac{(-1)\, (y - m\, x - c)}{m^{2} + 1}\right\rangle \\ =\; &\left\langle \frac{-m^{2}\, x + m\, y - m\, c }{m^{2} + 1},\, \frac{-y + m\, x + c}{m^{2} + 1}\right\rangle \end{aligned}[/tex].

Add twice the amount of this vector to position of [tex](x,\, y)[/tex] to find the position of the reflection, [tex]\langle x,\, y \rangle + 2\, \lambda \,\langle m,\, -1 \rangle[/tex].

[tex]x[/tex]-coordinate of the reflection:

[tex]\begin{aligned} & x + 2\, \lambda\, m \\ = \; & x + \frac{-2\, m^{2}\, x + 2\, m \, y - 2\, m \, c}{m^{2} + 1} \\ =\; & \frac{-(m^{2} - 1) \, x + 2\, m \, y - 2\, m \, c}{m^{2} + 1}\end{aligned}[/tex].

[tex]y[/tex]-coordinate of the reflection:

[tex]\begin{aligned} & y + (-2\, \lambda)\\ = \; & y + \frac{- 2\, y + 2\, m\, x + 2\, c}{m^{2} + 1} \\ =\; & \frac{(m^{2} - 1) \, y + 2\, m \, x + 2\, m \, c}{m^{2} + 1}\end{aligned}[/tex].