How many solutions to the equation 8x=58?

Write the name of the period that has the digits 486

kogti [31]

1. The name of period that has 486 in the given number which is 751 486 is hundreds.
period is considered to a group of places of the digits, like the given number 486 is the group of hundred because the highest number it contains is 400.
=> 400 + 80 + 6
=> 4 hundreds + 8 tens + 6 ones
=> 486
while 751 is the period of hundred thousands.
=> 700 000 + 50 000 + 1 000
=> 751 000

8 0

3 years ago

Explain the distance formula. Then use it to calculate the

Alex777 [14]

Answer:

$dAB=10\ units$

Step-by-step explanation:

we know that

The distance formula is derived by creating a triangle and using the Pythagorean theorem to find the length of the hypotenuse. The hypotenuse of the triangle is the distance between the two points

The formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

we have

A(1, 1) and B(7, −7)

Let

(x1,y1)=A(1, 1)

(x2,y2)=B(7, −7)

substitute the given values in the formula

$dAB=\sqrt{(-7-1)^{2}+(7-1)^{2}}$

$dAB=\sqrt{(-8)^{2}+(6)^{2}}$

$dAB=\sqrt{64+36}$

$dAB=\sqrt{100}$

$dAB=10\ units$

6 0

2 years ago

Read 2 more answers

Please help me on math

Eduardwww [97]

Answer:

45 inches

Step-by-step explanation:

85-40=45

b=45

how I know?

use the formula c^2-a^2=b^2

85 being c, a being 40 and b being 45

7 0

2 years ago

Read 2 more answers

NO LINKS OR FILES!

Archy [21]

(a) If the particle's position (measured with some unit) at time t is given by s(t), where

$s(t) = \dfrac{5t}{t^2+11}\,\mathrm{units}$

then the velocity at time t, v(t), is given by the derivative of s(t),

$v(t) = \dfrac{\mathrm ds}{\mathrm dt} = \dfrac{5(t^2+11)-5t(2t)}{(t^2+11)^2} = \boxed{\dfrac{-5t^2+55}{(t^2+11)^2}\,\dfrac{\rm units}{\rm s}}$

(b) The velocity after 3 seconds is

$v(3) = \dfrac{-5\cdot3^2+55}{(3^2+11)^2} = \dfrac{1}{40}\dfrac{\rm units}{\rm s} = \boxed{0.025\dfrac{\rm units}{\rm s}}$

(c) The particle is at rest when its velocity is zero:

$\dfrac{-5t^2+55}{(t^2+11)^2} = 0 \implies -5t^2+55 = 0 \implies t^2 = 11 \implies t=\pm\sqrt{11}\,\mathrm s \imples t \approx \boxed{3.317\,\mathrm s}$

(d) The particle is moving in the positive direction when its position is increasing, or equivalently when its velocity is positive:

$\dfrac{-5t^2+55}{(t^2+11)^2} > 0 \implies -5t^2+55>0 \implies -5t^2>-55 \implies t^2 < 11 \implies |t|$

In interval notation, this happens for t in the interval (0, √11) or approximately (0, 3.317) s.

(e) The total distance traveled is given by the definite integral,

$\displaystyle \int_0^8 |v(t)|\,\mathrm dt$

By definition of absolute value, we have

$|v(t)| = \begin{cases}v(t) & \text{if }v(t)\ge0 \\ -v(t) & \text{if }v(t)$

In part (d), we've shown that v(t) > 0 when -√11 < t < √11, so we split up the integral at t = √11 as

$\displaystyle \int_0^8 |v(t)|\,\mathrm dt = \int_0^{\sqrt{11}}v(t)\,\mathrm dt - \int_{\sqrt{11}}^8 v(t)\,\mathrm dt$

and by the fundamental theorem of calculus, since we know v(t) is the derivative of s(t), this reduces to

$s(\sqrt{11})-s(0) - s(8) + s(\sqrt{11)) = 2s(\sqrt{11})-s(0)-s(8) = \dfrac5{\sqrt{11}}-0 - \dfrac8{15} \approx 0.974\,\mathrm{units}$

7 0

2 years ago

The formula Y=X⁴-4X³+3X² is growing?

miskamm [114]

Answer:

The best way to know weather the formula y=x⁴-4x³+3x² is growing or not, is by graphing it.

As you can see in the attached picture:

For -inf<x< 0 the graph decreases.
For 0<x<0.634 the graph is growing
For 0,634<x<2.366 the graph decreases
For 2.366<x<+inf the graph is growing.

Therefore, the polynomial grows in the intervals stated before.

6 0

2 years ago