M is the midpoint of AB. find the coordinates of B if M= (-1,1) and A= (2,6)

25: 25, 50, 75, 100, 125, 150, 175, 200, 225, 250, 275, 300, 325, 350,…
60: 60, 120, 180, 240, 300, 360, 420, 480, 540, 600, 660, 720, 780, 840,…

LCM(25, 60) = 300

5 0

3 years ago

Figure RHOM is a rhombus. and are the diagonals of the rhombus, as well as angle bisectors of the vertex angles, and they create

Ganezh [65]

Answer:

∠MSR is right angle

Step-by-step explanation:

It is given that RHOM is a Rhombus

Also ΔHOM, ΔMHR, ΔRHO and ΔOMR are isosceles triangles

Let us take ΔMSR and ΔRSH

∠MRS =∠HRS ( since it is given that the diagonal RO bisects ∠R)

∠RMS =∠RHS ( since Δ MRH is isosceles triangle )

RS = RS ( common side )

By AAS congruency rule ΔMSR ≅ ΔHSR

so we have

∠MSR=∠RSH ( corresponding parts of congruent triangles are congruent)

also we have

∠MSR +∠RSH =180° ( supplementary angles)

∠MSR +∠MSR=180° ( since ∠MSR=∠RSH)

2∠MSR= 180°

∠MSR =90°

Hence ∠MSR is right angle

6 0

3 years ago

Read 2 more answers

Given s left parenthesis t right parenthesis equals 5 t squared plus 5 ts(t)=5t2+5t, find

finlep [7]

Answer:

The velocity function is $v(t)=10t+5$ .

The acceleration function is $a(t)=10$ .

When t = 44, the velocity is $v(44)=445 \:\frac{ft}{s}$ .

When t = 44, the acceleration is $a(44)=10\: \frac{ft}{s^2}$ .

Step-by-step explanation:

We know that the position function is given by

$s(t)=5t^2+5t$

Velocity is defined as the rate of change of position or the rate of displacement. If you take the derivative of the position function you get the instantaneous velocity function.

$v(t)=\frac{ds}{dt}$

Acceleration is defined as the rate of change of velocity. If you take the derivative of the instantaneous velocity function you get the instantaneous acceleration function.

$a(t)=\frac{dv}{dt}$

The instantaneous velocity function is given by

$v(t)=\frac{d}{dt} s(t)=\frac{d}{dt}(5t^2+5t)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\v(t)=\frac{d}{dt}\left(5t^2\right)+\frac{d}{dt}\left(5t\right)\\\\\mathrm{Apply\:the\:Power\:Rule}:\quad \frac{d}{dx}\left(x^a\right)=a\cdot x^{a-1}\\\\v(t)=10t+\frac{d}{dt}\left(5t\right)\\\\\mathrm{Apply\:the\:common\:derivative}:\quad \frac{d}{dt}\left(t\right)=1\\\\v(t)=10t+5$

The instantaneous acceleration function is given by

$a(t)=\frac{dv}{dt} =\frac{d}{dt}(10t+5)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\a(t)=\frac{d}{dt}\left(10t\right)+\frac{d}{dt}\left(5\right)\\\\a(t)=10$

To find the velocity and acceleration when t = 44, we substitute this value into the velocity and acceleration functions

$v(44)=10(44)+5\\v(44)=445 \:\frac{ft}{s}$

$a(44)=10\: \frac{ft}{s^2}$

6 0

3 years ago