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GarryVolchara [31]
3 years ago
10

What is the simplified expression for 3 y squared minus 6 y z minus 7 + 4 y squared minus 4 y z + 2 minus y squared z? y squared

minus y squared z minus 10 y z minus 5 y squared minus y squared z minus 5 y z minus 5 7 y squared minus y squared z minus 10 y z minus 5 7 y squared minus y squared z minus 2 y z minus 5
Mathematics
1 answer:
tresset_1 [31]3 years ago
4 0

Answer:

7y^2 - y^2z - 10yz-5

C. 7 y squared minus y squared z minus 10 y z minus 5

Step-by-step explanation:

A. y squared minus y squared z minus 10 y z minus 5

B. y squared minus y squared z minus 5 y z minus 5

C. 7 y squared minus y squared z minus 10 y z minus 5

D. 7 y squared minus y squared z minus 2 y z minus 5

Given:

3y^2 - 6yz - 7 + 4y^2 - 4yz + 2 - y^2z

Collect like terms in the expression

3y^2 + 4y^2 - 6yz - 4yz - y^2z - 7 + 2

Simplify

7y^2 - 10yz - y^2z - 5

Rewritten as

7y^2 - y^2z - 10yz-5

Option C. 7 y squared minus y squared z minus 10 y z minus 5

Is the answer

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3 years ago
1) Given: circle k(O), ED= diameter ,m∠OEF=32°, m(arc)EF=(2x+10)° Find: x
qaws [65]

1. The major arc ED has measure 180 degrees since ED is a diameter of the circle. The measure of arc EF is (2x+10)^\circ, so the measure of arc DF is

m\widehat{DF}=360^\circ-180^\circ-(2x+10)^\circ=(170-2x)^\circ

The inscribed angle theorem tells us that the central angle subtended by arc DF, \angle DOF, has a measure of twice the measure of the inscribed angle DEF (which is the same angle OEF) so

m\angle DOF=2m\angle OEF=64^\circ

so the measure of arc DF is also 64 degrees. So we have

170-2x=64\implies106=2x\implies\boxed{x=53}

###

2. Arc FE and angle EOF have the same measure, 56 degrees. By the inscribed angle theorem,

m\angle EOF=2m\angle EDF\implies56^\circ=2m\angle EDF\implies m\angle EDF=28^\circ

Triangle DEF is isosceles because FD and ED have the same length, so angles EFD and DEF are congruent. Also, the sum of the interior angles of any triangle is 180 degrees. It follows that

m\angle EFD+m\angle EDF+m\angle DEF=180^\circ\implies\boxed{m\angle EFD=76^\circ}

Triangle OFE is also isosceles, so angles EFO and FEO are congruent. So we have

m\angle EFO+m\angle FEO+m\angle EOF=180^\circ\implies\boxed{m\angle EFO=62^\circ}

7 0
2 years ago
A rock thrown vertically upward from the surface of the moon at a velocity of 36​m/sec reaches a height of s = 36t - 0.8 t^2 met
Verdich [7]

Answer:

a. The rock's velocity is v(t)=36-1.6t \:{(m/s)}  and the acceleration is a(t)=-1.6  \:{(m/s^2)}

b. It takes 22.5 seconds to reach the highest point.

c. The rock goes up to 405 m.

d. It reach half its maximum height when time is 6.59 s or 38.41 s.

e. The rock is aloft for 45 seconds.

Step-by-step explanation:

  • Velocity is defined as the rate of change of position or the rate of displacement. v(t)=\frac{ds}{dt}
  • Acceleration is defined as the rate of change of velocity. a(t)=\frac{dv}{dt}

a.

The rock's velocity is the derivative of the height function s(t) = 36t - 0.8 t^2

v(t)=\frac{d}{dt}(36t - 0.8 t^2) \\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\v(t)=\frac{d}{dt}\left(36t\right)-\frac{d}{dt}\left(0.8t^2\right)\\\\v(t)=36-1.6t

The rock's acceleration is the derivative of the velocity function v(t)=36-1.6t

a(t)=\frac{d}{dt}(36-1.6t)\\\\a(t)=-1.6

b. The rock will reach its highest point when the velocity becomes zero.

v(t)=36-1.6t=0\\36\cdot \:10-1.6t\cdot \:10=0\cdot \:10\\360-16t=0\\360-16t-360=0-360\\-16t=-360\\t=\frac{45}{2}=22.5

It takes 22.5 seconds to reach the highest point.

c. The rock reach its highest point when t = 22.5 s

Thus

s(22.5) = 36(22.5) - 0.8 (22.5)^2\\s(22.5) =405

So the rock goes up to 405 m.

d. The maximum height is 405 m. So the half of its maximum height = \frac{405}{2} =202.5 \:m

To find the time it reach half its maximum height, we need to solve

36t - 0.8 t^2=202.5\\36t\cdot \:10-0.8t^2\cdot \:10=202.5\cdot \:10\\360t-8t^2=2025\\360t-8t^2-2025=2025-2025\\-8t^2+360t-2025=0

For a quadratic equation of the form ax^2+bx+c=0 the solutions are

x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

\mathrm{For\:}\quad a=-8,\:b=360,\:c=-2025:\\\\t=\frac{-360+\sqrt{360^2-4\left(-8\right)\left(-2025\right)}}{2\left(-8\right)}=\frac{45\left(2-\sqrt{2}\right)}{4}\approx 6.59\\\\t=\frac{-360-\sqrt{360^2-4\left(-8\right)\left(-2025\right)}}{2\left(-8\right)}=\frac{45\left(2+\sqrt{2}\right)}{4}\approx 38.41

It reach half its maximum height when time is 6.59 s or 38.41 s.

e. It is aloft until s(t) = 0 again

36t - 0.8 t^2=0\\\\\mathrm{Factor\:}36t-0.8t^2\rightarrow -t\left(0.8t-36\right)\\\\\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}\\\\t=0,\:t=45

The rock is aloft for 45 seconds.

5 0
3 years ago
Math help please and thanks
Artyom0805 [142]

Answer:

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Step-by-step explanation:

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