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tatuchka [14]
3 years ago
13

Choose Yes or No to tell whether each value of p is a solution for this inequality.

Mathematics
2 answers:
Scilla [17]3 years ago
7 0

Answer:

No

No

Yes

Yes

Step-by-step explanation:

10(0)-8 = -8   -8 >= 12

10(1)-8 = 2   2 >= 12

10(2)-8 = 12   12 >= 12

Temka [501]3 years ago
5 0

Answer:

no,no,yes,yes

Step-by-step explanation:

10(0)-8=-8

10(1)-8=2

10(2)-8=12

10(3)-8=22

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Transpose then formula <br><br> F = (9/5)C + 32<br><br> So that C is the subject of the equation.
Phoenix [80]

Answer:

C = F - 32 /(9/5)

Step-by-step explanation:

F = (9/5)C + 32

Subtract 32 from both sides

F - 32 = (9/5)C + 32 - 32

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Divide both sides by (9/5)

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3 0
4 years ago
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Consider the planes 4x+3y+4z=1 and 4x+4z=0. (A) Find the unique point P on the y−axis which is on both planes. ( 0 equation edit
Nana76 [90]

Answer:

Step-by-step explanation:

A) From the order of the exercise we already know that the intersection points lies on the Y-axis, so its coordinates are P(0;y;0). In order to find it, we only need to substitute the equation 4x+4z=0 into the equation 4x+3y+4z=1. Then,

1=4x+3y+4z = 3y + (4x+4z)= 3y+0.

From the expression above it is easy to obtain that y=1/3, and the intersection point is P(0;1/3;0).

B) To obtain the parallel vector to both planes we use the cross product of the normal vector of the planes.

\left[\begin{array}{ccc}i&j&k\\4&3&4\\4&0&4\end{array}\right] = 12i-0j+12k

As we want a unit vector, we must calculate the modulus of u:

|u|=\sqrt{12^2+0^2+12^2} = \sqrt{2\cdot 12^2}=12\sqrt{2}.

Thus, the wanted vector is \frac{u}{12\sqrt{2}}. Therefore,

u = \frac{1}{\sqrt{2}}i-\frac{1}{\sqrt{2}}k = \frac{\sqrt{2}}{2}i-\frac{\sqrt{2}}{2}k.

C) In order to obtain the vector equation of the intersection line of both planes, we just need to put together the above results.

r = \frac{1}{3}j +\lambda \left( \frac{\sqrt{2}}{2}i- \frac{\sqrt{2}}{2}k \right)

where \lambda is a real number.

8 0
4 years ago
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