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Oliga [24]
3 years ago
10

What is the base....

Mathematics
1 answer:
olga2289 [7]3 years ago
3 0

Answer:

Step-by-step explanation:

-5x the base

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Consider f and c below. f(x, y, z) = (y2z + 2xz2)i + 2xyzj + (xy2 + 2x2z)k,
qaws [65]
\dfrac{\partial f}{\partial x}=y^2z+2xz^2
f(x,y,z)=xy^2z+x^2z^2+g(y,z)

\dfrac{\partial f}{\partial y}=2xyz+\dfrac{\partial g}{\partial y}=2xyz
\dfrac{\partial g}{\partial y}=0\implies g(y,z)=h(z)

\dfrac{\partial f}{\partial z}=xy^2+2x^2z+\dfrac{\mathrm dh}{\mathrm dz}=xy^2+2x^2z
\dfrac{\mathrm dh}{\mathrm dz}=0\implies h(z)=0

\implies f(x,y,z)=xy^2z+x^2z^2+C
8 0
3 years ago
42:28
gogolik [260]

Answer:

The statements about arcs and angles that are true in the figure are;

1) ∠EFD ≅ ∠EGD

2) \overline{ED}\cong \overline{FD}

3) mFD = 120°

Step-by-step explanation:

1) ∠ECD + ∠CEG + ∠CDG + ∠GDE = 360° (Sum of interior angle of a quadrilateral)

∠CEG = ∠CDG = 90° (Given)

∠GDE = 60° (Given)

∴ ∠ECD = 360° - (∠CEG + ∠CDG + ∠GDE)

∠ECD = 360° - (90° + 90° + 60°) = 120°

∠ECD = 2 × ∠EFD (Angle subtended is twice the angle subtended at the circumference)

120° = 2 × ∠EFD

∠EFD = 120°/2 = 60°

∠EFD ≅ ∠EGD

∠ECD = 120°

∠EGD = 60°

∴∠EGD ≠ ∠ECD

2) Given that arc mEF ≅ arc mFD

Therefore, ΔECF and ΔDCF are isosceles triangles having two sides (radii EC and CF in ΔECF and radii EF and CD in ΔDCF

∠ECF = mEF = mFD = ∠DCF (Given)

∴ ΔECF ≅ ΔDCF (Side Angle Side, SAS, rule of congruency)

\\ \overline{EF}\cong \overline{FD} (Corresponding Parts of Congruent Triangles are Congruent, CPCTC)

∠FED ≅ ∠FDE (base angles of isosceles triangle)

∠FED + ∠FDE + ∠EFD = 180° (sum of interior angles of a triangle)

∠FED + ∠FDE = 180° - ∠EFD = 180° - 60° = 120°

∠FED + ∠FDE = 120° = ∠FED + ∠FED (substitution)

2 × ∠FED  = 120°

∠FED = 120°/2 = 60° = ∠FDE

∴ ∠FED = ∠FDE = ∠EFD =  60°

ΔEFD  is an equilateral triangle as all interior angles are equal

\\ \overline{EF}\cong \overline{FD}\cong \overline{ED} (definition of equilateral triangle)

\overline{ED}\cong \overline{FD}

3) Having that ∠EFD = 60° and ∠CFE = ∠CFD (CPCTC)

Where, ∠EFD = ∠CFE + ∠CFD (Angle addition)

60° = ∠CFE + ∠CFD = ∠CFE + ∠CFE (substitution)

60° = 2 × ∠CFE

∠CFE =60°/2 = 30° = ∠CFD

\overline{CF}\cong \overline{CD} (radii of the same circle)

ΔFCD is an isosceles triangle (definition)

∠CFD ≅ ∠CDF (base angles of isosceles ΔFCD)

∠CFD + ∠CDF + ∠DCF = 180°

∠DCF = 180° - (∠CFD + ∠CDF) = 180° - (30° + 30°) = 120°

mFD = ∠DCF (definition)

mFD = 120°.

5 0
3 years ago
the temperature at noon was -3 degrees C by 10 p.m. on the same day the temperature decrease by 5.4 degrees what was the tempera
Anvisha [2.4K]

The answer is -8.4 degrees

5 0
3 years ago
Read 2 more answers
In the diagram, parallel lines P and Q are cut by transversal R.
Paraphin [41]

Answer:

x=125

Step-by-step explanation:

7 0
3 years ago
Find the distance between the points (2, 7) and (-6, -2).
maw [93]

Answer:

Approximately 12.04 units.

Step-by-step explanation:

The find the distance between any two points, we can use the distance formula, which is:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2

We have the points (2,7) and (-6,-2). Let's let (2,7) be (x₁, y₁) and let's let (-6, -2) be (x₂, y₂). Substitute:

d=\sqrt{(-6-2)^2+(-2-7)^2

Subtract:

d=\sqrt{(-8)^2+(-9)^2

Square:

d=\sqrt{64+81}

Add:

d=\sqrt{145}

Approximate

d=\sqrt{145}\approx12.04

So, the distance between (2,7) and (-6,-2) is approximately 12.04 units.

And we're done!

7 0
3 years ago
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