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Leni [432]
3 years ago
5

The difference of twice a number and 2 is at least −26.

Mathematics
2 answers:
natita [175]3 years ago
8 0

2x - 2 \geqslant  - 26
777dan777 [17]3 years ago
5 0

Answer:

2 (c-2) ≤ -26

Step-by-step explanation:

Let's look at the phrase "Twice the difference of a number and 8".

We're using c for the unknown number.

So "the difference of a number and 8" is written −c8.  

Taking twice this quantity, we get 2−c8.

The symbol for "is at most" is ≤. We get the following:

2 (c-2) ≤ -26

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(-5/2, 5)

Step-by-step explanation:

Apply x-y intercepts theorem

(x,y)=(0,y) and (x,0).

then substitutes.

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Help please! Need help solving
NISA [10]
For the perimeter you have to consider the EXTERNAL SHAPE ONLY:

a) the perimeter of half a circle WITHOUT ITS DIAMETER  is:
π.R - 2R → 3.14x1.5 - 2x1.5 → 1.71 cm

b) the perimeter of the rectangle WITHOUT THE UPPER SIDE is:
2.5+3+2.5 = 8 cm

Hence the perimeter of the figure is 1.71 + 8 = 9.71 cm

2) Area of the figure:
Area of half circle is π.R²/2 →3.14 x (1.5)²/2 → Area half circle = 3.5325 cm²
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3 0
3 years ago
Find the sum of 2/6 +7/6​
ladessa [460]

Answer:

your answer is 9/6, which simplifies to 3/2, or 1 1/2

Step-by-step explanation:

8 0
3 years ago
Change each of the following points from rectangular coordinates to spherical coordinates and to cylindrical coordinates.
FromTheMoon [43]

Answer and Step-by-step explanation: Spherical coordinate describes a location of a point in space: one distance (ρ) and two angles (Ф,θ).To transform cartesian coordinates into spherical coordinates:

\rho = \sqrt{x^{2}+y^{2}+z^{2}}

\phi = cos^{-1}\frac{z}{\rho}

For angle θ:

  • If x > 0 and y > 0: \theta = tan^{-1}\frac{y}{x};
  • If x < 0: \theta = \pi + tan^{-1}\frac{y}{x};
  • If x > 0 and y < 0: \theta = 2\pi + tan^{-1}\frac{y}{x};

Calculating:

a) (4,2,-4)

\rho = \sqrt{4^{2}+2^{2}+(-4)^{2}} = 6

\phi = cos^{-1}(\frac{-4}{6})

\phi = cos^{-1}(\frac{-2}{3})

For θ, choose 1st option:

\theta = tan^{-1}(\frac{2}{4})

\theta = tan^{-1}(\frac{1}{2})

b) (0,8,15)

\rho = \sqrt{0^{2}+8^{2}+(15)^{2}} = 17

\phi = cos^{-1}(\frac{15}{17})

\theta = tan^{-1}\frac{y}{x}

The angle θ gives a tangent that doesn't exist. Analysing table of sine, cosine and tangent: θ = \frac{\pi}{2}

c) (√2,1,1)

\rho = \sqrt{(\sqrt{2} )^{2}+1^{2}+1^{2}} = 2

\phi = cos^{-1}(\frac{1}{2})

\phi = \frac{\pi}{3}

\theta = tan^{-1}\frac{1}{\sqrt{2} }

d) (−2√3,−2,3)

\rho = \sqrt{(-2\sqrt{3} )^{2}+(-2)^{2}+3^{2}} = 5

\phi = cos^{-1}(\frac{3}{5})

Since x < 0, use 2nd option:

\theta = \pi + tan^{-1}\frac{1}{\sqrt{3} }

\theta = \pi + \frac{\pi}{6}

\theta = \frac{7\pi}{6}

Cilindrical coordinate describes a 3 dimension space: 2 distances (r and z) and 1 angle (θ). To express cartesian coordinates into cilindrical:

r=\sqrt{x^{2}+y^{2}}

Angle θ is the same as spherical coordinate;

z = z

Calculating:

a) (4,2,-4)

r=\sqrt{4^{2}+2^{2}} = \sqrt{20}

\theta = tan^{-1}\frac{1}{2}

z = -4

b) (0, 8, 15)

r=\sqrt{0^{2}+8^{2}} = 8

\theta = \frac{\pi}{2}

z = 15

c) (√2,1,1)

r=\sqrt{(\sqrt{2} )^{2}+1^{2}} = \sqrt{3}

\theta = \frac{\pi}{3}

z = 1

d) (−2√3,−2,3)

r=\sqrt{(-2\sqrt{3} )^{2}+(-2)^{2}} = 4

\theta = \frac{7\pi}{6}

z = 3

5 0
3 years ago
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