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PolarNik [594]
3 years ago
6

Sketch the angle in standard position and draw an arrow representing the correct amount of rotation. Find the measure of two oth

er​ angles, one positive and one​ negative, that are coterminal with the given angle and closer to the given angle than any other coterminal angle. Give the quadrant of the​ angle, if applicable. -90°
Mathematics
1 answer:
ser-zykov [4K]3 years ago
6 0

Answer:

The coterminal angles are : 270° , -450°

sketched angle is in third quadrant

Step-by-step explanation:

The sketch of the angle in standard position is attached  below

The coterminal angles :

- 90 + 360 = 270°

-90 - 360 = -450°

Quadrant of the angle( -90° )  = Third quadrant

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The diagram shows a 5 cm x 5 cm x 5 cm cube.
mylen [45]

Answer:

~8.66cm

Step-by-step explanation:

The length of a diagonal of a rectangular of sides a and b is

\sqrt{a^2+b^2}

in a cube, we can start by computing the diagonal of a rectangular side/wall containing A and then the diagonal of the rectangle formed by that diagonal and the edge leading to A. If the cube has sides a, b and c, we infer that the length is:

\sqrt{\sqrt{a^2+b^2}^2 + c^2} = \sqrt{a^2+b^2+c^2}

Using this reasoning, we can prove that in a n-dimensional space, the length of the longest diagonal of a hypercube of edge lengths a_1, a_2, a_3, \ldots, a_n is

\sqrt{a_1^2 + a_2^2 + a_3^2 + \ldots + a_n^2}

So the solution here is

\sqrt{(5cm)^2 + (5cm)^2 + (5cm)^2} = \sqrt{75cm^2} = 5\sqrt{3cm^2} \approx 5\cdot 1.732cm = 8.66cm

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3 years ago
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Explain how to work out these type of questions please!
bixtya [17]

x = 9 \times 5 \\ x = 45
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What is the range of the data in this stem-and-leaf plot?
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Which graph represents the solution set to the system of inequalities?<br><br> y≤1/4 x−2, y≥−5/4 x+2
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Solve the following equations;-<br><br> x - 9 /√x + 3 =1
fgiga [73]

Answer:

13 , 6.

Step-by-step explanation:

A equation is given to us and we need to find out the value of x . The given equation to us is ,

\sf\implies \dfrac{ x -9}{\sqrt{x+3}}= 1

Cross multiply ,

\sf\implies x - 9 =\sqrt{ x +3}

Squaring both sides ,

\sf\implies (x - 9)^2 = x + 3

Simplify the whole square ,

\sf\implies x^2+9^2-2.9.x = x +3 \\

\sf\implies x^2+81 - 18x = x -3

Add 3 and subtract x on both sides ,

\sf\implies x^2 -18x-x +81-3=0

Simplify ,

\sf\implies x^2 -19x +78=0

Split the middle term of the quadratic equation,

\sf\implies x^2-13x-6x+78=0

Take out common ,

\sf\implies x( x -13)-6(x-13)=0

Take out ( x -13 ) as common ,

\sf\implies ( x -13)(x-6)=0

Equate both factors to 0 ,

\sf\implies \boxed{\pink{\sf x = 13,6}}

<u>Hence</u><u> the</u><u> </u><u>value</u><u> of</u><u> </u><u>x</u><u> is</u><u> </u><u>1</u><u>3</u><u> </u><u>or </u><u>6</u><u> </u><u>.</u>

7 0
3 years ago
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