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Leno4ka [110]
4 years ago
6

What is the circumference of the circle below? (Round your answer to the nearest tenth.)

Mathematics
1 answer:
yuradex [85]4 years ago
3 0

Answer:

D

Step-by-step explanation:

Circumference of circle = 2πr

r = 9cm

Circumference of circle = 2×9×3.142

Circumference of circle = 56.5cm

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Subtract <br><img src="https://tex.z-dn.net/?f=12x%20-%20%289%20%20%2B%204x%29" id="TexFormula1" title="12x - (9 + 4x)" alt="12
KATRIN_1 [288]
I suggest Photomath it works really well
8 0
3 years ago
Simplify the expression 5+6i/4+6i<br> A) 56-6i/-20<br> B) -16-6i/-20<br> C)56-6i/52<br> D) -16-6i/52
IRISSAK [1]

Answer:

The\ simplified\ form\ of\ the\ \frac{5 + 6i}{4 + 6i}\ is\ \frac{56-6i}{52}.

Option (C) is correct .

Step-by-step explanation:

As the expression given in the question as follows.

= \frac{5 + 6i}{4 + 6i}

Now rationalize the expression.

= \frac{5 + 6i\times 4 - 6i}{4 + 6i\times 4 - 6i}

Using the formula

a² - b² = (a - b) (a + b)

Using in the above

= \frac{5 + 6i\times 4 - 6i}{(4)^{2} - (6i)^{2}}

= \frac{5\times 4-5\times 6i+6i\times 4-36\ i^{2}}{16 - 36(i)^{2}}

As i² = -1

= \frac{20-30i+24i+36}{16 + 36}

= \frac{56-6i}{52}

Therefore\ the\ simplified\ form\ of\ the\ \frac{5 + 6i}{4 + 6i}\ is\ \frac{56-6i}{52}.

Option (C) is correct .

8 0
3 years ago
Read 2 more answers
Find the zero of each function and state the multiplicity of each zero. Please show all steps.
vodka [1.7K]

Answer:

1. y=(x+3)^3. Zero: x=-3 multiplicity 3.

2. y=(x-2)^2 (x-1). Zeros: x=2 multiplicity 2; x=1 multiplicity 1.

3. y=(2x+3)(x-1)^2. Zeros: x=-3/2 multiplicity 1; x=1 multiplicity 2.


Step-by-step explanation:

1. y=(x+3)^3

y=0\\ (x+3)^3=0\\ \sqrt[3]{(x+3)^3}=\sqrt[3]{0}\\ x+3=0\\ x+3-3=0-3\\ x=-3

Zero: x=-3 multiplicity 3.


2. y=(x-2)^2 (x-1)

y=0\\ (x-2)^2(x-1)=0\\ \left \{ {{(x-2)^2=0} \atop {x-1=0}} \right\\ \left \{ {{\sqrt{(x-2)^2} =\sqrt{0} } \atop {x-1+1=0+1}} \right\\ \left \{ {{x-2=0} \atop {x=1}} \right\\ \left \{ {{x-2+2=0+2} \atop {x=1}} \right\\ \left \{ {{x=2} \atop {x=1}} \right.

Zeros: x=2 multiplicity 2; x=1 multiplicity 1


3. y=(2x+3)(x-1)^2

y=0\\ (2x+3)(x-1)^2=0\\ \left \{ {{2x+3=0} \atop {(x-1)^2=0}} \right\\ \left \{ {{2x+3-3=0-3} \atop {\sqrt{(x-1)^2} =\sqrt{0} }} \right\\ \left \{ {{2x=-3} \atop {x-1=0}} \right\\ \left \{ {{\frac{2x}{2} =\frac{-3}{2} } \atop {x-1+1=0+1}} \right\\ \left \{ {{x=-\frac{3}{2} } \atop {x=1}} \right.

Zeros: x=-3/2 multiplicity 1; x=1 multiplicity 2.

4 0
3 years ago
15 POINTS PLEASE HELP!!! WILL MARK BRAINLIEST
dangina [55]

Given: ∠ DEF

To construct: ∠TSZ ≅ ∠DEF

Construction: Consider the attachment

Step-01: Draw a line XY and choose a point S on it as a vertex of the required angle. Further marks point T such that DE = ST

Step-02: Take an arc AB from point E in ∠DEF of any length and draw at point S which cuts at point P on XY line.

Step-03: Take another arc of length AB from point B in ∠DEF and draw from point P which cuts to the previous arc at Q.

Step-04: Now, join the point SQ and extend up to Z such that EF = SZ

Hence, ∠ TSZ will be the required congruent constructed angle to∠DEF

6 0
3 years ago
What is the area of the following circle with radius of 7
seropon [69]

Answer:

153.94

Step-by-step explanation:

\pi r^{2}\\\pi  (7)^2\\=153.94--->about this amount

5 0
3 years ago
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