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chubhunter [2.5K]
4 years ago
5

An equation ____ has one solution. *always *sometimes *never

Mathematics
1 answer:
Scilla [17]4 years ago
4 0

Answer:

sometimes

Step-by-step explanation:

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Determine which statement is true about the zeros of the function graphed below.
Oksi-84 [34.3K]
The real solution occurs when the graph intersects the x axis,

In the problem shown, the graph does not intersect the x axis, therefore it has no real solution, this means that the answer must have a complex conjugate pair

The answer is
function f has exactly two complex solutions
4 0
3 years ago
Read 2 more answers
Chas build a circular pattern for his dog The radius of the pen is 9 feet how much fencing did Chaz use for the pen
kicyunya [14]

Answer:

The fencing Chaz used for the pen is 56.67 feet

Step-by-step explanation:

Fencing interprets to circumference

If the radius of the pen is 9 ft

circumference = 2\pi r

= 2*(22/7)*9

= 396/7

= 56.67 ft

3 0
3 years ago
Please help me!!!!!!!!!!!!!!!
Ainat [17]

r = 7.53 so d = 2r = 2(7.53) = 15.06 cm

Area of square = d^2 / 2 = (15.06)^2 / 2 = 113.41 cm^2

Area of circle = 3.14 (7.53)^2 = 178.04 cm^2

Area of yellow region = Area of circle - Area of square

Area of yellow region = 178.04 cm^2 - 113.41 cm^2

Area of yellow region =64.63 cm^2 = 64.6 cm^2 (nearest tenth)

Answer

64.6 cm^2

7 0
3 years ago
Evaluate 4 + (-2) - (-3) – 6.
fgiga [73]

Answer:

Step-by-step explanation:

-1

8 0
3 years ago
Read 2 more answers
What is the perimeter of this quadrilateral?<br> (5,5)<br> (2, 4)<br> (4, 1)<br> (6, 1)
lbvjy [14]

~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_1}{5}~,~\stackrel{y_1}{5})\qquad B(\stackrel{x_2}{2}~,~\stackrel{y_2}{4}) ~\hfill AB=\sqrt{[ 2- 5]^2 + [ 4- 5]^2} \\\\\\ AB=\sqrt{(-3)^2+(-1)^2}\implies \boxed{AB=\sqrt{10}} \\\\[-0.35em] ~\dotfill\\\\ B(\stackrel{x_1}{2}~,~\stackrel{y_1}{4})\qquad C(\stackrel{x_2}{4}~,~\stackrel{y_2}{1}) ~\hfill BC=\sqrt{[ 4- 2]^2 + [ 1- 4]^2}

BC=\sqrt{2^2+(-3)^2}\implies \boxed{BC=\sqrt{13}} \\\\[-0.35em] ~\dotfill\\\\ C(\stackrel{x_1}{4}~,~\stackrel{y_1}{1})\qquad D(\stackrel{x_2}{6}~,~\stackrel{y_2}{1}) ~\hfill CD=\sqrt{[ 6- 4]^2 + [ 1- 1]^2} \\\\\\ CD=\sqrt{2^2+0^2}\implies \boxed{CD=2} \\\\[-0.35em] ~\dotfill\\\\ D(\stackrel{x_1}{6}~,~\stackrel{y_1}{1})\qquad A(\stackrel{x_2}{5}~,~\stackrel{y_2}{5}) ~\hfill DA=\sqrt{[ 5- 6]^2 + [ 5- 1]^2}

DA=\sqrt{(-1)^2+4^2}\implies \boxed{DA=\sqrt{17}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\Large Perimeter}}{\sqrt{10}~~ + ~~\sqrt{13}~~ + ~~2~~ + ~~\sqrt{17}}~~ \approx ~~ 12.89

8 0
3 years ago
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