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leva [86]
2 years ago
8

Exercise 1.2 1 Round each of these numbers to the degree of accuracy shown in brackets. a) 7765 (nearest hundred). b) 1099 (near

est thousand). c) 487 (nearest ten) d) 34 766 (nearest hundred) e) 45620000 (nearest million please help me​
Mathematics
1 answer:
wlad13 [49]2 years ago
6 0

Answer:

a. 700

b. 1000

c. 80

d. 700

e.

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Which number line shows the solution to the inequality -2x-3> -11
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Answer:

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Step-by-step explanation:

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What are the coordinates of the center of a circle whose equation is (x + 7)2 + (y – 5)2 = 16?
Dafna11 [192]
Hi there! The answer is ( - 7, 5)

The standard form of the equation of a circle is the following:
(x - a) {}^{2}  + (y - b) {}^{2}  = r {}^{2}

In this equation r represents the radius of the circle and the point (a, b) is its centre.

Therefore, the centre of the circle
(x + 7) {}^{2}  + (y - 5) {}^{2}  = 16
is the point ( - 7, 5).
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3 years ago
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Kisachek [45]
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3 years ago
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jek_recluse [69]

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5 0
3 years ago
I need you to answer with a, b, c, d
solong [7]

To find the zeros of a quadratic fiunction given the equation you can use the next quadratic formula after equal the function to 0:

\begin{gathered} ax^2+bx+c=0 \\  \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \end{gathered}

For the given function:

f(x)=2x^2-10x-3x=\frac{-(-10)\pm\sqrt[]{(-10)^2-4(2)(-3)}}{2(2)}x=\frac{10\pm\sqrt[]{100+24}}{4}\begin{gathered} x=\frac{10\pm\sqrt[]{124}}{4} \\  \\ x=\frac{10\pm\sqrt[]{2\cdot2\cdot31}}{4} \\  \\ x=\frac{10\pm\sqrt[]{2^2\cdot31}}{4} \\  \\ x=\frac{10\pm2\sqrt[]{31}}{4} \\  \end{gathered}\begin{gathered} x_1=\frac{10}{4}+\frac{2\sqrt[]{31}}{4} \\  \\ x_1=\frac{5}{2}+\frac{\sqrt[]{31}}{2} \end{gathered}\begin{gathered} x_2=\frac{10}{4}-\frac{2\sqrt[]{31}}{4} \\  \\ x_2=\frac{5}{2}-\frac{\sqrt[]{31}}{2} \end{gathered}

Then, the zeros of the given quadratic function are:

\begin{gathered} x=\frac{5}{2}+\frac{\sqrt[]{31}}{2} \\  \\ x_{}=\frac{5}{2}-\frac{\sqrt[]{31}}{2} \end{gathered}

Answer: Third option

8 0
1 year ago
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