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Svetradugi [14.3K]
3 years ago
7

Here is a sketch of a curve.

Mathematics
1 answer:
Ugo [173]3 years ago
5 0

Answer:

The turning point of the graph is at (3,-16)

Step-by-step explanation:

The general equation of curve is given as:

y = x² + ax + b

The two point for which the equation satisfies is (0,-7) and (7,0).

Substitute (0,-7) in the general equation:

y = x² + ax + b

-7 = 0² + a(0) + b

b = -7

Substitute (7,0) in the general equation:

y = x² + ax + b

0 = 7² + 7a + b

Where b = -7

0 = 49 + 7a - 7

0 = 42 + 7a

a = -6

The equation of the curve is:

a = -6 and b = -7

y = x² + ax + b

y = x² - 6x - 7

Graph of the equation is attached below.

We can see that the the turning point of the graph is at (3,-16)

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What is the quotient of StartFraction 7 Superscript negative 1 Baseline Over 7 Superscript negative 2 Baseline EndFraction? Star
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The quotient of the number given number  7 Superscript negative 1 Baseline Over 7 Superscript negative 2 Baseline is 7.

<h3>What is the quotient?</h3>

Quotient is the resultant number which is obtain by dividing a number with another. Let a number a is divided by number b. Then the quotient of these two number will be,

q=\dfrac{a}{b}

Here, (<em>a, b</em>) are the real numbers.

The number StartFraction 7 Superscript negative 1 Baseline Over 7 Superscript negative 2 Baseline EndFraction, given can be written as,

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Let the quotient of this division is n. Therefore,

n=\dfrac{7^{-1}}{7^{-2}}

A number in numerator of a fraction with negative exponent can be written in the denominator with the same but positive exponent and vise versa. Therefore,

n=\dfrac{7^{2}}{7^{1}}\\n=7

Hence, the quotient of the number given number  7 Superscript negative 1 Baseline Over 7 Superscript negative 2 Baseline is 7.

Learn more about the quotient here;

brainly.com/question/673545

7 0
2 years ago
Find the area bounded by the curves x = 2y2 and x = 1 - y. Your work must include an integral in one variable.
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Hello,

in order to simplify, i have taken the inverses functions

Step-by-step explanation:

\int\limits^\frac{1}{2} _{-1} {(-2x^2-x+1)} \, dx \\\\=[\frac{-2x^3}{3} -\frac{x^2}{2} +x]^\frac{1}{2} _{-1}\\\\\\=\dfrac{-2-3+12}{24} -\dfrac{-5}{6} \\\\\boxed{=\dfrac{9}{8} =1.25}\\

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Step-by-step explanation:

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