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3 years ago

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A triangle has an area of 900m^2 . If a parallelogram has the same height and base as the triangle, what is the area of the para

llelogram?

1 answer:

igor_vitrenko [27]3 years ago

6 0

Answer:

area = 1800 m²

Step-by-step explanation:

area of one triangle = 900 m²

if a parallelogram has the same height and base as the triangle, then that means the area or the two triangle and shaped as a parallelogram

is twice the area given.

area = 900 * 2

area = 1800 m²

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Answer:

5.36,275,336,535

Step-by-step explanation:

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2 years ago

Three judges measure the lap times of a runner and each measurement is different from the others by a few hundredths of a second

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Answer:

Systematic error

Step-by-step explanation:

Assuming that none of the judges are biased, the most likely reason for this difference is the occurrence of systematic errors.

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In this case, the reaction time from observing the finish and stopping the clock for each judge might be different, which configures a systematic error.

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3 years ago

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3 years ago

Solve 5x^2 − 3x + 17 = 9.

Vedmedyk [2.9K]

Answer:

The roots of the equation are x = $\frac{3+\sqrt{151}i}{10}$ and x = $\frac{3-\sqrt{151}i}{10}$

and there are no real roots of the equation given above

Step-by-step explanation:

To solve:

5x² − 3x + 17 = 9

or

⇒ 5x² − 3x + 17 - 9 = 0

or

⇒ 5x² − 3x + 8 = 0

Now,

the roots of the equation in the form ax² + bx + c = 0 is given as:

x = $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

in the above given equation

a = 5

b = -3

c = 8

therefore,

x = $\frac{-(-3)\pm\sqrt{(-3)^2-4\times5\times8}}{2\times5}$

or

x = $\frac{3\pm\sqrt{9-160}}{10}$

or

x = $\frac{3+\sqrt{-151}}{10}$ and x = $\frac{3-\sqrt{-151}}{10}$

or

x = $\frac{3+\sqrt{151}i}{10}$ and x = $\frac{3-\sqrt{151}i}{10}$

here i = √(-1)

Hence,

The roots of the equation are x = $\frac{3+\sqrt{151}i}{10}$ and x = $\frac{3-\sqrt{151}i}{10}$

and there are no real roots of the equation given above

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2 years ago

The function f(x) = x2 + 6x + 3 is transformed such that g(x) = f(x − 2). Find the vertex of g(x).

garri49 [273]

Answer:

(-1,-6)

Step-by-step explanation:

$f(x)=x^2+6x+3$

$g(x)=f(x-2)$

$g(x)=(x-2)^2+6(x-2)+3$ (Replaced x in f with (x-2))

$g(x)=x^2-4x+4+6x-12+3$ (Used $(x+b)^2=x^2+2bx+b^2$ and distributive property)

$g(x)=x^2-4x+6x+4-12+3$ (Gathered like terms)

$g(x)=x^2+2x-5$ (Simplified)

The vertex of a parabola occurs at $(\frac{-b}{2a},g(\frac{-b}{2a})$ .

Let's find $\frac{-b}{2a}$ first.

$\frac{-2}{2(1)}=-1$

Now we can obtain $g(\frac{-b}{2a})$ which is $g(-1)$ in this case:

$g(x)=x^2+2x-5$

$g(-1)=(-1)^2+2(-1)-5$

$g(-1)=1-2-5$

$g(-1)=-1-5$

$g(-1)=-6$ .

The vertex is (-1,-6).

6 0

3 years ago