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3 years ago

What is .5 x 10 - 1.5 ×1.5

Mathematics

1 answer:

JulsSmile [24]3 years ago

3 0

0.5x10 =5

5-1.5=3.5

3.5x1.5=5.25

so the final answer is 5.25

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Which one of the following would prove that two triangles are congruenta

White raven [17]

Answer:

Corroponding side lengths are equal

Step-by-step explanation:

Two triangles may have the same areas, but the sides might be different or in different spots than each other. The same goes for perimeter, therefore it doesn't mean that they are congruent. If two triangles had the same angle measures, it could still have any size side lengths. When corrosponding side lengths are equal, they have the same angles as well, so they are only congruent in this example.

Hope this helps :)

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3 years ago

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3 years ago

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3 years ago

What is the area of the rectangle, in square centimeters? A rectangle with length 12 centimeters and width 5 centimeters. 17 34

sertanlavr [38]

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3 years ago

Read 2 more answers

Find the x-coordinates of any relative extrema and inflection point(s) for the function f(x)= 6x^1/3 + 3x^4/3. You must justify

irina [24]

Answer:

x-coordinates of relative extrema = $\frac{-1}{2}$

x-coordinates of the inflexion points are 0, 1

Step-by-step explanation:

$f(x)=6x^{\frac{1}{3}}+3x^{\frac{4}{3}}$

Differentiate with respect to x

$f'(x)=6\left ( \frac{1}{3} \right )x^{\frac{-2}{3}}+3\left ( \frac{4}{3} \right )x^{\frac{1}{3}}=\frac{2}{x^{\frac{2}{3}}}+4x^{\frac{1}{3}}$

$f'(x)=0\Rightarrow \frac{2}{x^{\frac{2}{3}}}+4x^{\frac{1}{3}}=0\Rightarrow x=\frac{-1}{2}$

Differentiate f'(x) with respect to x

$f''(x)=2\left ( \frac{-2}{3} \right )x^{\frac{-5}{3}}+\frac{4}{3}x^{\frac{-2}{3}}=\frac{-4x^{\frac{2}{3}}+4x^{\frac{5}{3}}}{3x^{\frac{2}{3}}x^{\frac{5}{3}}}\\f''(x)=0\Rightarrow \frac{-4x^{\frac{2}{3}}+4x^{\frac{5}{3}}}{3x^{\frac{2}{3}}x^{\frac{5}{3}}}=0\Rightarrow x=1$

At x = $\frac{-1}{2}$ ,

$f''\left ( \frac{-1}{2} \right )=\frac{4\left ( -1+4\left ( \frac{-1}{2} \right ) \right )}{3\left ( \frac{-1}{2} \right )^{\frac{5}{3}}}>0$

We know that if $f''(a)>0$ then x = a is a point of minima.

So, $x=\frac{-1}{2}$ is a point of minima.

For inflexion points:

Inflexion points are the points at which f''(x) = 0 or f''(x) is not defined.

So, x-coordinates of the inflexion points are 0, 1

7 0

3 years ago