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Arlecino [84]
3 years ago
10

Simplify the square root of 320

Mathematics
1 answer:
Pavlova-9 [17]3 years ago
5 0
17.89, is this an option?
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-2/5 - 4/5 + -9/10 = ???<br><br> Please help, will mark brainliest for answer with an explanation
NeX [460]

Answer:

Step-by-step explanation:

-2/5 - 4/5 + -9/10

= -2/5 - 4/5 - 9/10

= {-2(2) -4(2) - 9)/10

= (-4 - 8 - 9)/10

= -21/10

= -2 1/10

5 0
3 years ago
Find the area of the triangle that has a base of 4.2 mm and a height of 1.5 mm. 2.85 mm2 6.3 mm2 3.15 mm2 5.7 mm2
Advocard [28]

Answer: 3.15 mm²

Step-by-step explanation:

First, we multiply the height by the base: 4.2 x 1.5 = 6.3

But, since this is a triangle, we have to divide this number by 2: 6.3 / 2 = 3.15

Hope this helps you!

4 0
4 years ago
Read 2 more answers
Write a system of inequalities that defines a shaded region that looks like a<br> right triangle.
eduard

(1) \ x>0 \\ \\ (2) \ y>0 \\ \\ (3) \ y

<h2>Explanation:</h2>

A right triangle is a special triangle that has a right angle. In this case, we have to write a system of inequalities that defines a shaded region that looks like a  right triangle. First of all, let's say that at the origin the triangle will have a right angle. To do so, we'd need to set:

x>0 \\ \\ y>0

So the shaded region in this first part will be the First Quadrant as indicated in the first figure below. So if the opposite side lies on the x-axis the adjacent side will lie on the y-axis or if the adjacent side lies on the x-axis the opposite side will lie on the y-axis. Everything is ok up to this point. We just need to define the hypotenuse, so we'd need to define a line.  In order to have a right triangle, we need a line with negative slope and positive y-intercept shaded under the line. So, this inequality could be:

y

Finally, the system of inequalities would be:

(1) \ x>0 \\ \\ (2) \ y>0 \\ \\ (3) \ y

And the final shaded region is the one shown in the second figure.

<h2>Learn more:</h2>

Inequalities: brainly.com/question/2486051

#LearnWithBrainly

3 0
3 years ago
Find the critical numbers of the function f(x) = x6(x − 2)5.x = (smallest value)x = x = (largest value)(b) What does the Second
Marrrta [24]

Answer:

a) x=0, x=\frac{12}{11}, x=2 \: b) The 2nd Derivative test shows us the change of sign and concavity at some point. c) At which point the concavity changes or not. This is only possible with the 2nd derivative test.

Step-by-step explanation:

a) To find the critical numbers, or critical points of:

f(x)=x^{6}(x-2)^{5}

1) The procedure is to calculate the 1st derivative of this function. Notice that in this case, we'll apply the <em>Product Rule</em> since there is a product of two functions.

f(x)=x^{6}(x-2)^{5}\Rightarrow f'(x)=(f*g)'(x)\\=f'g+fg'\Rightarrow (fg)'(x)=6x^{5}(x-2)^{5}+5x^{6}(x-2)^{4} \Rightarrow 6x^{5}(x-2)^{5}+5x^{6}(x-2)^{4}=0\\f'(x)=6x^{5}(x-2)^{5}+5x^{6}(x-2)^{4}

2) After that, set this an equation then find the values for x.

x^{5}(x-2)^{4}[6(x-2)+5x]=0\Rightarrow x^{5}(x-2)^{4}[11x-12]=0\Rightarrow x_{1}=0\\(x-2)^{4}=0\Rightarrow \sqrt[4]{(x-2)}=\sqrt[4]{0}\Rightarrow x-2=0\Rightarrow x_{2}=2\\(11x-12)=0\Rightarrow x_{3}=\frac{12}{11}

x=0\:(smallest\:value)\:x_{3}=\frac{12}{11}\:x=2 (largest value)

b) The Second Derivative Test helps us to check the sign of given critical numbers.

Rewriting f'(x) factorizing:

f'(x)=(11x-12)(x-2)^4x^{5}

Applying product Rule to find the 2nd Derivative, similarly to 1st derivative:

f''(x)>0 \Rightarrow Concavity\: Up\\\\f''(x)

f''(x)=11\left(x-2\right)^4x^5+4\left(x-2\right)^3x^5\left(11x-12\right)+5\left(x-2\right)^4x^4\left(11x-12\right)\\f''(x)=10\left(x-2\right)^3x^4\left(11x^2-24x+12\right)

1) Setting this to zero, as an equation:

10\left(x-2\right)^3x^4\left(11x^2-24x+12\right)=0\\\\

10\left(x-2\right)^3x^4\left(11x^2-24x+12\right)=0\\(x-2)^{3}=0 \Rightarrow x_1=2\\x^{4}=0 \therefore x_2=0\\11x^{2}-24x+12=0 \Rightarrow x_3=\frac{12+2\sqrt{3}}{11}\:,x_4=\frac{12-2\sqrt{3}}{11}\cong 0.78

2) Now, let's define which is the inflection point, the domain is as a polynomial function:

D=(-\infty

Looking at the graph.

Plugging these inflection points in the original equationf(x)=x^{6}(x-2)^{5} to get y coordinate:

We have as Inflection Points and their respective y coordinates (Converting to approximate decimal numbers)

(1.09,-1.05) Inflection Point and Local Minimum

(2,0) Inflection Point and Saddle Point

(0,0) Inflection Point Local Maximum

(Check the graph)

c) At which point the concavity changes or not. This is only possible with the 2nd derivative test.

At

x=\frac{12}{11}\cong1.09 Local Minimum

At\:x=0,\:Local \:Maximum

At\:x=2, \:neither\:a\:minimum\:nor\:a\:maximum (Saddle Point)

5 0
3 years ago
Find the Area of the Shaded Region.
Sedbober [7]

Answer:

A of the shaded figure is 90cm

4 0
3 years ago
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