I need this in standard form slope intercept and point slope form.

Figure VWYX is a kite.<br><br> What is the value of x?

Alecsey [184]

Answer:

$x=6\°$

Step-by-step explanation:

we know that

The sum of the internal angles of a kite is equal to $360\°$

so

In this problem we have

$90\°+90\°+(18x-2)\°+(12x+2)\°=360\°$

$(18x-2)\°+(12x+2)\°=360\°-180\°$

$30x=180\°$

$x=6\°$

8 0

3 years ago

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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:

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

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