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3 years ago

7

The triangles are congruent by the SSS congruence theorem.

2 answers:

Vilka [71]3 years ago

6 0

Answer:

The answer is D

Step-by-step explanation:

Naily [24]3 years ago

5 0

I think it is C) i am sorry if i am wrong

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Hey so im also looking for the first answer and the choices were “ is not “ and “ is “

dexar [7]

so the investigator found the skid marks were 75 feet long hmmm what speed will that be?

$s=\sqrt{30fd}~~ \begin{cases} f=\stackrel{friction}{factor}\\ d=\stackrel{skid}{feet}\\[-0.5em] \hrulefill\\ f=\stackrel{dry~day}{0.7}\\ d=75 \end{cases}\implies s=\sqrt{30(0.7)(75)}\implies s\approx 39.69~\frac{m}{h}$

nope, the analysis shows that Charlie was going faster than 35 m/h.

now, assuming Charlie was indeed going at 35 m/h, then his skid marks would have been

$s=\sqrt{30fd}~~ \begin{cases} f=\stackrel{friction}{factor}\\ d=\stackrel{skid}{feet}\\[-0.5em] \hrulefill\\ f=\stackrel{dry~day}{0.7}\\ s=35 \end{cases}\implies 35=\sqrt{30(0.7)d} \\\\\\ 35^2=30(0.7)d\implies \cfrac{35^2}{30(0.7)}=d\implies 58~ft\approx d$

4 0

2 years ago

Differentiate y=x⁴(1-2x⁵)⁶(5-8x³)².

Liula [17]

Step-by-step explanation:

Let $y(x)=f(x)g(x)h(x)$ where

$f(x) = x^4$

$g(x)= (1 -2x^5)^6$

$h(x)= (5 - 8x^3)^2$

so that

$y(x) = x^4(1 -2x^5)^6(5 - 8x^3)^2$

Recall that the derivative of the product of functions is

$y'(x)=f'(x)g(x)h(x)+f(x)g'(x)h(x)+f(x)g(x)h'(x)$

so taking the derivatives of the individual functions, we get

$f'(x) = 4x^3$

$g'(x) = 6(1 - 2x^5)^5(-10x^4)$

$h'(x) = 2(5 - 8x^3)(-24x^2)$

So the derivative of y(x) is given by

$y'(x) = 4x^3(1 -2x^5)^6(5 - 8x^3)^2 + x^4 6(1 -2x^5)^5(-10x^4)(5 - 8x^3)^2 + x^4(1 -2x^5)^6 2(5 - 8x^3)(-24x^2)$

or

$y'(x) = 4x^3(1 -2x^5)^6(5 - 8x^3)^2$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:- 60x^8(1 -2x^5)^5(5 - 8x^3)^2$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:- 48x^6(1 -2x^5)^6 2(5 - 8x^3)$

3 0

3 years ago