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3 years ago

14

The second rate superhero and is Arch Enemy Melvin the minor menace started 700 miles apart and fleet street toward each other i

n their jet propulsion packs it took 5 hours for the two to meet if the superheroes speed was 10 mph slower than a minute what was the menaces speed

1 answer:

Papessa [141]3 years ago

8 0

Answer:75 mph

Step-by-step explanation:

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You need $40 $ 40 to buy a new pair of shoes. You have saved $15 $ 15 . Which inequality models how much more you must save?

Hoochie [10]

Answer:

40 < 15+x

Step-by-step explanation:

8 0

3 years ago

Exercise 10.4.1: Bayes' Theorem - detecting a biased coin. About (a) Sally has two coins. The first coin is a fair coin and the

8_murik_8 [283]

Answer:

0.026

Step-by-step explanation:

Given the result of 10 coin flips :

T,T,H,T,H,T,T,T,H,T

Number of Heads, H = 3

Number of tails, T = 7

Let :

B = biased coin

B' = non-biased coin

E = event

Probability that it is the biased coin:

P(E Given biased coin) / P(E Given biased coin) * P(E Given non-biased coin) * P(non-biased coin)

P(E|B)P(B) / (P(E|B)*P(B) + P(E|B')P(B')

([(0.75^3) * (0.25^7)] * 0.5) /([(0.75^3) * (0.25^7)] * 0.5) + (0.5^10) * 0.5

0.0000128746 / 0.00050115585

= 0.0263671875

7 0

3 years ago

In slope intercept form what is the equation of the line having a slope of 5 and passing through the point (-6,-26)

____ [38]

Slope/gradient = 5
y= MX + C
(-6,-26)
-6 is x1 and -26 is y1
so therefore y= 5x -26

5 0

4 years ago

In the triangle pictured, let A, B, C be the angles at the three vertices, and let a,b,c be the sides opposite those angles. Acc

Troyanec [42]

Answer:

Step-by-step explanation:

(a)

Consider the following:

$A=\frac{\pi}{4}=45°\\\\B=\frac{\pi}{3}=60°$

Use sine rule,

$\frac{b}{a}=\frac{\sinB}{\sin A}
\\\\=\frac{\sin{\frac{\pi}{3}}
}{\sin{\frac{\pi}{4}}}\\\\=\frac{[\frac{\sqrt{3}}{2}]}{\frac{1}{\sqrt{2}}}\\\\=\frac{\sqrt{2}}{2}\times \frac{\sqrt{2}}{1}=\sqrt{\frac{3}{2}}$

Again consider,

$\frac{b}{a}=\frac{\sin{B}}{\sin{A}}
\\\\\sin{B}=\frac{b}{a}\times \sin{A}\\\\\sin{B}=\sqrt{\frac{3}{2}}\sin {A}\\\\B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

Thus, the angle B is function of A is, $B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

Now find $\frac{dB}{dA}$

Differentiate implicitly the function $\sin{B}=\sqrt{\frac{3}{2}}\sin{A}$ with respect to A to get,

$\cos {B}.\frac{dB}{dA}=\sqrt{\frac{3}{2}}\cos A\\\\\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos A}{\cos B}$

b)

When $A=\frac{\pi}{4},B=\frac{\pi}{3}$ , the value of $\frac{dB}{dA}$ is,

$\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos {\frac{\pi}{4}}}{\cos {\frac{\pi}{3}}}\\\\=\sqrt{\frac{3}{2}}.\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}\\\\=\sqrt{3}$

c)

In general, the linear approximation at x= a is,

$f(x)=f'(x).(x-a)+f(a)$

Here the function $f(A)=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

At $A=\frac{\pi}{4}$

$f(\frac{\pi}{4})=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{\frac{\pi}{4}}]\\\\=\sin^{-1}[\sqrt{\frac{3}{2}}.\frac{1}{\sqrt{2}}]\\\\\=\sin^{-1}(\frac{\sqrt{2}}{2})\\\\=\frac{\pi}{3}$

And,

$f'(A)=\frac{dB}{dA}=\sqrt{3}$ from part b

Therefore, the linear approximation at $A=\frac{\pi}{4}$ is,

$f(x)=f'(A).(x-A)+f(A)\\\\=f'(\frac{\pi}{4}).(x-\frac{\pi}{4})+f(\frac{\pi}{4})\\\\=\sqrt{3}.[x-\frac{\pi}{4}]+\frac{\pi}{3}$

d)

Use part (c), when $A=46°$ , B is approximately,

$B=f(46°)=\sqrt{3}[46°-\frac{\pi}{4}]+\frac{\pi}{3}\\\\=\sqrt{3}(1°)+\frac{\pi}{3}\\\\=61.732°$

8 0

3 years ago

The equation for line s can be written as y = x + 3. Line t, which is parallel to line s,

arlik [135]

Answer:

y=x-1

Step-by-step explanation:

6 0

3 years ago