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3 years ago

HELP please, see the picture below thank you.I have a test, please

Mathematics

2 answers:

vladimir1956 [14]3 years ago

8 0

Answer:

5 - 10g + 20b

Step-by-step explanation:

(5 times 1) plus (5 times -3g) plus (5 times 4h)

5 - 10g + 20h.

This is the answer.

#teamtrees #WAP (Water And Plant)

Zolol [24]3 years ago

7 0

5(1 - 2g + 4h)
5 - 10g + 20h

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Find the unit tangent vector T and the principal unit normal vector N for the following parameterized curve.

const2013 [10]

Answer:

$T(t) = ( -sin(t^2), cos(t^2) )\\\\N(t) = T'(t) / |T'(t) | = (-cos(t^2) , -sin(t^2))$

$T(t) =r'(t)/|r'(t)| = (2t/ \sqrt{4t^2 + 36} , -6/\sqrt{4t^2 + 36},0)$

$N(t) =T(t)/|T'(t)| = (3/(9 + t^2)^{1/2} , t/(9 + t^2)^{1/2},0)$

Step-by-step explanation:

Remember that for any curve $r(t)$

The tangent vector is given by

$T(t) = \frac{r'(t) }{| r'(t)| }$

And the normal vector is given by

$N(t) = \frac{T'(t)}{|T'(t)|}$

For this case, using the chain rule

$r'(t) = ( -10*2tsin(t^2) , 102t cos(t^2) )\\$

And also remember that

$|r'(t)| = \sqrt{(-10*2tsin(t^2))^2 + ( 10*2t cos(t^2) )^2} \\\\ = \sqrt{400 t^2*( sin(t^2)^2 + cos(t^2) ^2 })\\=\sqrt{400t^2} = 20t$

Therefore

$T(t) = r'(t) / |r'(t) | = ( -10*2tsin(t^2) , 10*2t cos(t^2) )/ 20t\\\\ = ( -10*2tsin(t^2)/ 20t , 10*2t cos(t^2) / 20t )\\= ( -sin(t^2), cos(t^2) )$

Similarly, using the quotient rule and the chain rule

$T'(t) = ( -2t cos(t^2) , -2t sin(t^2))$

And also

$|T'(t)| = \sqrt{ ( -2t cos(t^2))^2 + (-2t sin(t^2))^2} = \sqrt{ 4t^2 ( ( cos(t^2))^2 + ( sin(t^2))^2)} = \sqrt{4t^2} \\ = 2t$

Therefore

$N(t) = T'(t) / |T'(t) | = (-cos(t^2) , -sin(t^2))$

Notice that

1. $|N(t)| = |T(t) | = \sqrt{ cos(t^2)^2 + sin(t^2)^2 } = \sqrt{1} = 1$

2. $N(t)*T(T) = cos(t^2) sin(t^2 ) - cos(t^2) sin(t^2 ) = 0$

Simlarly

$r'(t) = (2t,-6,0) \\$

and

$|r'(t)| = \sqrt{(2t)^2 + 6^2} = \sqrt{4t^2 + 36}$

Therefore

$T(t) =r'(t)/|r'(t)| = (2t/ \sqrt{4t^2 + 36} , -6/\sqrt{4t^2 + 36},0)$

Then

$T'(t) = (9/(9 + t^2)^{3/2} , (3 t)/(9 + t^2)^{3/2},0)$

and also

$|T'(t)| = \sqrt{ ( (9/(9 + t^2)^{3/2} )^2 + ( (3 t)/(9 + t^2)^{3/2})^2 + 0^2 }\\= 3/(t^2 + 9 )$

And since

$N(t) =T(t)/|T'(t)| = (3/(9 + t^2)^{1/2} , t/(9 + t^2)^{1/2},0)$

6 0

3 years ago

Graph x^2+60x+256in^2

Softa [21]

Answer:

Step-by-step explanation:

there you goooooo oooooooooooooooooo

8 0

3 years ago

A teacher’s monthly gross pay is $5,321. The federal government withholds 15% of the teacher’s monthly gross pay for income taxe

lubasha [3.4K]

Answer:

798.15

Step-by-step explanation:

heres a neat trick you can do instead of doing that long process to get 15% of 5321 just use a calculator and do 5321 x .15

3 0

3 years ago

P/4+=16<br> Pls show work!!

Dima020 [189]

Answer:

P = 64

Step-by-step explanation:

brainliest question

3 0

3 years ago

The lengths of nails produced in a factory are normally distributed with a mean of 3.34 centimeters and a standard deviation of

Shtirlitz [24]

Answer:

3.47 and 3.21

Step-by-step explanation:

Let us assume the nails length be X

$X \sim N(3.34,0.07^2)$

Value let separated the top 3% is T and for bottom it would be B

$P(X < T)= 0.97$

Now converting, we get

$P(Z < \frac{T-3.34}{0.07})= 0.97$

Based on the normal standard tables, we get

$P(Z < 1.881)= 0.97$

Now compare these two above equations

$\frac{T-3.34}{0.07} = 1.881 \\\\ T = 1.881 \times 0.07 + 3.34 \\\\ = 3.47$

So for top 3% it is 3.47

Now for bottom we applied the same method as shown above

$P(Z < \frac{B-3.34}{0.07})= 0.03$

Based on the normal standard tables, we get

$P(Z < -1.881)= 0.03$

Now compare these two above equations

$\frac{B-3.34}{0.07} = -1.881$

$= -1.881 \times 0.07 + 3.34 \\\\ = 3.21$

hence, for bottom it would be 3.21

7 0

3 years ago