All categories

2 years ago

Plssssssssssssssssssssssssssssss

Mathematics

2 answers:

user100 [1]2 years ago

7 0

Answer:

Correct answer is 4/7:10

coldgirl [10]2 years ago

5 0

Answer:

the last one 7/4 _10

Step-by-step explanation:

You might be interested in

Which one do I put? Pick out of the 3 options.

SpyIntel [72]

The answer is -6 is less than 3

3 0

3 years ago

Find the coordinates of point P

Drupady [299]

Here is the formula for finding a partitioning point:
x=x1+k(x2-x1), y=y1+k(y2-y1)
k is the ratio of the segment from the beginning point to the partitioning : the whole segment. In this case, k=AP:AB=5/16
so x=1+(5/16)*(-2-1)=1/16
y=6+(5/16)(-3-6)=51/16
so the answer is (-1/16, 51/16)

Please double check my calculation by yourself.

refer to this website for the formula and how to find k:

"This ratio is called k, and is determined by writing the numerator over the sum of the numerator and the denominator of the original ratio."
https://cobbk12.blackboard.com/bbcswebdav/institution/eHigh%20School/Courses/CCVA%20CCGPS%20Coordina....

6 0

3 years ago

5(8n+ 7) < 5-5(n+3)

svet-max [94.6K]

Answer:

n<-1

Step-by-step explanation:

Distribute 5 and -5 into each parenthesis respectively: 40n+35<5-5n-15

Combine like terms: 40n+35<-5n-10

Add 5n to both sides: 45n+35<-10

Subtract 35 from both sides: 45n<-45

Divide both sides by 45: n<-1

3 0

2 years ago

(10 points) Consider the initial value problem y′+3y=9t,y(0)=7. Take the Laplace transform of both sides of the given differenti

Rashid [163]

Answer:

The solution

$Y (s) = 9( -1 +3 t + e^{-3 t} ) + 7 e ^{-3 t}$

Step-by-step explanation:

Explanation:-

Consider the initial value problem y′+3 y=9 t,y(0)=7

Step(i):-

Given differential problem

y′+3 y=9 t

Take the Laplace transform of both sides of the differential equation

L( y′+3 y) = L(9 t)

Using Formula Transform of derivatives

L(y¹(t)) = s y⁻(s)-y(0)

By using Laplace transform formula

$L(t) = \frac{1}{S^{2} }$

Step(ii):-

Given

L( y′(t)) + 3 L (y(t)) = 9 L( t)

$s y^{-} (s) - y(0) + 3y^{-}(s) = \frac{9}{s^{2} }$

$s y^{-} (s) - 7 + 3y^{-}(s) = \frac{9}{s^{2} }$

Taking common y⁻(s) and simplification, we get

$( s + 3)y^{-}(s) = \frac{9}{s^{2} }+7$

$y^{-}(s) = \frac{9}{s^{2} (s+3}+\frac{7}{s+3}$

Step(iii):-

By using partial fractions , we get

$\frac{9}{s^{2} (s+3} = \frac{A}{s} + \frac{B}{s^{2} } + \frac{C}{s+3}$

$\frac{9}{s^{2} (s+3} = \frac{As(s+3)+B(s+3)+Cs^{2} }{s^{2} (s+3)}$

On simplification we get

9 = A s(s+3) +B(s+3) +C(s²) ...(i)

Put s =0 in equation(i)

9 = B(0+3)

B = 9/3 = 3

Put s = -3 in equation(i)

9 = C(-3)²

C = 1

Given Equation 9 = A s(s+3) +B(s+3) +C(s²) ...(i)

Comparing 'S²' coefficient on both sides, we get

9 = A s²+3 A s +B(s)+3 B +C(s²)

0 = A + C

put C=1 , becomes A = -1

$\frac{9}{s^{2} (s+3} = \frac{-1}{s} + \frac{3}{s^{2} } + \frac{1}{s+3}$

Step(iv):-

$y^{-}(s) = \frac{9}{s^{2} (s+3}+\frac{7}{s+3}$

$y^{-}(s) =9( \frac{-1}{s} + \frac{3}{s^{2} } + \frac{1}{s+3}) + \frac{7}{s+3}$

Applying inverse Laplace transform on both sides

$L^{-1} (y^{-}(s) ) =L^{-1} (9( \frac{-1}{s}) + L^{-1} (\frac{3}{s^{2} }) + L^{-1} (\frac{1}{s+3}) )+ L^{-1} (\frac{7}{s+3})$

By using inverse Laplace transform

$L^{-1} (\frac{1}{s} ) =1$

$L^{-1} (\frac{1}{s^{2} } ) = \frac{t}{1!}$

$L^{-1} (\frac{1}{s+a} ) =e^{-at}$

Final answer:-

Now the solution , we get

$Y (s) = 9( -1 +3 t + e^{-3 t} ) + 7 e ^{-3t}$

5 0

3 years ago

Did Tyson use the order of operations to correctly evaluate the expression? *

nlexa [21]

Answer:

Yes

Step-by-step explanation:

Just trust

6 0

3 years ago

Read 2 more answers