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Vlad [161]
3 years ago
5

HURRY!! BRAINLIEST IF RIGHT!!​

Mathematics
1 answer:
Dvinal [7]3 years ago
6 0
I believe it’s be sorry if I’m wrong
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If Sophie gets $26.00 and Jack gets four times the amount as Sophie. How much do Jack get?​
alina1380 [7]

Answer:

he gets 104$

Step-by-step explanation:

104 dollars

26x4=104

5 0
3 years ago
Read 2 more answers
A function h(x) has zeroed at x=-6 and x=12 where does the function g(x) =h(3x) have zeroed?
saul85 [17]

Answer:.

Step-by-step explanation:.

7 0
3 years ago
The Baldwin family walked four over five of a mile in four over nine of an hour. What is their unit rate in miles per hour?
sineoko [7]
So walked 4/5 mile in a 4/9 h  so just use the rule of 3 simple 

60 minutes ------ 1 hour
x min. ----------- 4/9 h
--------------------------------
x = 4/9 *60/1 = 240/9 =  26,6 min. 

4/5 mile -------in 26,6 min.
1 mile --------   x min.
-----------------------------
x = 1*26,6/(4/5) = 26,6 /(0,8) = 33,25 min. 

1 mile ..... 33,25 min.
x miles ----- 60 min.
---------------------------
 x = 60/33,25 = 1,8 miles 

so their unit rate in miles per hour will be 1,8 miles / hour 

hope this will help you 
4 0
3 years ago
) For what values of k does the function y = cos(kt) satisfy the differential equation 4y 00 = −25y? (1) (b) For those values of
nordsb [41]

Answer:

a.k=\pm\frac{5}{2}

Step-by-step explanation:

We are given that a solution

y= cos (kt) satisfied the differential equation

4y''=-25y

We have to find  the value of k

a.y=coskt

Differentiate w.r.t x

Then we get

y'=-k sinkt

\frac{d(cos ax)}{dx}=-asin ax

Again differentiate w.r.t x

y''=-k^2 cos kt (\frac{d(sinax)}{dx}=a cos ax

Substitute the value in given differential equation

-4 k^2 coskt=-25 coskt

coskt cancel on both sides then we get

4k^2=25

k^2=\frac{25}{4}

a.k=\sqrt{\frac{25}{4}}=\pm\frac{5}{2}

b.We have to show that y=A sin kt + B cos kt is a solution to given differential equation for k=\pm\frac{5}{2}

Substitute the values of k

Then we get

y=A cos \frac{5}{2} kt+ B sin \frac{5}{2} kt

Differentiate w.r.t x

y'=-\frac{5}{2} A sin \frac{5}{2}t+ \frac{5}{2} B cos \frac{5}{2}t

Again differentiate w.r.t x

Then we get

y''=-\frac{25}{4}A cos \frac{5}{2} t+\frac{25}{4} B sin\frac{5}{2} t

Substitute the value of y'' and y in given differential equation

-25 A cos \frac{5}{2} t +25 B sin \frac{5}{2} t=-25(A cos \frac{5}{2} t+ B sin \frac{5}{2} t)

-25 A cos \frac{5}{2} t +25 B sin \frac{5}{2} t=-25 y

LHS=RHS

Hence, every function of the form

y=A cos kt +B sin kt is a s solution of given differential equation for k=\pm\frac{5}{2}

Where A and B are constants

5 0
3 years ago
Please someone help
Digiron [165]
Is it 0.2 repeating? I'm not sure sorry hon...
3 0
3 years ago
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