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diamong [38]
3 years ago
6

The table below shows the distance d(t) in meters that an object travels in t seconds:

Mathematics
1 answer:
mixas84 [53]3 years ago
6 0
Average change in d(t)   =    (350-126) / (5-3) =  112 m/s

This represents the average speed of the object during the 2 second period between  t = 3 and t = 5.
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Can 15/8 be simplified
Olegator [25]

No it can't

In mixed number form it's 1 7/8

3 0
3 years ago
Can someone please help me with this question? I don't know how? What is the volume of a rectangular prism with a length of 2x –
lesantik [10]
Volume is legnth times widht times height
lenght=2x-1
width=x-2
height=x+1
multiply all together
use mass distributive property
distributive=a(b+c)=ab+ac so extending that
(a+c)(c+c)=(a+b)(c)+(a+b)(d) then keep distributing so
(2x-1)(x-2)(x+1)
do each one seperately
do the first two first and put the other one (x+1) to the side for later
(2x-1)(x-2)=(2x-1)(x)+(2x-1)(-2)=(2x^2-x)+(-4x+2)=2x^2-5x+2
then do the other one
(x+1)(2x^2-5x+2)=(x)(2x^2-5x+2)+(1)(2x^2-5x+2)=(2x^3-5x^2+2x)+(2x^2-5x+2)=2x^3-3x^2-3x+2

the lasst form is 2x^3-3x^2-3x+2





3 0
3 years ago
Solve 5y'' + 3y' – 2y = 0, y(0) = 0, y'(0) = 2.8 y(t) = 0 Preview
mario62 [17]

Answer:  The required solution is

y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.

Step-by-step explanation:   We are given to solve the following differential equation :

5y^{\prime\prime}+3y^\prime-2y=0,~~~~~~~y(0)=0,~~y^\prime(0)=2.8~~~~~~~~~~~~~~~~~~~~~~~~(i)

Let us consider that

y=e^{mt} be an auxiliary solution of equation (i).

Then, we have

y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.

Substituting these values in equation (i), we get

5m^2e^{mt}+3me^{mt}-2e^{mt}=0\\\\\Rightarrow (5m^2+3y-2)e^{mt}=0\\\\\Rightarrow 5m^2+3m-2=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow 5m^2+5m-2m-2=0\\\\\Rightarrow 5m(m+1)-2(m+1)=0\\\\\Rightarrow (m+1)(5m-1)=0\\\\\Rightarrow m+1=0,~~~~~5m-1=0\\\\\Rightarrow m=-1,~\dfrac{1}{5}.

So, the general solution of the given equation is

y(t)=Ae^{-t}+Be^{\frac{1}{5}t}.

Differentiating with respect to t, we get

y^\prime(t)=-Ae^{-t}+\dfrac{B}{5}e^{\frac{1}{5}t}.

According to the given conditions, we have

y(0)=0\\\\\Rightarrow A+B=0\\\\\Rightarrow B=-A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)

and

y^\prime(0)=2.8\\\\\Rightarrow -A+\dfrac{B}{5}=2.8\\\\\Rightarrow -5A+B=14\\\\\Rightarrow -5A-A=14~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{Uisng equation (ii)}]\\\\\Rightarrow -6A=14\\\\\Rightarrow A=-\dfrac{14}{6}\\\\\Rightarrow A=-\dfrac{7}{3}.

From equation (ii), we get

B=\dfrac{7}{3}.

Thus, the required solution is

y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.

7 0
3 years ago
Which of the following statements is true for the quadratic equation x2 – 16x + 64 = 0?
irinina [24]

Answer:

Please provide the statements.

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Use compatible numbers to find two estimates 73÷4,858
zvonat [6]
Compatible numbers to find two estimates 73÷4,858

70÷4,850 = 0.014
You sure it's not 4,858÷73
4,850÷70 =

______ _____ 69 2/7
70)4850. 7)485
42
65
63
8 0
4 years ago
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