All categories

2 years ago

Mal spends 7 hours in school each day. Her lunch period is 1 hour long, and she spends a

Mathematics

1 answer:

Alex787 [66]2 years ago

8 0

Answer:

z = 5*(1/2)

z = 5/10

---

time switching classes:

w = 7/10

---

y - 6x - z - w = 0

6x = y - z - w

x = (y - z - w)/6

x = (76/10 - 5/10 - 7/10)/6

x = (76 - 5 - 7)/(10*6)

x = (64)/(10*6)

x = (2*2*2*2*2*2)/(2*5*2*3)

x = (2*2*2*2)/(5*3)

x = 16/15

x = 1.0666666666

---

check:

y = 7 + 3/5

y = 7.6

z = 1/2

z = 0.5

w = 7/10

w = 0.7

y - 6x - z - w = 0

6x = y - z - w

x = (y - z - w)/6

x = (7.6 - 0.5 - 0.7)/6

x = 1.0666666666

---

answer:

z = 5*(1/2)

z = 5/10

---

time switching classes:

w = 7/10

---

y - 6x - z - w = 0

6x = y - z - w

x = (y - z - w)/6

x = (76/10 - 5/10 - 7/10)/6

x = (76 - 5 - 7)/(10*6)

x = (64)/(10*6)

x = (2*2*2*2*2*2)/(2*5*2*3)

x = (2*2*2*2)/(5*3)

x = 16/15

x = 1.0666666666

---

check:

y = 7 + 3/5

y = 7.6

z = 1/2

z = 0.5

w = 7/10

w = 0.7

y - 6x - z - w = 0

6x = y - z - w

x = (y - z - w)/6

x = (7.6 - 0.5 - 0.7)/6

x = 1.0666666666

---

answer:

each class is 1.07 hours

Step-by-step explanation:

You might be interested in

A bag is filled with green and blue marbles. There are 125 marbles in the bag. If there are 29 more green

ValentinkaMS [17]

Answer:

89 green marbles and 46 blue marbles

Step-by-step explanation:

3 0

2 years ago

Seven plus seven equal to

vazorg [7]

Answer:

the answer is 14

Step-by-step explanation:

7+7=14

I hope this help

3 0

2 years ago

Read 2 more answers

Solve y'' + 10y' + 25y = 0, y(0) = -2, y'(0) = 11 y(t) = Preview

svetlana [45]

Answer: The required solution is

$y=(-2+t)e^{-5t}.$

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

$y^{\prime\prime}+10y^\prime+25y=0,~~~~~~~y(0)=-2,~~y^\prime(0)=11~~~~~~~~~~~~~~~~~~~~~~~~(i)$

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

$m^2e^{mt}+10me^{mt}+25e^{mt}=0\\\\\Rightarrow (m^2+10y+25)e^{mt}=0\\\\\Rightarrow m^2+10m+25=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow m^2+2\times m\times5+5^2=0\\\\\Rightarrow (m+5)^2=0\\\\\Rightarrow m=-5,-5.$

So, the general solution of the given equation is

$y(t)=(A+Bt)e^{-5t}.$

Differentiating with respect to t, we get

$y^\prime(t)=-5e^{-5t}(A+Bt)+Be^{-5t}.$

According to the given conditions, we have

$y(0)=-2\\\\\Rightarrow A=-2$

and

$y^\prime(0)=11\\\\\Rightarrow -5(A+B\times0)+B=11\\\\\Rightarrow -5A+B=11\\\\\Rightarrow (-5)\times(-2)+B=11\\\\\Rightarrow 10+B=11\\\\\Rightarrow B=11-10\\\\\Rightarrow B=1.$

Thus, the required solution is

$y(t)=(-2+1\times t)e^{-5t}\\\\\Rightarrow y(t)=(-2+t)e^{-5t}.$

6 0

3 years ago

Enter the value of p so that the expression 4 (n+8) is equivalent to (n+p)4

Vladimir [108]

Sick my dock is the answer

7 0

2 years ago

HCF AND LCM of 32 and 48 using the prime factorisation

torisob [31]

Answer:

Below.

Step-by-step explanation:

32 = 2*2*2*2*2

48 = 2*2*2*2*3

So The GCF = 2*2*2*2 = 16.

LCM = 2*2*2*2*2*3 = 96

3 0

2 years ago