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3 years ago

Giving 50 points to anyone who answers these 2 questions im having trouble with! (3 and 4)

Mathematics

2 answers:

Zielflug [23.3K]3 years ago

8 0

Problem 3:

We basically find the Least Common Multiple of the numbers 5 and 11 and add the number of minutes to 6:00.

LCM of (5, 11)= 55

Answer: 6:55

Problem 4:

$\frac{7}{8}$ divided by $\frac{4}{5}$ equals $\frac{7}{10}$

denis-greek [22]3 years ago

4 0

Answer:

1. 6:55 2. 7/10

Step-by-step explanation:

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cable has 2 1/5 yard of rope. he uses 3/4 of the rope to make a dog leash. which expression can be used to represent 3/4 of 21/5

Greeley [361]

Answer:

Step-by-step explanation:

Thank me later

7 0

3 years ago

PLEASE HELP! NEED TO SUBMIT SOON! NOT SURE IF I GOT THE RIGHT ANSWERS! THANK YOU IN ADVANCE!!!

Eddi Din [679]

Nice

4s=amount of grams of protine in straberry for s ounces
3b=amount of grams of protine in banana for b ouces

total ounces is 6
s+b=6

total grams is 22
4s+3b=22

s+b=6
4s+3b=22
those are the 2 equations

a.
s+b=6
4s+3b=22

b. elmination, we can elminate alll the b's and have 1s left

c.
multiply first equaiton by -3 and add to second

4s+3b=22
<u>-3s-3b=-18 +</u>
1s+0b=4
s=4
sub
s+b=6
4+b=6
minus 4
b=2

4 oz straberry
2 oz banana

8 0

3 years ago

Point M belongs to AN , point N belongs to BM , AB = 18 in and AM : MN : NB = 1:2:3. Find MN.

yulyashka [42]

MN = 6 in

add the parts of the ratio, 1 + 2 + 3 = 6

divide AB by 6 to find 1 part of the ratio

$\frac{18}{6}$ = 3 ← 1 part

AM = 3 in ← 1 part

MN = 2 × 3 = 6 in ← 2 parts

NB = 3 × 3 = 9 in ← 3 parts

and 3 + 6 + 9 = 18 in = AB

3 0

3 years ago

Pete walks 6 miles during each trip to school. How many trips does he need to make in order to make in a total of 48 miles?

Diano4ka-milaya [45]

Answer:

8 trips

Step-by-step explanation:

ALl you have to do is divide 48 by 6, which will give you 8, and there's your answer!

8 0

3 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