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umka2103 [35]
3 years ago
6

A student solved this problem and said the answer was 8 3/4 hours.

Mathematics
1 answer:
aleksklad [387]3 years ago
7 0
No way is this correct!
5-3/2-9/4=1 1/4
The answer is C.
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Help plz!
julia-pushkina [17]
She uses four pounds because 4/5 of five pounds is four.
4 0
4 years ago
WILL GIVE BRAINLIEST 4 FIRST ANSWER.
Fofino [41]

Answer:

c

Step-by-step explanation:

you should divide the distance on time

so

33/6=5.5

26/4=6.5

60/8=7.5

17/2=8.5

you can see answer in this order in c

5 0
3 years ago
Read 2 more answers
Write the equations of three distinct lines that have the same y-intercept,-1.
guajiro [1.7K]
Three possible answers are:
y = 2x-1
y = 5x-1
y = -3x-1

All of them have the same y intercept of -1. The slopes are the only thing that differ.

Any linear equation can be written in the form y = mx+b where
m = slope
b = y intercept
7 0
4 years ago
x = c1 cos(t) + c2 sin(t) is a two-parameter family of solutions of the second-order DE x'' + x = 0. Find a solution of the seco
igomit [66]

Answer:

x=-cos(t)+2sin(t)

Step-by-step explanation:

The problem is very simple, since they give us the solution from the start. However I will show you how they came to that solution:

A differential equation of the form:

a_n y^n +a_n_-_1y^{n-1}+...+a_1y'+a_oy=0

Will have a characteristic equation of the form:

a_n r^n +a_n_-_1r^{n-1}+...+a_1r+a_o=0

Where solutions r_1,r_2...,r_n are the roots from which the general solution can be found.

For real roots the solution is given by:

y(t)=c_1e^{r_1t} +c_2e^{r_2t}

For real repeated roots the solution is given by:

y(t)=c_1e^{rt} +c_2te^{rt}

For complex roots the solution is given by:

y(t)=c_1e^{\lambda t} cos(\mu t)+c_2e^{\lambda t} sin(\mu t)

Where:

r_1_,_2=\lambda \pm \mu i

Let's find the solution for x''+x=0 using the previous information:

The characteristic equation is:

r^{2} +1=0

So, the roots are given by:

r_1_,_2=0\pm \sqrt{-1} =\pm i

Therefore, the solution is:

x(t)=c_1cos(t)+c_2sin(t)

As you can see, is the same solution provided by the problem.

Moving on, let's find the derivative of x(t) in order to find the constants c_1 and c_2:

x'(t)=-c_1sin(t)+c_2cos(t)

Evaluating the initial conditions:

x(0)=-1\\\\-1=c_1cos(0)+c_2sin(0)\\\\-1=c_1

And

x'(0)=2\\\\2=-c_1sin(0)+c_2cos(0)\\\\2=c_2

Now we have found the value of the constants, the solution of the second-order IVP is:

x=-cos(t)+2sin(t)

3 0
3 years ago
Please help ill give brainiest answer
trapecia [35]

Answer:

1: AAS, RQC 2: ASA, SRP

Step-by-step explanation:

(1) We are shown that two angles and one side are congruent in the order of AAS. Make sure you write the letters in terms of the corresponding angles. The angles and sides are congruent because the problem labels it for us. For example, B,A,C=R,Q,S. Answers: AAS, RQC.

(2) We are shown that two angles and one side are congruent in the order of ASA. Make sure you write the letters in terms of the corresponding angles again. For example, P,Q,R=P,S,R. The angles are congruent because the problem labels it for us. Side PR is congruent to side PR by reflexive property. Answers: ASA, SRP.

I hope this helped :) Good luck

4 0
3 years ago
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