Ask question

Ask question

All categories

VashaNatasha [74]

3 years ago

13

How can proportional relationships be identified in tables and graphs?

1 answer:

makkiz [27]3 years ago

3 0

Answer:

A proportional relationship can be represented in different ways: a ratio table, a graph of a straight line through the origin, or an equation of the form y = mx, where m is the constant of proportionality

You might be interested in

47 of the birthday cake was eaten on your birthday. The next day your dad ate 23 of what was left. You are given what your dad l

Lyrx [107]

I think the best approach is assuming in percentage. if the cake= 100, then just add the 47 and 23. that would equal 70. then, subtract the 70 from the 100. that would be 30 as your final answer.
« 47+23+?=100 »
« 70+?=100 »
-70 -70
?=30

5 0

2 years ago

Who would be taller kim 56.03 or alexa 56.3

erma4kov [3.2K]

I think Alexa would be since .3 is a larger number than .03

8 0

3 years ago

Read 2 more answers

A walk in interview 12% of Candidates can be selected and 28% of Canada can be put on hold for the next hiring date if 75 candid

Vikentia [17]

Answer:

45 candidates are expected to be rejected

Step-by-step explanation:

Here, we want to calculate the number of candidates to be rejected

The number of candidates to be rejected can be calculated from;

100% - (28% + 12%) = 60%

So 60% of the 75 interviewed are to be rejected

Thus, we have

60/100 * 75

= 45

4 0

3 years ago

Read 2 more answers

How would you describe the slice that created the triangular cross section on this cake

ahrayia [7]

perpendicular to the base

5 0

3 years ago

Read 2 more answers

Find the general solution of<br><br> dz/dt=5ze^(sint)cost+2e^(sint)cost

gizmo_the_mogwai [7]

Answer:

The general solution of given differential equation is $\frac{1}{5}ln|5z+2|=e^{\sin t}+C$ .

Step-by-step explanation:

The given differential equation is

$\frac{dz}{dt}=5ze^{\sin t}\cos t+2e^{\sin t}\cos t$

Taking out common factors.

$\frac{dz}{dt}=(5z+2)e^{\sin t}\cos t$

Using variable separable method, we get

$\frac{dz}{5z+2}=e^{\sin t}\cos t dt$

Integrate both sides.

$\int\frac{dz}{5z+2}=\int e^{\sin t}\cos t dt$

$I_1=I_2$ .... (1)

First solve LHS,

$I_1=\int\frac{dz}{5z+2}$

Substitute $5z+2=u$

$5\frac{dz}{du}=1$

$dz=\frac{1}{5}du$

$I_1=\frac{1}{5}\int\frac{du}{u}$

$I_1=\frac{1}{5}ln|u|+C_1$

$I_1=\frac{1}{5}ln|5z+2|+C_1$

Now, solve RHS,

$I_2=\int e^{\sin t}\cos t dt$

Substitute $\sin t=v$ ,

$\cos t\frac{dt}{dv}=1$

$\cos tdt=dv$

$I_2=\int e^{v}dv$

$I_2=e^{v}+C_2$

$I_2=e^{\sin t}+C_2$

Subtitle the values of I₁ and I₂ in equation (1).

$\frac{1}{5}ln|5z+2|+C_1=e^{\sin t}+C_2$

$\frac{1}{5}ln|5z+2|=e^{\sin t}+C_2-C_1$

$\frac{1}{5}ln|5z+2|=e^{\sin t}+C$

Therefore the general solution of given differential equation is $\frac{1}{5}ln|5z+2|=e^{\sin t}+C$ .

4 0

4 years ago