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3 years ago

Another Math Question!

Mathematics

1 answer:

Tresset [83]3 years ago

3 0

I did this equation and it’s A

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6 3/10 divided by 9/5

ankoles [38]

60/3 ÷ 9/5
60/3 × 5/9
300/27
= 100/9
= $11 \frac{1}{9}$

6 0

3 years ago

Read 2 more answers

(Differential Equations) Put these equations in explicit form:

Nuetrik [128]

Answer:

1. $y= C(t*exp(-\frac{t^{2} }{2(2!)}+\frac{t^{4} }{4(4!)}-\frac{t^{6} }{6(6!)}+... ))}$

2. $t+C=ln(csc(y)-cot(y)$ $23. [tex]Assuming t as independent variable:F(r,t)=t+\frac{1}{m} exp(m+r)+\frac{r^{2} }{2} =C\\Step-by-step explanation:1. Separable variables:[tex]\frac{dy}{dt}=\frac{y*cos(t) }{t}\\ \frac{dy}{y}= \frac{cos(t) }{t}dt\\ \int {\frac{dy }{y}} \, dt=\int {\frac{cos(t) }{t}} \, dt \\ln(y)-ln(C)=ln(t)-\frac{t^{2} }{2(2!)} +\frac{t^{4} }{4(4!)} -\frac{t^{6} }{6(6!)}+... \\y=C(t*exp(\frac{t^{2} }{2(2!)} +\frac{t^{4} }{4(4!)} -\frac{t^{6} }{6(6!)}+...))$

2. Separable variables

\frac{dy}{sin(y)}=dt\\ \int\ \frac{1}{sin(y)}} \, dy = \int\ 1} \, dt\\t+C=ln(csc(y)-cot(y))[/tex]

3. Homogeneous D.E

Rewriting:

$dr+(\frac{1}{m} exp(m+r)+r)dt=0\\\frac{dF}{dt}=1 -> F(r,t)=t+C(r)\\\frac{dF}{dy}=0+C'(r)= \frac{1}{m} exp(m+r)+r -> C(r)=\frac{1}{m} exp(m+r)+\frac{r^{2} }{2} \\F(r,t)=t+\frac{1}{m} exp(m+r)+\frac{r^{2} }{2} =C\\$

7 0

3 years ago

What is the sum of the first 27 terms of the following sequence?

FromTheMoon [43]

Answer choices............................

5 0

3 years ago

What is the area of the figure?

Paladinen [302]

What you would need to do, is add the area of all the rectangles. The formula is wl or width * length. So find the are of all 3 rectangles, starting from the top:

5*2 = 10

3*2 = 6

5(because the rectangles are the same length)*2=10

10+10+6=26. The area of your rectangle is 26 units

3 0

3 years ago

Determine whether the equation represents a direct variation 2y=5x+1

olga nikolaevna [1]

<span>2y/2=5x+1/2
y=5x/2+1/2
no it's not a direct variation because of 1/2</span>

8 0

3 years ago