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3 years ago

1/29+12/58+12/29=

1 answer:

5 0

Just find a common denominator and find the smallest number that goes into it, which in this case it is 29, so
1/29+6/29+12/29=19/29

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Write an equation for the line that passes through (4,-6) and (-7,-6). Is the line parallel or perpendicular to the y-axis?

alisha [4.7K]

Answer:

y = -6, perpendicular

Step-by-step explanation:

You can use the points to find the slope of the line:

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} =\frac{-6-(-6)}{-7-4} =\frac{-6+6}{-7-4}=\frac{0}{-11}=0$

Then, using point-slope form, choose one set of coordinates to use for y1 and x1:

$y-y_{1}=m(x-x_{1})$

$y-(-6)=0(x-4)$

$y+6=0$

$y=-6$

Since the slope of the line is zero, it's a horizontal line. The y-axis is a vertical line, which means that the line is perpendicular to the y-axis.

7 0

3 years ago

LaToya has $113.94 in her checking account. During the week she goes to an ATM and withdraws $40. She is charged a usage fee of

Scilla [17]

113.94-42.50+189.73-22.50-70.18= $168.49

LaToya has $168.49 in her checking account

8 0

3 years ago

Read 2 more answers

Could yall help me PLEASE

klasskru [66]

Answer: 3.6

Step-by-step explanation:

(Not the best at math but I am an above level student)

C= pi x diameter

C= pi x 1.205

C= 3.7

3 0

2 years ago

Y''+y'+y=0, y(0)=1, y'(0)=0

mars1129 [50]

Answer:

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Step-by-step explanation:

A second order linear , homogeneous ordinary differential equation has form $ay''+by'+cy=0$ .

Given: $y''+y'+y=0$

Let $y=e^{rt}$ be it's solution.

We get,

$\left ( r^2+r+1 \right )e^{rt}=0$

Since $e^{rt}\neq 0$ , $r^2+r+1=0$

{ we know that for equation $ax^2+bx+c=0$ , roots are of form $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ }

We get,

$y=\frac{-1\pm \sqrt{1^2-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}$

For two complex roots $r_1=\alpha +i\beta \,,\,r_2=\alpha -i\beta$ , the general solution is of form $y=e^{\alpha t}\left ( c_1\cos \beta t+c_2\sin \beta t \right )$

i.e $y=e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Applying conditions y(0)=1 on $e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$ , $c_1=1$

So, equation becomes $y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

On differentiating with respect to t, we get

$y'=\frac{-1}{2}e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )+e^{\frac{-t}{2}}\left ( \frac{-\sqrt{3}}{2} \sin \left ( \frac{\sqrt{3}t}{2} \right )+c_2\frac{\sqrt{3}}{2}\cos\left ( \frac{\sqrt{3}t}{2} \right )\right )$

Applying condition: y'(0)=0, we get $0=\frac{-1}{2}+\frac{\sqrt{3}}{2}c_2\Rightarrow c_2=\frac{1}{\sqrt{3}}$

Therefore,

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

3 0

3 years ago

Which equation could have been used to create this graph?

pishuonlain [190]

B. The y-intercept is 0 so it can't be C or D and the slope is 5/1. How I remember this is rise/run. Hope this helps!

3 0

3 years ago