Good job, you got the equations! XD
I'll just help you solve
Multiply the 2nd row by 9
<span>9x+9y=117</span>
<span><span><span>9x+27y=207
</span></span></span>Subtract the 2nd row from the 1st row
<span><span><span>−18y=−90
</span></span></span>Divide both sides by <span>−18</span>
<span><span><span>y=<span><span>−90/</span><span>−18
</span></span></span></span>Two negatives make a positive
<span><span>y=<span><span>90/</span><span>18</span></span></span><span></span></span><span>
y = 5
Substitute 5 into an equation
9x+9(5)=11<span>7
</span>9x+45=11<span>7
</span>Subtract <span>45</span> from both sides
<span><span><span>9x=117−45
</span></span></span>9x=7<span>2
</span>Divide both sides by <span><span>9</span></span>
x = ___
Hope this will help</span></span>
Y= 5x
x+y= 245
X+ (5X) = 245
6X = 245
divide both sides by six and their is your answer :)
good luck! and message me if you have any more questions and you want to learn and not cheat you way out of it
Answer:
WHAT DOES BRAINLEST MEAN CAN SOMEONE TELL ME???
Step-by-step explanation:
Answer:
12 cm
Step-by-step explanation:
To calculate the length of a spring with a 2 kg load, compare the displacement of a 1 kg load and adjust accordingly.
When a 1 kg load is suspended from the spring, the spring which is 6 cm stretches to 9 cm. This is 3 cm longer due to the weight. If you attach a weight which is twice as much then the displacement will be twice as much. Instead of stretching an additional 3 cm, it will stretch 2*3 = 6 cm. Add this to the length of the spring and it stretches in total 6 + 6 = 12 cm.
The acceleration of the particle is given by the formula mentioned below:
Differentiate the position vector with respect to t.
![\begin{gathered} \frac{ds(t)}{dt}=\frac{d}{dt}\sqrt[]{\mleft(t^3+1\mright)} \\ =-\frac{1}{2}(t^3+1)^{-\frac{1}{2}}\times3t^2 \\ =\frac{3}{2}\frac{t^2}{\sqrt{(t^3+1)}} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Cfrac%7Bds%28t%29%7D%7Bdt%7D%3D%5Cfrac%7Bd%7D%7Bdt%7D%5Csqrt%5B%5D%7B%5Cmleft%28t%5E3%2B1%5Cmright%29%7D%20%5C%5C%20%3D-%5Cfrac%7B1%7D%7B2%7D%28t%5E3%2B1%29%5E%7B-%5Cfrac%7B1%7D%7B2%7D%7D%5Ctimes3t%5E2%20%5C%5C%20%3D%5Cfrac%7B3%7D%7B2%7D%5Cfrac%7Bt%5E2%7D%7B%5Csqrt%7B%28t%5E3%2B1%29%7D%7D%20%5Cend%7Bgathered%7D)
Differentiate both sides of the obtained equation with respect to t.
![\begin{gathered} \frac{d^2s(t)}{dx^2}=\frac{3}{2}(\frac{2t}{\sqrt[]{(t^3+1)}}+t^2(-\frac{3}{2})\times\frac{1}{(t^3+1)^{\frac{3}{2}}}) \\ =\frac{3t}{\sqrt[]{(t^3+1)}}-\frac{9}{4}\frac{t^2}{(t^3+1)^{\frac{3}{2}}} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Cfrac%7Bd%5E2s%28t%29%7D%7Bdx%5E2%7D%3D%5Cfrac%7B3%7D%7B2%7D%28%5Cfrac%7B2t%7D%7B%5Csqrt%5B%5D%7B%28t%5E3%2B1%29%7D%7D%2Bt%5E2%28-%5Cfrac%7B3%7D%7B2%7D%29%5Ctimes%5Cfrac%7B1%7D%7B%28t%5E3%2B1%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%29%20%5C%5C%20%3D%5Cfrac%7B3t%7D%7B%5Csqrt%5B%5D%7B%28t%5E3%2B1%29%7D%7D-%5Cfrac%7B9%7D%7B4%7D%5Cfrac%7Bt%5E2%7D%7B%28t%5E3%2B1%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%20%5Cend%7Bgathered%7D)
Substitute t=2 in the above equation to obtain the acceleration of the particle at 2 seconds.
![\begin{gathered} a(t=1)=\frac{3}{\sqrt[]{2}}-\frac{9}{4\times2^{\frac{3}{2}}} \\ =1.32ft/sec^2 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20a%28t%3D1%29%3D%5Cfrac%7B3%7D%7B%5Csqrt%5B%5D%7B2%7D%7D-%5Cfrac%7B9%7D%7B4%5Ctimes2%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%20%5C%5C%20%3D1.32ft%2Fsec%5E2%20%5Cend%7Bgathered%7D)
The initial position is obtained at t=0. Substitute t=0 in the given position function.
