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2 years ago

Help i don’t get it

Mathematics

1 answer:

Sliva [168]2 years ago

3 0

Answer:

2.2+3(3.4)

=2.2+10.2

=12.4

Step-by-step explanation:

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Simplify the expression:<br> 10q–3q+2q2+3

saw5 [17]

If the 2q is raised to the second power the answer is 2q^2+7q+3

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2 years ago

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Kate, Alexia, and Trina just completed the 400-meter dash at the track meet. Kate finished the race 6 seconds faster than Alexia

Digiron [165]

Answer:

Kate = 52 seconds, Alexia = 58 seconds, Trina = 49 seconds

Step-by-step explanation:

A = K + 6

T = K - 3

K + A + T = 2 min 39 sec or 159 sec

K + (K + 6) + (K - 3) = 159sec

3K + 3 = 159sec

3K = 156sec

K = 52 sec

A = 52 + 6 = 58 sec

T = 52 - 3 = 49 sec

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2 years ago

Including an 18% tip, Ms. Guy and her friends spent $74.93 on dinner. What was the cost of the dinner excluding the tip?

12345 [234]

Answer:

61.45

Step-by-step explanation:

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2 years ago

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Evaluate the integral. (remember to use absolute values where appropriate. Use c for the constant of integration.) 5 cot5(θ) sin

ozzi

$I=5\int \frac{cos^{4}\theta }{sin\theta }\times cos\theta d\theta \\\\I=5\int \left ( 1-sin^{2}\theta \right )^{2}\times \frac{cos\theta }{sin\theta }d\theta \\put\ \sin\theta =t\\\\dt=cos\theta d\theta \\\\I=5\int\frac{t^{4}+1-2t^{2}}{t}dt\ \ \ \ \ \ \ \ \ \ \because (a-b)^2=a^2+b^2-2ab\\\\I=5\left ( \int t^{3}dt + \int \frac{1}{t} -2\int t \right )dt$

by using the integration formula

we get,

$\\I=5\left ( \frac{t^{4}}{4} +logt -t^{2}\right )\\\\I=\frac{5}{4}t^{4}+5\log t-5t^{2}+c$

now put the value of t=\sin\theta in the above equation

we get,

$\int 5\cot^5\theta \sin^4\theta d\theta=\frac{5}{4}sin^{4}\theta+5\log \sin\theta - 5sin^{2} \theta+c$

hence proved

7 0

2 years ago

Suppose z equals f (x comma y ), where x (u comma v )space equals space 2 u plus space v squared, y (u comma v )space equals spa

barxatty [35]

$z=f(x(u,v),y(u,v)),\begin{cases}x(u,v)=2u+v^2\\y(u,v)=3u-v\end{cases}$

We're given that $f_x(6,1)=3$ and $f_y(6,1)=-1$ , and want to find $\frac{\partial z}{\partial v}(1,2)$ .

By the chain rule, we have

$\dfrac{\partial z}{\partial v}=\dfrac{\partial z}{\partial x}\dfrac{\partial x}{\partial v}+\dfrac{\partial z}{\partial y}\dfrac{\partial y}{\partial v}$

and

$\dfrac{\partial x}{\partial v}=2v$

$\dfrac{\partial y}{\partial v}=-1$

Then

$\dfrac{\partial z}{\partial v}(1,2)=\dfrac{\partial z}{\partial x}(6,1)\dfrac{\partial x}{\partial v}(1,2)+\dfrac{\partial z}{\partial y}(6,1)\dfrac{\partial y}{\partial v}(1,2)$

(because the point $(x,y)=(6,1)$ corresponds to $(u,v)=(1,2)$ )

$\implies\dfrac{\partial z}{\partial v}(1,2)=3\cdot2\cdot2+(-1)\cdot(-1)=\boxed{13}$

4 0

2 years ago