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3 years ago

Someone help and please make sure the answer is right

Mathematics

1 answer:

yarga [219]3 years ago

8 0

Answer:

-10

Step-by-step explanation:

Hope this helps you!

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The value of a car is $30,000. It loses 6.5% of its value each year. What will the value of the car be after 5 years?

Kitty [74]

Decay equation= P*(1-r)^t
P= Principal amount= $30,000
r= rate=0.065
t= time (in years) =5
x=30,000(1-0.065)^5
The value of the car in 5 years is approx. $21,438

3 0

4 years ago

LINEAR ALGEBRA

kenny6666 [7]

Answer:

The value of the constant $k$ so that $\vec u_{3}$ is a linear combination of $\vec u_{1}$ and $\vec u_{2}$ is $\frac{7}{10}$ .

Step-by-step explanation:

Let be $\vec u_{1} = [2,3,1]$ , $\vec u_{2} = [4,1,0]$ and $\vec u_{3} = [1, 2,k]$ , $\vec u_{3}$ is a linear combination of $\vec u_{1}$ and $\vec u_{3}$ if and only if:

$\alpha_{1} \cdot \vec u_{1} + \alpha_{2} \cdot \vec u_{2} +\alpha_{3}\cdot \vec u_{3} = \vec O$ (Eq. 1)

Where:

$\alpha_{1}$ , $\alpha_{2}$ , $\alpha_{3}$ - Scalar coefficients of linear combination, dimensionless.

By dividing each term by $\alpha_{3}$ :

$\lambda_{1}\cdot \vec u_{1} + \lambda_{2}\cdot \vec u_{3} = -\vec u_{3}$

$\vec u_{3}=-\lambda_{1}\cdot \vec u_{1}-\lambda_{2}\cdot \vec u_{2}$ (Eq. 2)

$\vec O$ - Zero vector, dimensionless.

And all vectors are linearly independent, meaning that at least one coefficient must be different from zero. Now we expand (Eq. 2) by direct substitution and simplify the resulting expression:

$[1,2,k] = -\lambda_{1}\cdot [2,3,1]-\lambda_{2}\cdot [4,1,0]$

$[1,2,k] = [-2\cdot\lambda_{1},-3\cdot \lambda_{1},-\lambda_{1}]+[-4\cdot \lambda_{2},-\lambda_{2},0]$

$[0,0,0] = [-2\cdot \lambda_{1},-3\cdot \lambda_{1},-\lambda_{1}]+[-4\cdot \lambda_{2},-\lambda_{2},0]+[-1,-2,-k]$

$[-2\cdot \lambda_{1}-4\cdot \lambda_{2}-1,-3\cdot \lambda_{1}-\lambda_{2}-2,-\lambda_{1}-k] =[0,0,0]$

The following system of linear equations is obtained:

$-2\cdot \lambda_{1}-4\cdot \lambda_{2}= 1$ (Eq. 3)

$-3\cdot \lambda_{1}-\lambda_{2}= 2$ (Eq. 4)

$-\lambda_{1}-k = 0$ (Eq. 5)

The solution of this system is:

$\lambda_{1} = -\frac{7}{10}$ , $\lambda_{2} = \frac{1}{10}$ , $k = \frac{7}{10}$

The value of the constant $k$ so that $\vec u_{3}$ is a linear combination of $\vec u_{1}$ and $\vec u_{2}$ is $\frac{7}{10}$ .

4 0

4 years ago

Alisha bought a shirt as a gift for russell's birthday. The shirt normally costs $39 and it was on sale for $19 what was the per

lara31 [8.8K]

it would be about 51 or 52 percent exact but if you have to round about 50 percent was taken off the shirt.

EXPLANATION: 39 times 0.5 = 19.5

39- 19.5 = 19.50

HOPE THIS HELPS

6 0

3 years ago

Find the X intercept and the Y intercept of the graph of the equation. 9/8x+8y=18

aleksandr82 [10.1K]

Answer:

$\large\boxed{x-intercept=16\to(16,\ 0)}\\\boxed{y-intercept=\dfrac{9}{4}\to\left(0,\ \dfrac{9}{4}\right)}$

Step-by-step explanation:

$\dfrac{9}{8}x+8y=18\\\\x-intercept\ \text{is for}\ y=0:\\\\\dfrac{9}{8}x+8(0)=18\\\\\dfrac{9}{8}x+0=18\\\\\dfrac{9}{8}x=18\qquad\text{multiply both sides by}\ \dfrac{8}{9}\\\\\dfrac{8\!\!\!\!\diagup^1}{9\!\!\!\!\diagup_1}\cdot\dfrac{9\!\!\!\!\diagup^1}{8\!\!\!\!\diagup_1}x=\dfrac{8}{9\!\!\!\!\diagup_1}\cdot18\!\!\!\!\!\diagup^2\\\\x=16\\\\y-intercept\ \text{is for}\ x=0:\\\\\dfrac{9}{8}(0)+8y=18\\\\0+8y=18\\\\8y=18\qquad\text{divide both sides by 8}\\\\y=\dfrac{18}{8}\\\\y=\dfrac{9}{4}$

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3 years ago

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Step-by-step explanation:

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3 years ago