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3 years ago

13

rding to linear pattern. If the initial value of the vehicle is $26,000, and the value ten years later is $0, what will the depr

eciated value be after 5 years

1 answer:

Verdich [7]3 years ago

3 0

The answer will be 5,200

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Help me. I don't understand. Please help me find X and IK

sertanlavr [38]

Answer:

$x=4\\\\y=6\\\\IK=47$

Step-by-step explanation:

Two Tangent Theorem: Tangents which meet at same point are equal in length.

Here tangents at J and H meet at I.

$Hence\ HI=IJ\\\\y^2-10=4y+2\\\\y^2-4y-12=0\\\\y^2-6y+2y-12=0\\\\y(y-6)+2(y-6)=0\\\\(y-6)(y+2)\\\\y\ can\ not\ be\ negative\ as\ in\ that\ case\ IJ\ will\ be\ negative.\\\\Hence\ y=6\\\\IJ=4y+2=4\times 6+2=26\\\\\\\\Tangents\ at\ J\ and\ L\ meet\ at\ K.\ Use\ two\ tangent\ theorem.\\\\JK=KL\\\\5x+1=6x-3\\\\6x-5x=1+3\\\\x=4\\\\JK=5x+1=5\times 4+1=21\\\\\\IK=IJ+JK=26+21\\\\IK=47$

6 0

3 years ago

Which expression correctly represents “three less than the product of a number and two, increased by five”?

Tju [1.3M]

The answer is 3-n2-5

5 0

3 years ago

Read 2 more answers

Use the Laplace transform to solve differential equation

Inessa [10]

Using Laplace transform we have:L(x')+7L(x) = 5L(cos(2t))sL(x)-x(0) + 7L(x) = 5s/(s^2+4)(s+7)L(x)- 4 = 5s/(s^2+4)(s+7)L(x) = (5s - 4s^2 -16)/(s^2+4)
=> L(x) = -(4s^2 - 5s +16)/(s^2+4)(s+7)
now the boring part, using partial fractions we separate 1/(s^2+4)(s+7) that is:(7-s)/[53(s^2+4)] + 1/53(s+7). So:
L(x)= (1/53)[(-28s^2+4s^3-4s^2+35s-5s^2+5s)/(s^2+4) + (-4s^2+5s-16)/(s+7)]L(x)= (1/53)[(4s^3 -37s^2 +40s)/(s^2+4) + (-4s^2+5s-16)/(s+7)]
denoting T:= L^(-1)and x= (4/53) T(s^3/(s^2+4)) - (37/53)T(s^2/(s^2+4)) +(40/53) T(s^2+4)-(4/53) T(s^2/s+7) +(5/53)T(s/s+7) - (16/53) T(1/s+7)

4 0

3 years ago

Two architects bid for a small government project, both

blondinia [14]

Answer:

10 hours

Step-by-step explanation:

let x be the total number of hours

Amazing design total charge = 45x+250

super structures total charge = 60x+100

both companies charge the same amount when both equation equals

45x+250=60x+100

60x-45x=250-100

15x=150

x=10

therefore at 10 hours both companies will charge the same amount

4 0

2 years ago

Help!! 50 points and brainliest!

Viktor [21]

Answer:

Second choice:

$x=2t$

$y=4t^2+4t-3$

Fifth choice:

$x=t+1$

$y=t^2+4t$

Step-by-step explanation:

Let's look at choice 1.

$x=t+1$

$y=t^2+2t$

I'm going to subtract 1 on both sides for the first equation giving me $x-1=t$ . I will replace the $t$ in the second equation with this substitution from equation 1.

$y=(x-1)^2+2(x-1)$

Expand using the distributive property and the identity $(u+v)^2=u^2+2uv+v^2$ :

$y=(x^2-2x+1)+(2x-2)$

$y=x^2+(-2x+2x)+(1-2)$

$y=x^2+0+-1$

$y=x^2$

So this not the desired result.

Let's look at choice 2.

$x=2t$

$y=4t^2+4t-3$

Solve the first equation for $t$ by dividing both sides by 2:

$t=\frac{x}{2}$ .

Let's plug this into equation 2:

$y=4(\frac{x}{2})^2+4(\frac{x}{2})-3$

$y=4(\frac{x^2}{4})+2x-3$

$y=x^2+2x-3$

This is the desired result.

Choice 3:

$x=t-3$

$y=t^2+2t$

Solve the first equation for $t$ by adding 3 on both sides:

$x+3=t$ .

Plug into second equation:

$y=(x+3)^2+2(x+3)$

Expanding using the distributive property and the earlier identity mentioned to expand the binomial square:

$y=(x^2+6x+9)+(2x+6)$

$y=(x^2)+(6x+2x)+(9+6)$

$y=x^2+8x+15$

Not the desired result.

Choice 4:

$x=t^2$

$y=2t-3$

I'm going to solve the bottom equation for $t$ since I don't want to deal with square roots.

Add 3 on both sides:

$y+3=2t$

Divide both sides by 2:

$\frac{y+3}{2}=t$

Plug into equation 1:

$x=(\frac{y+3}{2})^2$

This is not the desired result because the $y$ variable will be squared now instead of the $x$ variable.

Choice 5:

$x=t+1$

$y=t^2+4t$

Solve the first equation for $t$ by subtracting 1 on both sides:

$x-1=t$ .

Plug into equation 2:

$y=(x-1)^2+4(x-1)$

Distribute and use the binomial square identity used earlier:

$y=(x^2-2x+1)+(4x-4)$

$y=(x^2)+(-2x+4x)+(1-4)$

$y=x^2+2x+-3$

$y=x^2+2x-3$ .

This is the desired result.

3 0

3 years ago

Read 2 more answers