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3 years ago

15

Como se hace este trabajo

1 answer:

Orlov [11]3 years ago

8 0

14. 11
15. 1
16. 6
17. 0

I'm not exactly sure what your exact question was so i just answered the document in the question. It's simple math really. I'm an AP Math student. Then again, I could still be wrong since I do not know what it is you're learning.

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I’m on a test so I need help

sladkih [1.3K]

The answer is 2 y2 -9

4 0

3 years ago

Expand this question: 6(x - 4y)

user100 [1]

6x-24y is the answer

8 0

2 years ago

How many solutions does the equation have?

Salsk061 [2.6K]

Answer: there is only one solution

Step-by-step explanation:

Combine like terms by performing the opposite operation of subtracting 4x on both sides of the equation

The 4x's will cross out on the right

4x - 4x = 0x = 0

On the left:

2x - 4x = -2x

Now the equation looks like:

-2x + 3 = 2

Continue to combine like terms by subtracting 3 on both sides of the equation

On the left:

3 - 3 = 0

On the right:

2 - 3 = -1

Equation:

-2x = -1

Isolate x by performing the opposite operation of dividing -2 on both sides of the equation

On the left:

-2x ÷ -2 = 1

On the right:

-1 ÷ -2 = 1/2

x= 1/2

4 0

3 years ago

Read 2 more answers

Solve 9x^2- 17x- 85 = 0 Give your solutions correct to 3 significant figures.

dybincka [34]

Answer:

$x =\frac{17}{18} +\frac{1}{18} \sqrt{3349}$ OR $x =\frac{17}{18} +\frac{-1}{18} \sqrt{3349}$

Step-by-step explanation:

9x2−17x−85=0

For this equation: a=9, b=-17, c=-85

9x2+−17x+−85=0

Step 1: Use quadratic formula with a=9, b=-17, c=-85.

$x =\frac{-b + \sqrt{b^{2}-4ac } }{2a}$

$x =\frac{-(-17)+\sqrt{(-17)^{2} -4(9)(-85)} }{2(9)}$

$x =\frac{17}{18} +\frac{-1}{18} \sqrt{3349}$

4 0

2 years ago

Consider a series circuit consisting of a resistor of R ohms, an inductor of L henries, and variable voltage source of V(t) volt

ser-zykov [4K]

Answer:

$I(t)=\frac{1}{3}(1-e^{-30t})$

Step-by-step explanation:

We are given that

$\frac{dI}{dt}+\frac{R}{L}I=\frac{V(t)}{L}$

R=150 ohm

L=5 H

V(t)=10 V

$P=\frac{R}{L}=\frac{150}{5}=30$

$I.F=e^{\int Pdt}=e^{\int 30 dt}=e^{30 t}$

$I(t)\times I.F=\int e^{30 t}\times 10 dt+C$

$I(t)\times e^{30 t}=\frac{10}{30}e^{30 t}+C$

$I(t)=\frac{1}{3}+Ce^{-30 t}$

I(0)=0

Substitute t=0

$0=\frac{1}{3}+C$

$C=-\frac{1}{3}$

Substitute the values

$I(t)=\frac{1}{3}-\frac{1}{3}e^{-30 t}$

$I(t)=\frac{1}{3}(1-e^{-30t})$

5 0

3 years ago