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klemol [59]
3 years ago
14

Paul opens a savings account with $350. He saves $150 per month. Assume that he does not withdraw money or make any additional d

eposits. what is a linear model equation?
Mathematics
1 answer:
Verizon [17]3 years ago
8 0
I believe the equation is y=350+150x with y being the total amount of money in his account and x being the number of months..Paul starts with $350, which is the y-intercept (starting value) of the equation then the slope is 150 because his total savings increases by $150 for every month he saves without making any withdrawals. I apologize if I'm wrong but I hope this helps.
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Rania is playing with her friend Arun, and they are setting up dominoes across the room. Each domino is 2 inches long. They have
gogolik [260]

Answer:

no

Step-by-step explanation:

It should be less then or equal too, seeing as they want to reach all the way across the room.

6 0
2 years ago
#10 using right angle below find the tangent of angle A.
just olya [345]

Answer: the first option is the correct answer.

Step-by-step explanation:

Triangle ABC is a right angle triangle.

From the given right angle triangle,

AB represents the hypotenuse of the right angle triangle.

With m∠A as the reference angle,

AC represents the adjacent side of the right angle triangle.

BC represents the opposite side of the right angle triangle.

To determine the tangent of angle A, we would apply the Tangent trigonometric ratio. It is expressed as

Tan θ, = opposite side/adjacent side. Therefore,

Tan A = 5/5√3 = 1/√3

Rationalizing the surd, it becomes

1/√3 × √3/√3

Tan A = √3/3

6 0
3 years ago
Answer the question about the image.
Anton [14]

Answer:

4

Step-by-step explanation:

hope this helps and have a great day

3 0
2 years ago
Let alpha and beta be conjugate complex numbers such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}
miskamm [114]

Answer:

-3+i\sqrt{3} , 1+\sqrt{3}

Step-by-step explanation:

Given that alpha and beta be conjugate complex numbers

such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}.

Let

\alpha = x+iy\\\beta = x-iy

since they are conjugates

\alpha-\beta = x+iy-(x-iy)\\= 2iy= 2i\sqrt{3} \\y =\sqrt{3}

\frac{\alpha}{\beta^2} }\\=\frac{x+i\sqrt{3} }{(x-i\sqrt{3})^2} \\=\frac{x+i\sqrt{3}}{x^2-3-2i\sqrt{3}} \\=\frac{x+i\sqrt{3}((x^2-3+2i\sqrt{3}) }{(x^2-3-2i\sqrt{3)}(x^2-3-2i\sqrt{3})}

Imaginary part of the above =0

i.e. \sqrt{3} (x^2-3)+2x\sqrt{3} =0\\x^2+2x-3=0\\(x+3)(x-1) =0\\x=-3,1

So the value of alpha = -3+i\sqrt{3} , 1+\sqrt{3}

3 0
3 years ago
Can someone help me with the questions in the picture?
OverLord2011 [107]

Answer:

y = 8

Step-by-step explanation:

Equation of the line that passes through (-2, 8) with a slope of 0, can be written in point-slope form, y - y_1 = m(x - x_1) and also in slope-intercept form, y = mx + b.

Using a point, (-2, 8) and the slope (m), 0, substitute x1 = -2, y1 = 8 and m = 0 in y - y_1 = m(x - x_1).

Thus:

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Rewrite in slope-intercept form

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y = 8

8 0
3 years ago
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