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3 years ago

HELP ASAP !!

Mathematics

1 answer:

m_a_m_a [10]3 years ago

4 0

Answer:

A = 5000e^(0.10*10)

A = 5000e^1

A = 5000*e

A = $13,591.41

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A system of equations is shown below.

prisoha [69]

Given:

The system of equations is

$y=1.5x-3$

$y=1.5x+4$

To find:

The true statement about the given system of equations.

Solution:

The slope intercept form of a line is

$y=mx+b$

Where, m is the slope and b is the y-intercept.

We have,

$y=1.5x-3$

$y=1.5x+4$

On comparing these two lines with slope intercept form, we get

$m_1=1.5,b_1=-3$

$m_2=1.5,b_2=4$

Since the slopes of the lines are equal but the y-intercepts are different, therefore, the two lines are parallel and the system has no solution.

Therefore, the correct option is A.

5 0

2 years ago

Kit bought a soft drink and a sandwich for $9.00. What was the price of each if the sandwich cost 3.5 times as much as the soft

Aleks04 [339]

The answer woukd be 3.00 u guess math is my worst subject dont rage if it wrong

6 0

2 years ago

Solve the equation for the indicated variable.<br><br> x=2y-2 for y

stira [4]

Answer:

2y - 2 = x

2y = x + 2

y = (x + 2)/2

3 0

3 years ago

Write an equation that is parallel to y=-1/4x+2

AlexFokin [52]

Answer:

y = -1/4 -5

Step-by-step explanation:

Any equation will be parallel to your base equation as long as the y intercept is different. To write a parallel equation, you just have to change the y intercept.

For example, I changed it from 2 to negative 5.

5 0

2 years ago

Use implicit differentiation to find an equation of the tangent line to the curve at the given point. x2/3 + y2/3 = 4 (−3 3 , 1)

vovikov84 [41]

Answer with Step-by-step explanation:

We are given that an equation of curve

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

We have to find the equation of tangent line to the given curve at point $(-3\sqrt3,1)$

By using implicit differentiation, differentiate w.r.t x

$\frac{2}{3}x^{-\frac{1}{3}}+\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=0$

Using formula : $\frac{dx^n}{dx}=nx^{n-1}$

$\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=-\frac{2}{3}x^{-\frac{1}{3}}$

$\frac{dy}{dx}=\frac{-\frac{2}{3}x^{-\frac{1}{3}}}{\frac{2}{3}y^{-\frac{1}{3}}}$

$\frac{dy}{dx}=-\frac{x^{-\frac{1}{3}}}{y^{-\frac{1}{3}}}$

Substitute the value x= $-3\sqrt3,y=1$

Then, we get

$\frac{dy}{dx}=-\frac{(-3\sqrt3)^{-\frac{1}{3}}}{1}$

$\frac{dy}{dx}=-(-3^{\frac{3}{2}})^{-\frac{1}{3}}=-\frac{1}{-(3)^{\frac{3}{2}\times \frac{1}{3}}}=\frac{1}{\sqrt3}$

Slope of tangent=m= $\frac{1}{\sqrt3}$

Equation of tangent line with slope m and passing through the point $(x_1,y_1)$ is given by

$y-y_1=m(x-x_1)$

Substitute the values then we get

The equation of tangent line is given by

$y-1=\frac{1}{\sqrt3}(x+3\sqrt3)$

$y-1=\frac{x}{\sqrt3}+3$

$y=\frac{x}{\sqrt3}+3+1$

$y=\frac{x}{\sqrt3}+4$

This is required equation of tangent line to the given curve at given point.

8 0

3 years ago