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3 years ago

6

Marissa works in a bread shop every hour

1 answer:

anzhelika [568]3 years ago

4 0

Answer:

She's getting paid minimum wage, in debt, and struggling in college.

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Sin x cos x = -cos x<br> Find all angles that satisfy the equation below

n200080 [17]

Answer:

$x=90^\circ,270^\circ$

Step-by-step explanation:

$sinxcosx=-cosx$

$sinxcosx+cosx=0$

$cosx(sinx+1)=0$

$cosx=0$

$x=90,270$

$sinx+1=0$

$sinx=-1$

$x=270$

Therefore, the angles that satisfy the equation are $x=90^\circ,270^\circ$

8 0

3 years ago

a sack of potatoes weighs 14 pounds 9 ounces. After Wendy makes potato salad for a picnic, the sack weighs 9 pounds 14 ounces. w

Aleksandr [31]

5.76 is the correct answer to your problem

3 0

4 years ago

Solve the initial value problems.

slavikrds [6]

Both equations are linear, so I'll use the integrating factor method.

The first ODE

$xy' + (x+1)y = 0 \implies y' + \dfrac{x+1}x y = 0$

has integrating factor

$\exp\left(\displaystyle \int\frac{x+1}x \, dx\right) =\exp\left(x+\ln(x)\right) = xe^x$

In the original equation, multiply both sides by eˣ :

$xe^x y' + (x+1) e^x y = 0$

Observe that

d/dx [xeˣ] = eˣ + xeˣ = (x + 1) eˣ

so that the left side is the derivative of a product, namely

$\left(xe^xy\right)' = 0$

Integrate both sides with respect to x :

$\displaystyle \int \left(xe^xy\right)' \, dx = \int 0 \, dx$

$xe^xy = C$

Solve for y :

$y = \dfrac{C}{xe^x}$

Use the given initial condition to solve for C. When x = 1, y = 2, so

$2 = \dfrac{C}{1\cdot e^1} \implies C = 2e$

Then the particular solution is

$\boxed{y = \dfrac{2e}{xe^x} = \dfrac{2e^{1-x}}x}$

The second ODE

$(1+x^2)y' - 2xy = 0 \implies y' - \dfrac{2x}{1+x^2} y = 0$

has integrating factor

$\exp\left(\displaystyle \int -\frac{2x}{1+x^2} \, dx\right) = \exp\left(-\ln(1+x^2)\right) = \dfrac1{1+x^2}$

Multiply both sides of the equation by 1/(1 + x²) :

$\dfrac1{1+x^2} y' - \dfrac{2x}{(1+x^2)^2} y = 0$

and observe that

d/dx[1/(1 + x²)] = -2x/(1 + x²)²

Then

$\left(\dfrac1{1+x^2}y\right)' = 0$

$\dfrac1{1+x^2}y = C$

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

When x = 0, y = 3, so

$3 = C(1+0^2) \implies C=3$

$\implies \boxed{y = 3(1 + x^2) = 3 + 3x^2}$

7 0

2 years ago

Solve equation<br> 24a-22=-4 (1-6a)

iren2701 [21]

$24a-22=-4(1-6a) \\
24a-22=-4+24a \\
24a-24a=-4+22 \\
0=18 \\
not \ true \\ \\
\boxed{\hbox{no solution}}$

7 0

4 years ago

What point was used to write the equation y - 3=2(x +4)

Gnoma [55]

Answer:

(3, -4)

General Formulas and Concepts:

<u>Alg I</u>

Point-Slope Form: y - y₁ = m(x - x₁)

x₁ - x coordinate

y₁ - y coordinate

m - slope

Step-by-step explanation:

<u>Step 1: Define function</u>

y - 3 = 2(x + 4)

<u>Step 2: Break function</u>

Coordinate (3, -4)

Slope <em>m</em> = 2

7 0

4 years ago