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2 years ago

8

Omar saved the same amount of

1 answer:

xz_007 [3.2K]2 years ago

3 0

Answer:

Omar saved 10 dollars in total

Step-by-step explanation:

total= 2.5*4

total= $10

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Factoring with Algebra Tiles Represent the quadratic polynomial 2x2 + x – 6 using algebra tiles and determine the equivalent fac

Nezavi [6.7K]

Answer:

Two zero pair

$f(x)=(2x-3)(x+2)$

Step-by-step explanation:

We are given the following quadratic equation:

$f(x) = 2x^2+x-6$

Since, this a quadratic equation we would have two zero pairs for this model.

Factored form of polynomial:

$f(x) = 2x^2+x-6\\f(x)=2x^2+ 4x -3x -6\\f(x)=2x(x+2)-3(x+2)\\f(x)=(2x-3)(x+2)$

is the required factored form of the given polynomial.

6 0

3 years ago

The actual distance between two villages is 120 kilometers. On a map of the area, 0.5 centimeter represents 2 kilometers.

stira [4]

Answer:

30 centimeters

Step-by-step explanation:

120 ÷2 =60 now you must use only .5 of that number so you have thirty

3 0

2 years ago

Choose the equations in which d = 1000 will make the equation true

Brilliant_brown [7]

Answer:

there are no equations I can choose from

Step-by-step explanation:

...

5 0

2 years ago

What is the volume of a hemisphere with a radius of 2.3m round to the nearest tenth of a cubic meter

kodGreya [7K]

Answer: $25.5 m^{3}$

Step-by-step explanation:

If the volume of a sphere is

$V=\frac{4}{3} \pi r^{3}$

Where $r=2.3 m$ is the radius

The volume of a hemisphere is half the volume of the total sphere:

$V_{hemisphere}=\frac{V}{2}=\frac{\frac{4}{3} \pi r^{3}}{2}$

$V_{hemisphere}=\frac{2}{3} \pi r^{3}$

Solving this equation:

$V_{hemisphere}=\frac{2}{3} \pi (2.3 m)^{3}$

Finally:

$V_{hemisphere}=25.48m^{3} \approx 25.5 m^{3}$

5 0

3 years ago

Problem 4: Solve the initial value problem

pishuonlain [190]

Separate the variables:

$y' = \dfrac{dy}{dx} = (y+1)(y-2) \implies \dfrac1{(y+1)(y-2)} \, dy = dx$

Separate the left side into partial fractions. We want coefficients a and b such that

$\dfrac1{(y+1)(y-2)} = \dfrac a{y+1} + \dfrac b{y-2}$

$\implies \dfrac1{(y+1)(y-2)} = \dfrac{a(y-2)+b(y+1)}{(y+1)(y-2)}$

$\implies 1 = a(y-2)+b(y+1)$

$\implies 1 = (a+b)y - 2a+b$

$\implies \begin{cases}a+b=0\\-2a+b=1\end{cases} \implies a = -\dfrac13 \text{ and } b = \dfrac13$

So we have

$\dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = dx$

Integrating both sides yields

$\displaystyle \int \dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = \int dx$

$\dfrac13 \left(\ln|y-2| - \ln|y+1|\right) = x + C$

$\dfrac13 \ln\left|\dfrac{y-2}{y+1}\right| = x + C$

$\ln\left|\dfrac{y-2}{y+1}\right| = 3x + C$

$\dfrac{y-2}{y+1} = e^{3x + C}$

$\dfrac{y-2}{y+1} = Ce^{3x}$

With the initial condition y(0) = 1, we find

$\dfrac{1-2}{1+1} = Ce^{0} \implies C = -\dfrac12$

so that the particular solution is

$\boxed{\dfrac{y-2}{y+1} = -\dfrac12 e^{3x}}$

It's not too hard to solve explicitly for y; notice that

$\dfrac{y-2}{y+1} = \dfrac{(y+1)-3}{y+1} = 1-\dfrac3{y+1}$

Then

$1 - \dfrac3{y+1} = -\dfrac12 e^{3x}$

$\dfrac3{y+1} = 1 + \dfrac12 e^{3x}$

$\dfrac{y+1}3 = \dfrac1{1+\frac12 e^{3x}} = \dfrac2{2+e^{3x}}$

$y+1 = \dfrac6{2+e^{3x}}$

$y = \dfrac6{2+e^{3x}} - 1$

$\boxed{y = \dfrac{4-e^{3x}}{2+e^{3x}}}$

7 0

2 years ago