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2 years ago

2 - 1/4 in simplest form PLS HURRYYYY

Mathematics

1 answer:

gavmur [86]2 years ago

8 0

Answer:

The answer is 1.75

Step-by-step explanation:

2-1/5 or (.25)

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If nico spends $5.76 on tomatoes,how many pounds did he purchase?

tia_tia [17]

The amount of in pound of tomatoes purchased is $\frac{5.76}{p}$

To obtain the number of pounds of tomatoes purchased :

Let the price per pound of tomato = p

The Number of tomatoes purchased equals :

Total amount spent ÷ price per pound

Therefore, the Number of tomatoes purchased will be :

5.76 ÷ p

The Number of pounds of tomatoes purchased = $\frac{5.76}{p}$

Learn more :brainly.com/question/18109354

4 0

2 years ago

Set up a double integral for calculating the flux of the vector field ????⃗ (????⃗ )=????⃗ , where ????⃗ =⟨x,y,z⟩, through the p

Tema [17]

Answer:

-937.5π

Step-by-step explanation:

F (r) = r = (x, y, z) the surface equation z = 3(x^2 + y^2) z_x = 6x, z_y = 6y the normal vector n = (- z_x, - z_y, 1) = (- 6x, - 6y, 1)

Thus, flux ∫∫s F · dS is given as;

∫∫ <x, y, z> · <-z_x, -z_y, 1> dA

=∫∫ <x, y, 3x² + 3y²> · <-6x, -6y, 1>dA , since z = 3x² + 3y²

Thus, flux is;

= ∫∫ -3(x² + y²) dA.

Since the region of integration is bounded by x² + y² = 25, let's convert to polar coordinates as follows:

∫(θ = 0 to 2π) ∫(r = 0 to 5) -3r² (r·dr·dθ)

= 2π ∫(r = 0 to 5) -3r³ dr

= -(6/4)πr^4 {for r = 0 to 5}

= -(6/4)5⁴π - (6/4)0⁴π

= -937.5π

4 0

3 years ago

Please anyone can you guys help me it would mean the world

lapo4ka [179]

It says 404 not found. :(

4 0

3 years ago

Anderson won $10000 in a Lotto competition.

kozerog [31]

Answer:

jimmy got 30,000

Step-by-step explanation:

10,000

the wife got 60% = 60,000

jimmy got 3x prince = 30,000

prince = 10,000

8 0

3 years ago

Let $f(x) = x^2$ and $g(x) = \sqrt{x}$. Find the area bounded by $f(x)$ and $g(x).$

Anna [14]

Answer:

$\large\boxed{1\dfrac{1}{3}\ u^2}$

Step-by-step explanation:

Let's sketch graphs of functions f(x) and g(x) on one coordinate system (attachment).

Let's calculate the common points:

$x^2=\sqrt{x}\qquad\text{square of both sides}\\\\(x^2)^2=\left(\sqrt{x}\right)^2\\\\x^4=x\qquad\text{subtract}\ x\ \text{from both sides}\\\\x^4-x=0\qquad\text{distribute}\\\\x(x^3-1)=0\iff x=0\ \vee\ x^3-1=0\\\\x^3-1=0\qquad\text{add 1 to both sides}\\\\x^3=1\to x=\sqrt[3]1\to x=1$

The area to be calculated is the area in the interval [0, 1] bounded by the graph g(x) and the axis x minus the area bounded by the graph f(x) and the axis x.

We have integrals:

$\int\limits_{0}^1(\sqrt{x})dx-\int\limits_{0}^1(x^2)dx=(*)\\\\\int(\sqrt{x})dx=\int\left(x^\frac{1}{2}\right)dx=\dfrac{2}{3}x^\frac{3}{2}=\dfrac{2x\sqrt{x}}{3}\\\\\int(x^2)dx=\dfrac{1}{3}x^3\\\\(*)=\left(\dfrac{2x\sqrt{x}}{2}\right]^1_0-\left(\dfrac{1}{3}x^3\right]^1_0=\dfrac{2(1)\sqrt{1}}{2}-\dfrac{2(0)\sqrt{0}}{2}-\left(\dfrac{1}{3}(1)^3-\dfrac{1}{3}(0)^3\right)\\\\=\dfrac{2(1)(1)}{2}-\dfrac{2(0)(0)}{2}-\dfrac{1}{3}(1)}+\dfrac{1}{3}(0)=2-0-\dfrac{1}{3}+0=1\dfrac{1}{3}$

6 0

3 years ago