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Veronika [31]
3 years ago
10

Find the center and radius of (x + 0.2)2 + (y + 0.1)2 = 0.3 (with an explanation please)

Mathematics
1 answer:
pshichka [43]3 years ago
8 0
Hope this helps you!

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A semi-trailer truck stops at a weigh station before crossing over a bridge. The weight limit on the bridge is75,500 pounds. The
Studentka2010 [4]
Part I

The weight of the truck without cargo is 22150+11300= 33450 pound
The weight of the cargo is given as c and altogether with the weight of the truck, the sum must not exceed 75500

The inequality is given 33450+c \leq 75500

Part II
33450+c \leq 75500
c \leq 75500-33450
c \leq 42050

Hence, the maximum weight allowed for the cargo is 42050 pounds
6 0
3 years ago
Rational Expression<br> Simplify: x/x+2
kirza4 [7]
3 is the answer cuz x/x cancels out and gives 1
5 0
4 years ago
sammy's bucket holds 2,530 milliliters of water. Marie's bucket holds 2 liters 30 milliliters of water. Katie's bucket holds 2 l
Zanzabum

Answer:

Marie's bucket holds the least amount of water.

Step-by-step explanation:

Sammy's bucket holds 2.53 liters of water. (2530 / 1000 = 2.530)

Marie's bucket holds 2.03 liters of water. (2 + 0.030 = 2.030)

Katie's bucket holds 2.35 liters of water. (2 + 0.350 = 2.350)

If you compare all these values, 2.03 is the least out of all of them.

8 0
3 years ago
Bella has 6.3 kilograms of berries. She packs 0.35 kilograms of berries into each container.she then sells each container for $2
Alex
Answer: $53.82


work:
6.3/.35= 18
18 • 2.99= 53.82
$53.82
4 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
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