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sweet [91]
2 years ago
10

What is 20 percent of 576

Mathematics
1 answer:
ryzh [129]2 years ago
5 0
115.20 is 20% of 576
576 x .20 = 115.20
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Kate had a sum of money,she spent 1/8 of the money on a skirt and the remaining amount on a handbag,the skirt cost $16,how much
tiny-mole [99]

Answer:

$18.28

Step-by-step explanation:

Step one:

given data

Kate had a sum of money,she spent 1/8 of the money on a skirt and the remaining amount on a handbag

let the total money be x

so  mathematically

x-1/8*x= 16

solve for x

x-x/8= 16

8x-x/8= 16

7x/8=16

cross multiply

7x= 8*16

7x=128

divide both sides by 7

x=128/7

x=$18.28

7 0
2 years ago
What is the 7th term in the sequence Below - 11,-4,3,10,17,24​
MAXImum [283]

Answer:

31

Step-by-step explanation:

Increases by 7 each time

8 0
3 years ago
Read 2 more answers
The sum of twice a<br> number and 31
Andreyy89

Answer:

Let the unknown number be x ;

2(x) +31\\=2x+31

Step-by-step explanation:

3 0
2 years ago
What is the equation of the parabola?
irina1246 [14]

Because the parabola opens down and the vertex is at (0, 5), we conclude that the correct option is:

y  = -(1/8)*x² + 5.

<h3>Which is the equation of the parabola?</h3>

The relevant information is that we have the vertex at (0, 5), and that the parabola opens downwards.

Remember that the parabola only opens downwards if the leading coefficient is negative. Then we can discard the two middle options.

Now, because the parabola has the point (0, 5), we know that when we evaluate the parabola in x = 0, we should get y = 5.

Then the constant term must be 5.

So the correct option is the first one:

y  = -(1/8)*x² + 5.

If you want to learn more about parabolas:

brainly.com/question/4061870

#SPJ1

3 0
1 year 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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