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3 years ago

What answer to this question?

Mathematics

1 answer:

horrorfan [7]3 years ago

4 0

Answer:

221

Step-by-step explanation:

13 x 17 = 221

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What is the value of A when we rewrite<br> (2/3)^x+4 - (2/3)^x as A • (2/3)^x

horsena [70]

Answer:

A= –65/81

Step-by-step explanation:

hope this explains it

5 0

3 years ago

Find the length of the arc.

Ket [755]

This is the answer in different units
Millimeters-7660.46
Centimeters-766.046
Meters-7.66047
Kilometers- 0.00766046
Inches-301.593
Feet-25.13
Yards- 8.37758

6 0

1 year ago

Least to greatest 0.5, 0.05, 5/8

NISA [10]

Well change 5/8 into a decimal which is so the least is 0.05 then 0.5 then 5/8

8 0

3 years ago

Can anyone help me ?

taurus [48]

Answer:

66,6% = ⅔

33,3% = ⅓

85% = 17⁄20

80% = ⅘

75% = ¾

90% = 9⁄10

20% = ⅕

10% = ⅒

Step-by-step explanation:

To convert from a fraction to a percentage, divide both the denominator and numerator to get a decimal point, then move the decimal point twice to the right, then attach the percentage symbol:

⅒ = 0,1 = 10%

⤻⤻

⅕ = 0,2 = 20%

⤻⤻

9⁄10 = 0,9 = 90%

⤻⤻

¾ = 0,75 = 75%

⤻⤻

⅘ = 0,8 = 80%

⤻⤻

17⁄20 = 0,85 = 85%

⤻⤻

_ _

⅓ = 0,3 = 33,3%

⤻⤻

_ _

⅔ = 0,6 = 66,6

All those bars above those digits are what are known as bar notation, indicating that digits repeat.

I am joyous to assist you anytime.

6 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

2 years ago