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3 years ago

14

Write the number in standard form. 2 x 102

1 answer:

NARA [144]3 years ago

5 0

Answer:

210

Step-by-step explanation:

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If f(x) and its inverse function, fx), are both plotted on the same coordinate plane, what is their point of intersection?

Lubov Fominskaja [6]

Answer: (3, 3)

Step-by-step explanation:

the graphs of $f$ and $f^{-1$ are flipped: if the point $(x, y)$ is on the graph of $f$ , then the point $(y, x)$ is on the graph of $f^{-1$ .

Any points of intersection will satisfy $(x, y) = (y, x)$ . A point that satisfies this property must have the same x and y coordinate. (Think about it.) The only point on $f$ like that is $(3, 3)$ , so that is the point of intersection.

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2 years ago

This figure is made up of a quadrilateral and a semicircle.

vodomira [7]

Here the figure is made up of a quadrilateral and a semi circle.

ABCD is the quadrilateral here. We will find the sides of the quadrilateral by using the distance formula.

If (x₁, y₁) and (x₂, y₂) are two points given, then the distance between two points by using distance formula is,

$d =\sqrt({ x_{1}-x_{2})^2 +( y_{1} -y_{2} )^2}$

The co-ordinate of A is (-1,2) and co-ordinate of B is (-2,-1).

So the length of side AB = $\sqrt{(-1-(-2))^2+(2-(-1))^2}$

= $\sqrt{(-1+2)^2+(2+1)^2}$ (As negative times negative is positive)

= $\sqrt{(1)^2+(3)^2} =\sqrt{1+9} =\sqrt{10}$

The co-ordinate of C is (4,-3) and D is (5,0)

The length of side CD

= $\sqrt(4-5)^2+(-3-0)^2}$

= $\sqrt{(-1)^2+(-3)^2}$

= $\sqrt{1+9} =\sqrt{10}$

So the sides AB and CD are equal.

The length of side AD

= $\sqrt{(-1-5)^2+(2-0)^2}$

= $\sqrt{(-6)^2+(2)^2} =\sqrt{36+4} =\sqrt{40}$

The length of side BC

= $\sqrt{(-2-4)^2+(-1-(-3))^2}$

= $\sqrt{(-2-4)^2+(-1+3)^2}$

= $\sqrt{(-6)^2+(2)^2}=\sqrt{36+4} =\sqrt{40}$

So the lengths of the sides AD and BC are equal.

So the quadrilateral is a rectangle whose length is $\sqrt{40}$ and width is $\sqrt{10}$ .

Area of a rectangle = length × width

= $(\sqrt{40}) (\sqrt{10})$

= $\sqrt{(40)(10)}=\sqrt{400}$

= $20 unit^2$

Now the diameter of the semicircle is the side AD = $\sqrt{40}$

So, the radius of the semi-circle = $\frac{\sqrt{40}}{2}$

= $\frac{\sqrt{(4)(10)}}{2}$

= $\frac{(\sqrt{4})(\sqrt{10})}{2}$

= $\frac{2\sqrt{10}}{2}$ = $\sqrt{10}$

Area of semi-circle = $\frac{1}{2} \pi r^2$ , where r is the radius.

= $\frac{1}{2} \pi (\sqrt{10})^2$

= $\frac{1}{2} \pi (10)$

= $\frac{(\pi)(10)}{2}$

= $\frac{10\pi}{2} = 5\pi$ = $15.7 unit^2$ ( Approximately taken to the nearest tenth)

Total area of the figure = $(20+15.7) unit^2$ = $35.7 unit^2$

We have got the required answer.

Option a is correct here.

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