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2 years ago

Gavyn was thinking of a number.gavyn halves it and gets an answer of 9.4 what was the original number?

Mathematics

1 answer:

kupik [55]2 years ago

3 0

Answer:

18.80

Explanation:

All you have to do is add 9.4 to 9.4, or multiply 9.4 times 2, and you get 18.80

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Find all rational zeros of P(x)=6x^4-x^3-55x^2+9x+9

o-na [289]

Answer: x = -3, -1/3, 1/2, 3

Step-by-step explanation:

6 0

2 years ago

A 3 cm x 3 cm rectangle sits inside a circle with radius of 4 cm.

Ivanshal [37]

Step-by-step explanation:

you did not show us the picture or described what the shaded area is.

I assume it is the area of the circle without the inner square (as the "rectangke" is actually a square, because its sides are equally long).

so, we need to calculate the area of the circle and subtract the area of the square.

the area of the square is 3×3 = 9 cm².

the area of the circle is

pi×r²

with r being the radius.

so,

pi×4² = pi×16 = 16pi

the shaded area is then

16pi - 9 = 41.26548246... cm² ≈ 41.27 cm²

nothing of the answer options seems to be in that area, so maybe my assumption about the shaded area is wrong. or the specified answer options are mistyped. or both.

anyway, without additional information that is the best I can do.

8 0

1 year ago

Pls answer quick it's gr 9 academic math lol

SVETLANKA909090 [29]

Answer:

1. (3^3 + 3^2)^2 actually equals (27 + 9)^2

which is the first mistake

2. (27 + 9)^2 does not equal (3^5)2, so (36)^2 does not equal 3^7

3. 3^7 DOES NOT EQUAL 21

Step-by-step explanation:

when you add powered numbers together, it does not multiply it, as your example:

1. (3^3 + 3^2)^2 actually equals (27 + 9)^2

which is the first mistake

2. (27 + 9)^2 does not equal (3^5)2, so (36)^2 does not equal 3^7

3. 3^7 DOES NOT EQUAL 21

4 0

2 years ago

Read 2 more answers

if there are 450 campers and 2/3 decides to come back next year how many campers will be coming back next year

DedPeter [7]

Answer:

So 452.3 campers will be coming back next year

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

The length l, width w, and height h of a box change with time. At a certain instant the dimensions are l = 3 m and w = h = 6 m,

Gemiola [76]

Answer:

a) The rate of change associated with the volume of the box is 54 cubic meters per second, b) The rate of change associated with the surface area of the box is 18 square meters per second, c) The rate of change of the length of the diagonal is -1 meters per second.

Step-by-step explanation:

a) Given that box is a parallelepiped, the volume of the parallelepiped, measured in cubic meters, is represented by this formula:

$V = w \cdot h \cdot l$

Where:

$w$ - Width, measured in meters.

$h$ - Height, measured in meters.

$l$ - Length, measured in meters.

The rate of change in the volume of the box, measured in cubic meters per second, is deducted by deriving the volume function in terms of time:

$\dot V = h\cdot l \cdot \dot w + w\cdot l \cdot \dot h + w\cdot h \cdot \dot l$

Where $\dot w$ , $\dot h$ and $\dot l$ are the rates of change related to the width, height and length, measured in meters per second.

Given that $w = 6\,m$ , $h = 6\,m$ , $l = 3\,m$ , $\dot w =3\,\frac{m}{s}$ , $\dot h = -6\,\frac{m}{s}$ and $\dot l = 3\,\frac{m}{s}$ , the rate of change in the volume of the box is:

$\dot V = (6\,m)\cdot (3\,m)\cdot \left(3\,\frac{m}{s} \right)+(6\,m)\cdot (3\,m)\cdot \left(-6\,\frac{m}{s} \right)+(6\,m)\cdot (6\,m)\cdot \left(3\,\frac{m}{s}\right)$

$\dot V = 54\,\frac{m^{3}}{s}$

The rate of change associated with the volume of the box is 54 cubic meters per second.

b) The surface area of the parallelepiped, measured in square meters, is represented by this model:

$A_{s} = 2\cdot (w\cdot l + l\cdot h + w\cdot h)$

The rate of change in the surface area of the box, measured in square meters per second, is deducted by deriving the surface area function in terms of time:

$\dot A_{s} = 2\cdot (l+h)\cdot \dot w + 2\cdot (w+h)\cdot \dot l + 2\cdot (w+l)\cdot \dot h$

Given that $w = 6\,m$ , $h = 6\,m$ , $l = 3\,m$ , $\dot w =3\,\frac{m}{s}$ , $\dot h = -6\,\frac{m}{s}$ and $\dot l = 3\,\frac{m}{s}$ , the rate of change in the surface area of the box is:

$\dot A_{s} = 2\cdot (6\,m + 3\,m)\cdot \left(3\,\frac{m}{s} \right) + 2\cdot (6\,m+6\,m)\cdot \left(3\,\frac{m}{s} \right) + 2\cdot (6\,m + 3\,m)\cdot \left(-6\,\frac{m}{s} \right)$

$\dot A_{s} = 18\,\frac{m^{2}}{s}$

The rate of change associated with the surface area of the box is 18 square meters per second.

c) The length of the diagonal, measured in meters, is represented by the following Pythagorean identity:

$r^{2} = w^{2}+h^{2}+l^{2}$

The rate of change in the surface area of the box, measured in square meters per second, is deducted by deriving the surface area function in terms of time before simplification:

$2\cdot r \cdot \dot r = 2\cdot w \cdot \dot w + 2\cdot h \cdot \dot h + 2\cdot l \cdot \dot l$

$r\cdot \dot r = w\cdot \dot w + h\cdot \dot h + l\cdot \dot l$

$\dot r = \frac{w\cdot \dot w + h \cdot \dot h + l \cdot \dot l}{\sqrt{w^{2}+h^{2}+l^{2}}}$

Given that $w = 6\,m$ , $h = 6\,m$ , $l = 3\,m$ , $\dot w =3\,\frac{m}{s}$ , $\dot h = -6\,\frac{m}{s}$ and $\dot l = 3\,\frac{m}{s}$ , the rate of change in the length of the diagonal of the box is:

$\dot r = \frac{(6\,m)\cdot \left(3\,\frac{m}{s} \right)+(6\,m)\cdot \left(-6\,\frac{m}{s} \right)+(3\,m)\cdot \left(3\,\frac{m}{s} \right)}{\sqrt{(6\,m)^{2}+(6\,m)^{2}+(3\,m)^{2}}}$

$\dot r = -1\,\frac{m}{s}$

The rate of change of the length of the diagonal is -1 meters per second.

6 0

2 years ago