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Gnoma [55]
3 years ago
6

How can you tell the difference between additive comparison and multiplicative comparison?

Mathematics
1 answer:
Ann [662]3 years ago
4 0
When you are adding you are ,say you have the number 4 and you need to add it to 5 you are basically mushing those two numbers together and seeing what you get, and I got 9. with multiplying you are multiplying it by itself however many times you need to, to find out the answer. :)
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A rectangle has a length that is 12 ft longer
Reil [10]
4x+24=84 4x=60 x=15 length =27 width = 15
7 0
3 years ago
A piece of cardboard measures 10 ft by 10 ft. Four equal squares of size x are removed from the corners. After removing the squa
Gnesinka [82]

Answer:

The value of x would be \frac{5}{3}

Step-by-step explanation:

Given,

The dimension of the cardboard = 10 ft by 10 ft,

∵ After removing four equal squares of size x ( in ft ) from the corners,

The dimension of the resultant box would be,

Length = ( 10 - 2x ) ft,

Width = ( 10 - 2x ) ft,

Height = x ft,

The volume of box,

V=(10-2x)\times (10 - 2x)\times x=x(10-2x)^2 = x(100 - 40x + 4x^2)=100x - 40x^2 + 4x^3

Differentiating with respect to x,

V'=100 - 80x + 12x^2

Again differentiating with respect to x,

V''=-80 + 24x

For maxima or minima,

V'=0

\implies 100 - 80x + 12x^2 = 0

\implies 3x^2 - 20x + 25=0

By quadratic formula,

x=\frac{20\pm \sqrt{20^2-4\times 3\times 25}}{6}

x=\frac{20\pm \sqrt{400 - 300}}{6}

x=\frac{20\pm \sqrt{100}}{6}

x=\frac{20\pm 10}{6}

\implies x = \frac{10}{6}=\frac{5}{3}\text{ or } x = 5

For x = 5/3, V'' = negative,

While for x = 5, V'' = Positive,

Hence, the value of x would be 5/3 ft for maximising the volume.

8 0
3 years ago
Rectangle A has 7 times the area of rectangle
krok68 [10]
Wa=0.5Wb, Ha=20, Hb is unkown
the area of rectangle a is : Wa*20
the area of rectangle b is : Wb*Hb
Rectangle a has 7 times the area of rectangle b: Wa*20=7*Wb*Hb
Replace Wa with 0.5Wb: 0.5Wb*20=7Wb*Hb => Hb=10/7
7 0
3 years ago
Help!!! please and thank you :)
Andrews [41]

Answer:

c

d

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Nadia is investigating rotations about the center of regular polygons that carry the regular polygon onto itself. She claims tha
GalinKa [24]

Answer:

\theta_1 \ n\ \theta_2 = 120, 240

Step-by-step explanation:

The question is incomplete, as the angles of rotation are not stated.

However, I will list the angles less than 360 degrees that will carry the hexagon and the nonagon onto itself

We have:

Nonagon = 9\ sides

Hexagon = 6\ sides

Divide 360 degrees by the number of sides in each angle, then find the multiples.

<u>Nonagon</u>

\theta = \frac{360}{9} =40

List the multiples of 40

\theta_1 = 40, 80, 120, 160, 200, 240, 280, 320

<u>Hexagon</u>

\theta = \frac{360}{6} =60

List the multiples of 60

\theta_2 = 60, 120, 180, 240, 300

List out the common angles

\theta_1 = 40, 80, 120, 160, 200, 240, 280, 320

\theta_2 = 60, 120, 180, 240, 300

\theta_1 \ n\ \theta_2 = 120, 240

This means that, only a rotation of 120, 240 will lift both shapes onto themselves, when applied to both shapes.

The other angles will only work on one of the shapes, but not both at the same time.

7 0
3 years ago
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