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3 years ago

Madilyn has 2 triangular vegetable gardens. One measures

Mathematics

1 answer:

Zepler [3.9K]3 years ago

7 0

Answer:

Gamma

Step-by-step explanation:

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Which of the answer choices could be the one missing data item for the following set, if the mode is 12?

crimeas [40]

Answer:

Step-by-step explanation:

first of all, what is mode: it is the highest occuring number in a set of numbers.

now since 12 is supposed to be the mean and it occurs twice but 9 also occurs twice so for 12 to be the mode, we must add one more 12.

therefore the answer is 12

8 0

3 years ago

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Find the domain of the function f of x equals the cube root of the quantity x plus three, minus one.

kifflom [539]

The domain of the function f of x equals the cube root of the quantity x plus three, minus one would be:

ALL REAL NUMBERS

I hope my answer has come to your help. God bless and have a nice day ahead!

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3 years ago

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A rectangular classroom has an area of 300 square feet. The length of the classroom is 5 feet longer than the width, w. Create a

weqwewe [10]

Answer:

w^2 + 5w = 300

Step-by-step explanation:

A= length x width

A =300

Width = w

length = w + 5

Therefore

300 = (w) x (w + 5)

300 = w^2 + 5w

w^2 + 5w = 300

6 0

3 years ago

Help on this please i need help so much

attashe74 [19]

The correct answer for each is
a= 390.8
b= 136.8

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3 years ago

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A rectangle has width that is 6 meters less than the length. The area of the rectangle is 280 square meters. Find the dimensions

salantis [7]

Dimensions are length 20 meter and width 14 meter

Solution:

Let "a" be the length of rectangle

Let "b" be the width of rectangle

Given that,

A rectangle has width that is 6 meters less than the length

Width = length - 6

b = a - 6

The area of the rectangle is 280 square meters

The area of the rectangle is given by formula:

$Area = length \times width$

Substituting the values we get,

$Area = a \times (a-6)\\\\280 = a^2-6a\\\\a^2-6a -280=0$

Solve the above equation by quadratic formula

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\\quad x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=1,\:b=-6,\:c=-280:\quad a_{1,\:2}=\frac{-\left(-6\right)\pm \sqrt{\left(-6\right)^2-4\cdot \:1\left(-280\right)}}{2\cdot \:1}$

$a =\frac{6 \pm \sqrt{36+1120}}{2}\\\\a = \frac{6 \pm \sqrt{1156}}{2}\\\\a = \frac{6 \pm 34}{2}\\\\Thus\ we\ have\ two\ solutions\\\\a = \frac{6+34}{2} \text{ or } a = \frac{6-34}{2}\\\\a = 20 \text{ or } a = -14$

Since, length cannot be negative, ignore a = -14

Thus solution of length is a = 20

Therefore,

width = length - 6

width = 20 - 6 = 14

Thus dimensions are length 20 meter and width 14 meter

6 0

3 years ago