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3 years ago

The quotient of 3,200 and 4 will have -blank- zeros.

Mathematics

2 answers:

mote1985 [20]3 years ago

4 0

B 2 hope I helped :)

Nostrana [21]3 years ago

3 0

B. It will have two zeros. Hope this helps!

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The reading speed of second grade students in a large city is approximately normal, with a mean of 89 words per minute

laila [671]

Try using
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3 years ago

Make x the subject of the formula y = mx + b

zubka84 [21]

Step-by-step explanation:

y=mx+c

Take c from both sides:

y-c = mx

Divide both sides by m:

(y-c) / m =x

Now swap sides to make it prettier:

x = (y-c) / m

6 0

2 years ago

Read 2 more answers

Solve the problem.When exposed to ethylene gas, green bananas will ripen at an accelerated rate. The number of days for ripening

Alex Ar [27]

The table and the plotted data are in the attachment.

Answer and Step-by-step explanation: A line of best fit is the best approximation of the given data. It is used to show the relation of two variables and it can be determined, more accurately, by the Least Square Method.

To use the method, follow the steps:

1) Calculate the mean of the two variables:

$X = \frac{\Sigma x_{i}}{n}$

$Y=\frac{\Sigma y_{i}}{n}$

2) The Slope of the line is given by:

$m=\frac{\Sigma (x_{i}-X)(y_{i}-Y)}{\Sigma (x_{i}-X)^{2}}$

3) The y-intercept is found by the formula:

$b=Y-mX$

For the ripening of bananas, the least square method gives the line:

y = -0.142x + 5.56

The rate of change of a line equation is its slope, so

Rate of change = -0.14

The rate of change of ripening time with respect to exposure time is -0.14, which means the more expose the fruit is, the less time it needs to ripe.

Download pdf

7 0

3 years ago

#11 i

Paraphin [41]

The area of a shape is the amount of space a shape can occupy, while the perimeter is the sum of its lengths.

The perimeter is $18 + 3\sqrt 5 + 3\sqrt 2$ units
The area of the sanctuary is 27 square units

We have:

$A = (5,7)\\B=(8, 7)\\C=(8, 1)\\D=( 1, 1)\\E=( 1, 4)\\F= ( 5,4)$

Perimeter

See attachment for the layout of the sanctuary

$BC = \sqrt{(8 - 8)^2 + (7 - 1)^2} = 6$

$EF = \sqrt{(1 - 5)^2 + (4 - 4)^2} = 4$

$FA = \sqrt{(5 - 5)^2 + (4 - 7)^2} = 3$

From the attachment, the sides are

AB, BF, FC, CD, DE and EA

Start by calculating the length of each side using the following distance formula:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_2)^2}$

So, we have:

$AB = \sqrt{(5 - 8)^2 + (7 - 7)^2} = 3$

$BF = \sqrt{(8 - 5)^2 + (7 -1)^2} = \sqrt{45} = 3\sqrt 5$

$FC = \sqrt{(8 - 5)^2 + (1 -4)^2} = \sqrt{18} = 3\sqrt 2$

$CD = \sqrt{(8 - 1)^2 + (1 - 1)^2} = 7$

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

$EA = \sqrt{(1 - 5)^2 + (4 -7)^2} = 5$

So, the perimeter (P) is:

$P = AB + BF + FC + CD + DE + EA$

$P = 3 + 3\sqrt 5 + 3\sqrt 2 + 7 + 3 + 5$

$P = 18 + 3\sqrt 5 + 3\sqrt 2$

Hence, the perimeter is $18 + 3\sqrt 5 + 3\sqrt 2$ units

Area

To calculate the area, we need to divide the sanctuary into three.

Triangle ABF
Triangle EFA
Trapezium CDEF

The area of ABF is:

$Area = \frac 12 \times AB \times FA$

Where:

$AB = 3$

$FA = \sqrt{(5 - 5)^2 + (4 - 7)^2} = 3$

$Area = \frac 12 \times 3 \times 3$

$Area = \frac 92$

The area of EFA is:

$Area = \frac 12 \times EF \times FA$

Where:

$FA = 3$

$EF = \sqrt{(1 - 5)^2 + (4 - 4)^2} = 4$

So:

$Area = \frac 12 \times 4 \times 3$

$Area = 6$

The area of CDEF is:

$Area = \frac 12(CD + EF) \times DE$

Where

$CD = 7$

$EF = 4$

$DE = 3$

So, we have:

$Area = \frac 12 (7 + 4) \times 3$

$Area = \frac 12 \times 11 \times 3$

$Area = \frac {33}2$

So, the area of the sanctuary is:

$Area = \frac 92 + 6 + \frac {33}2$

$Area = \frac {9 +12 + 33}2$

$Area = \frac {54}2$

$Area = 27$

Hence, the area of the sanctuary is 27 square units

Read more about areas and perimeters at:

brainly.com/question/11957651

8 0

3 years ago

Convert 6 2/3 cups to pints. Express your answer in simplest form.

Dmitry_Shevchenko [17]

First off, let's convert the mixed fraction to "improper", keeping in mind that, there are 2 cups in 1 pint.

$\bf \stackrel{mixed}{6\frac{2}{3}}\implies \cfrac{6\cdot 3+2}{3}\implies \stackrel{improper}{\cfrac{20}{3}}
\\\\\\
\begin{array}{ccll}
cups&pints\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
2&1\\\\
\frac{20}{3}&p
\end{array}\implies \cfrac{\quad 2\quad }{\frac{20}{3}}=\cfrac{1}{p}\implies \cfrac{\quad \frac{2}{1}\quad }{\frac{20}{3}}=\cfrac{1}{p}$

$\bf \cfrac{2}{1}\cdot \cfrac{3}{20}=\cfrac{1}{p}\implies \cfrac{3}{10}=\cfrac{1}{p}\implies p=\cfrac{10\cdot 1}{3}\implies p=\cfrac{10}{3}\implies p=3\frac{1}{3}$

5 0

3 years ago