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Rzqust [24]
3 years ago
15

Please help me with these geometry questions.

Mathematics
1 answer:
sashaice [31]3 years ago
7 0

Answer:

1)2(lb+bh+hl)

=2(17*12+12*8+8*17)

=2(204+96+136)

=2*436

=872 in^2

2)2*pi*r(r+h)

=2*3.14*14(14+29)

=6.28*14(43)

=6.28*14*43

=3780.56 mm^2

3)A=11*4.3+3*6 +3*8+11*3(Area of each plane figure -triangle and rectangle)

=47.3+18+24+33

=122.3cm^3

4)h*(sum of II sides)+14*8+19*8+14*8+25(Area of plane surface-trapezium,rectangle,square)

=12.1(19+5)+112+152+112+25

=12.1*24+401

=290.4+401

=691.4ft^2

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slavikrds [6]

Answer:

  • 2 groups of 6
  • 6 groups of 2
  • 4 groups of 3
  • 3 groups of 4
  • 1 group of 12
  • 12 groips of 1

Step-by-step explanation:

To figure out the possible groups find the factors of 12; in other words what two numbers would multiply to 12. These are: 1, 2, 3, 4, 6, and 12

  • 1 × 12 = 12
  • 2 × 6 = 12
  • 3 × 4 = 12
  • 4 × 3 = 12
  • 6 × 2 = 12
  • 12 × 1 = 12

So, your possible groupings are;

  • 2 groups of 6
  • 6 groups of 2
  • 4 groups of 3
  • 3 groups of 4
  • 1 group of 12
  • 12 groips of 1
8 0
3 years ago
Read 2 more answers
Pepsi is on sale at competing stores in your neighborhood. Store "A" gives you a $2 rebate if you buy three 6-packs at $3 each.
aliina [53]

Answer:

B.) is the best deal

Step-by-step explanation:

A.) three six packs is $3 for each pack minus $2

$3 + $3 + $3 = $9

$9 - $2 = $7

B.) guessing the three six packs is $3 then minusing 25%

$3 - 25% = $2.25

5 0
4 years ago
A 41 gram sample of a substance that's used to detect explosives has a k-value of 0.1392. Find the substance's half life, in day
-BARSIC- [3]

Answer:

The substance half-life is of 4.98 days.

Step-by-step explanation:

Equation for an amount of a decaying substance:

The equation for the amount of a substance that decay exponentially has the following format:

A(t) = A(0)e^{-kt}

In which k is the decay rate, as a decimal.

k-value of 0.1392.

This means that:

A(t) = A(0)e^{-0.1392t}

Find the substance's half life, in days.

This is t for which A(t) = 0.5A(0). So

A(t) = A(0)e^{-0.1392t}

0.5A(0) = A(0)e^{-0.1392t}

e^{-0.1392t} = 0.5

\ln{e^{-0.1392t}} = \ln{0.5}

-0.1392t = \ln{0.5}

t = -\frac{\ln{0.5}}{0.1392}

t = 4.98

The substance half-life is of 4.98 days.

5 0
3 years ago
Why do we square the residuals when using the least-squares line method to find the line of best fit?
Mekhanik [1.2K]

We square the residuals when using the least-squares line method to find the line of best fit because we believe that huge negative residuals (i.e., points well below the line) are just as harmful as large positive residuals (i.e., points that are high above the line).

<h3>What do you mean by Residuals?</h3>

We treat both positive and negative disparities equally by squaring the residual values.  We cannot discover a single straight line that concurrently minimizes all residuals. The average (squared) residual value is instead minimized.

We might also take the absolute values of the residuals rather than squaring them. Positive disparities are viewed as just as harmful as negative ones under both strategies.

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2 years ago
Drag tiles to match each fraction, decimal, or percent to an equivalent fraction, decimal, or percent.
yanalaym [24]

Answer:

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Step-by-step explanation:

1. 0.002

2. 308%

3. 2 6/10

7 0
3 years ago
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