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4 years ago

Can someone help me real quick PLEASE

Mathematics

1 answer:

suter [353]4 years ago

7 0

Answer:

The hexagon has an area of 220.6 square units.

Step-by-step explanation:

You can solve this by realising that the hexagon can be divided into 6 equilateral triangles of the same area.

The area of the triangle is the same as a rectangle of the same width and height, but divided by two, and we're told that the width of the side is 9.2, and the height is eight. So the total area of the hexagon is:

6 × (9.2 × 8) / 2

= 9.2 × 24

= 220.6

So the hexagon has an area of 220.6 square units.

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3 years ago

A substance with a half life is decaying exponentially. If there are initially 12 grams of the substance and after 70 minutes th

77julia77 [94]

Answer: 233 min

Step-by-step explanation:

This problem can be solved by the following equation:

$A=A_{o} e^{-kt}$ (1)

Where:

$A=7 g$ is the quantity left after time $t$

$A_{o}=12 g$ is the initial quantity

$t=70 min$ is the time elapsed

$k$ is the constant of decay for the material

So, firstly we need to find the value of $k$ from (1) in order to move to the next part of the problem:

$\frac{A}{A_{o}}=e^{-kt}$ (2)

Applying natural logarithm on both sides of the equation:

$ln(\frac{A}{A_{o}})=ln(e^{-kt})$ (3)

$ln(\frac{A}{A_{o}})=-kt$ (4)

$k=-\frac{ln(\frac{A}{A_{o}})}{t}$ (5)

$k=-\frac{ln(\frac{7 g}{12 g})}{70 min}$ (6)

$k=0.00769995 min^{-1}$ (7) Now that we have the value of $k$ we can solve the other part of this problem: Find the time $t$ for $A=2 g$ .

In this case we need to isolate $t$ from (1):

$t=-\frac{ln(\frac{A}{A_{o}})}{k}$ (8)

$t=-\frac{ln(\frac{2 g}{12 g})}{0.00769995 min^{-1}}$ (9)

Finally:

$t=232.697 min \approx 233 min$

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3 years ago

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lina2011 [118]

Answer:

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Step-by-step explanation:

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3 years ago

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Artemon [7]

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3 years ago

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Kruka [31]

Given:

$\Delta MAT\cong \Delta MHT$ and $m\angle 2=82^\circ$ .

To find:

Whether $m\angle 1=47^\circ$ is possible or not.

Solution:

We have,

$\Delta MAT\cong \Delta MHT$

$\angle AMT\cong \angle HMT$ (CPCTC)

$m\angle 3=m\angle 8$ ...(i)

And,

$m\angle 1=m\angle 8$ ...(ii) (Vertically opposite angles)

From (i) and (ii), we get

$m\angle 1=m\angle 3$ ...(iii)

Now,

$m\angle 1+m\angle 2+m\angle 3=180^\circ$ (Linear pair)

$m\angle 1+82^\circ+m\angle 1=180^\circ$ (Using (iii))

$2m\angle 1=180^\circ-82^\circ$

$2m\angle 1=98^\circ$

Divide both sides by 2.

$m\angle 1=\dfrac{98^\circ}{2}$

$m\angle 1=49^\circ$

The measure of angle 1 is 49 degrees so it cannot be equal to 47 degrees.

Therefore, the required answer is "no", the given statement $m\angle 1=47^\circ$ is not possible.

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3 years ago