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Solnce55 [7]
4 years ago
11

Look at the figure. What is another name for CD?

Mathematics
1 answer:
igor_vitrenko [27]4 years ago
8 0

Answer:

Correct choice is D

Step-by-step explanation:

Notion \overrightarrow{CD} means that you have to consider RAY with initial point C that goes in direction to point D (to the right).

Option A is false, because ray \overrightarrow{DC} has  initial point D and goes in direction to point C (to the left).

Option B is false, because ray \overrightarrow{CA} has  initial point C and goes in direction to point A (to the left).

Option C is false, because ray \overrightarrow{EC} has  initial point E and goes in direction to point C (to the left).

Option D is true, because ray \overrightarrow{CE} has  initial point C and goes in direction to points D and E (to the right).

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What is the area of triangle pqr on the grid? a triangle pqr is shown on a grid. the vertex p is on ordered pair 7 and 6, vertex
horsena [70]

P(7,6), Q(1,6), R(4,2)

We have PQ parallel to the x axis. We'll call that the base,

b = 7 - 1 = 6

The altitude is then the y difference h = 6 - 2 = 4

The area is \frac 1 2 bh = \frac 1 2 (6)(4) = 12

Answer: 12 square units


In general we can use the shoelace formula for the area of any polygon given coordinates. We write the points like this:

(7,6), (1,6), (4,2)

(1,6), (4,2), (7,6)

The area is then half the absolute value of the sum of the cross products:

A = \frac 1 2 | 7(6)-6(1) + 1(2)-6(4) + 4(6)-2(7) | = \frac 1 2 |24| = 12


6 0
3 years ago
Read 2 more answers
Radioactive Decay
SpyIntel [72]

Answer:

Percentage of (226Ra) after 900 years is 68%

Step-by-step explanation:

Let P(t) be the amount of (226Ra) present at any time t

Half life of (226 Ra) = 1599 years

If P₀ is initial amount of (226 Ra) then after 1599 years

P(1599)=P₀/2

Decay i amount of radioactive substance is related to time t as

\frac{dP}{dt}=kP(t)\\\\\frac{1}{P}\,dP=kdt\\\\Integrating\,\, both\,\,sides\\\\ln|P|=kt+c\\\\P(t)=Ce^{kt}\\\\at \,\, t=0\,\, P(0)=P_{o}\\\\P(0)=Ce^{k0}\\\\P_{o}=C\\\\then\\\\P(t)=P_{o}e^{kt}

To find value of k

at\,\, t=1599\,years\\\\P(1599)=\frac{P_{o}}{2}\\\\then\\\\\frac{P_{o}}{2} =P_{o}e^{k(1599)}\\\\\frac{1}{2} =e^{k(1599)}\\\\ln|\frac{1}{2}|=k(1599)\\\\k=\frac{ln|\frac{1}{2}|}{1599}=-4.3\times 10^{-3}\\\\\implies P(t)=P_{o}e^{-4.3\times 10^{-3}t}\\\\at\,\, t=900 \\\\P(900)=P_{o}e^{-4.3\times 10^{-3}(900)}\\\\P(900)=0.68P_{o}

Percentage of radioactive element is:

Amount after 900 years=\frac{P(900)}{P_{o}}\times 100\\\\=\frac{0.68P_{o}}{P_{o}}\times 100\%\\\\=68\%

3 0
3 years ago
3/ [(5)^(1/2) – (3)^(1/2)] simpfily
love history [14]
Usually the easiest way to solve this is by using a calculator. Of course only if your teacher allows you too.
8 0
3 years ago
Jordin went shopping for strawberries at a organic foods store. He saw that 12 out of 13 strawberries were about the size of his
ivann1987 [24]

Answer:

His inference is valid.

Step-by-step explanation:

The actual percentage of 12/13 is 92.31%, which rounds to 92%.

Hope this helps!

8 0
3 years ago
Can yall help me out here? (see photo)
pentagon [3]

Total customers: 26+42+33+24+40 =

165

The number of customers that bought 5-6 = 33

The probability would be 33/165

Total customers that bought fewer than 5 = 26+42 = 68

The probability is 68/165

8 0
3 years ago
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