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3 years ago

9

Triangle 1 has vertices at (A,B), (C,D), and (E,F). Triangle 2 has vertices at (A,3B), (C,3D), and (E,3F). What can you conclude

about triangle 2?

1 answer:

MissTica3 years ago

4 0

Answer:

A

Step-by-step explanation:

tell me if corect

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an artist is selling original paintings and prints pf the original at an art show. he has a total of 240 paintings and prints. t

RSB [31]

Given the the total number of paintings and prints that an artist has which is 240, and the ratio is 3 of paintings is to 5 prints, here is the answer as to what the ratio is of the remaining paintings to prints. Given that the total number is 240 with the ratio of 3 : 5, this would mean that there are a total of 90 paintings and 150 prints. Since 15 paintings and 100 prints are sold, this would leave us 75 paintings and 50 prints. The new ratio would be 3 : 2. Hope this answer helps.

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4 years ago

Bars of soap come in packages of 2 and packages of 8. The 2-bar pack costs $1.98, and the 8-bar pack costs $8.88. Which is the b

Sunny_sXe [5.5K]

Answer:

the answer is 2-bar pack that costs 1.98, it is 0.99 per bar.

Step-by-step explanation:

divide $1.98 and 2 which comes to $0.99 then divide $8.88 and 8 which comes to $1.11 so 0.99 is cheaper and the better deal

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2 years ago

What is the answer to this problem?

yarga [219]

Try this:
V=a³=6.2³=238,328 in.³
Answer: A.

3 0

3 years ago

The runner covers a 100 meters in 9.7 seconds. What is her speed?

mestny [16]

Answer:

10.3 m/s

Step-by-step explanation:

Speed = Distance / Time

100 metres travelled in 9.7 seconds means that her speed was 100/9.7 metres per second, which is 10.30927... rounded to 10.3 metres per second.

5 0

3 years ago

The half-life of a certain radioactive substance is 45 days. There are 6.2 grams present initially. On what day

kvasek [131]

Answer:

There will be less than 1 gram of the radioactive substance remaining by the elapsing of 118 days

Step-by-step explanation:

The given parameters are;

The half life of the radioactive substance = 45 days

The mass of the substance initially present = 6.2 grams

The expression for evaluating the half life is given as follows;

$N(t) = N_0 \left (\dfrac{1}{2} \right )^{\dfrac{t}{t_{1/2}}$

Where;

N(t) = The amount of the substance left after a given time period = 1 gram

N₀ = The initial amount of the radioactive substance = 6.2 grams

$t_{1/2}$ = The half life of the radioactive substance = 45 days

Substituting the values gives;

$1 = 6.2 \left (\dfrac{1}{2} \right )^{\dfrac{t}{45}$

$\dfrac{1}{6.2} = \left (\dfrac{1}{2} \right )^{\dfrac{t}{45}$

$ln\left (\dfrac{1}{6.2} \right ) = {\dfrac{t}{45} \times ln \left (\dfrac{1}{2} \right )$

$t = 45 \times \dfrac{ln\left (\dfrac{1}{6.2} \right ) }{ln \left (\dfrac{1}{2} \right )} \approx 118.45 \ days$

The time that it takes for the mass of the radioactive substance to remain 1 g ≈ 118.45 days

Therefore, there will be less than 1 gram of the radioactive substance remaining by the elapsing of 118 days.

3 0

3 years ago