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3 years ago

How do you find the base and height?

Mathematics

1 answer:

dusya [7]3 years ago

8 0

A triangle is 180 degrees. Depending on their angles: Right Angle - 90 degrees
Acute Angle - less than 90 degrees and Obtuse Angle greater than 90 degrees and under 180 degrees. Hope that helps.

You might be interested in

Is the LCM of a pair of numbers ever equal to one of the numbers?

den301095 [7]

Yes.

For example, the LCM of 12 and 36 is 36, because factoring these numbers gives

12 = 2^2 * 3

36 = 2^2 * 3^2

So, to match them, just multiply the twelve by 3, then they're both 36.

6 0

3 years ago

How much heat in kJ is released when nitrogen and hydrogen react as below to form 100.g ammonia heat equals-92.2kj

tresset_1 [31]

100g of ammonia releases 271.18 kJ of heat.

<u>Explanation:</u>

Given:

N₂ reacts with H₂

The reaction is:

2N₂ + 3H₂ → 2NH₃ + 92.2 kJ

According to the balanced reaction:

2 X 17g of ammonia releases 92.2 kJ of heat

34g of ammonia releases 92.2 kJ of heat

100g of ammonia will release = $\frac{92.2}{34} X 100$ kJ of heat

= 271.18 kJ

Therefore, 100g of ammonia releases 271.18 kJ of heat.

8 0

2 years ago

Help me, please? I don't get it

Korvikt [17]

Answer:

1/10 = 2/20

1/3 = 3/9

5/7 = 15/21

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Suppose after 2500 years an initial amount of 1000 grams of a radioactive substance has decayed to 75 grams. What is the half-li

krok68 [10]

Answer:

The correct answer is:

Between 600 and 700 years (B)

Step-by-step explanation:

At a constant decay rate, the half-life of a radioactive substance is the time taken for the substance to decay to half of its original mass. The formula for radioactive exponential decay is given by:

$A(t) = A_0 e^{(kt)}\\where:\\A(t) = Amount\ left\ at\ time\ (t) = 75\ grams\\A_0 = initial\ amount = 1000\ grams\\k = decay\ constant\\t = time\ of\ decay = 2500\ years$

First, let us calculate the decay constant (k)

$75 = 1000 e^{(k2500)}\\dividing\ both\ sides\ by\ 1000\\0.075 = e^{(2500k)}\\taking\ natural\ logarithm\ of\ both\ sides\\In 0.075 = In (e^{2500k})\\In 0.075 = 2500k\\k = \frac{In0.075}{2500}\\ k = \frac{-2.5903}{2500} \\k = - 0.001036$

Next, let us calculate the half-life as follows:

$\frac{1}{2} A_0 = A_0 e^{(-0.001036t)}\\Dividing\ both\ sides\ by\ A_0\\ \frac{1}{2} = e^{-0.001036t}\\taking\ natural\ logarithm\ of\ both\ sides\\In(0.5) = In (e^{-0.001036t})\\-0.6931 = -0.001036t\\t = \frac{-0.6931}{-0.001036} \\t = 669.02 years\\\therefore t\frac{1}{2} \approx 669\ years$

Therefore the half-life is between 600 and 700 years

5 0

3 years ago

I need help asap thanks

Sunny_sXe [5.5K]

$V=hb$

$h= \frac V b$

$h = \dfrac{ x^3 - 3x^2 + 5x - 3}{x^2 -2}$

Expressing that as a quotient and remainder,

$h =x - 3 = \dfrac{ 7 x - 9} { x^2 - 2}$

First choice.

7 0

2 years ago