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Lemur [1.5K]
3 years ago
6

Help me find the answer please need help asap

Mathematics
1 answer:
pychu [463]3 years ago
5 0
The answer would be 50 degrees
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Sebuah benda bergerak lurus dengan kecepatan 4km/jam selama 1 jam pertama.Pada jam kedua kecepatan dikurangi menjadi setengahnya
never [62]
A = 4 km/jam
r = 1/2
jarak terjauh
s = a/(1 - r)
s = 4/(1 - 1/2)
<span>s = 8 km</span>
3 0
3 years ago
A yogurt shop offers 5 different flavors and11 different topping how many choices are possible for a single serving of yogurt wi
GalinKa [24]
If you have 5 flavors and 11 toppings, but you can only have 1 of each you take 5×11 to get 55
3 0
4 years ago
A construction worker drops a hammer from a height of 25 m.
Snezhnost [94]

Answer:

Time at which the hammer reaches the ground is 2.3 seconds.

Step-by-step explanation:

We are given the formula as, h(t)=-4.9t^2+v_{o}+h_{o}.

It is known that, when height i.e. h_{o}=25 meter, then velocity v_{o}=0 m/sec.

So, we get the formula as, h(t)=-4.9t^2+25.

Now, when the hammer hits the ground, the height h_{o}=0 meter.

Thus, we have,

h(t)=-4.9t^2+25

⇒ 0=-4.9t^2+25

⇒ 4.9t^2=25

⇒ t^2=5.1

i.e. t= ±2.3 sec

Since, time cannot be negative.

So, t= 2.3 sec

Hence, the time at which the hammer reaches the ground is 2.3 seconds.

4 0
3 years ago
Read 2 more answers
2 points
ipn [44]
Discount: 0,15 * 25 = 3,75$

Price after discount: 25 - 3,75 = 21,25$

The game will cost less than 22$, which is Jade’s budget for a board game.
4 0
3 years ago
Recall that m(t) = (1/2)^t/h for radioactive decay, where h is the half-life. Suppose that a 500g sample of phosphorus-32 decays
katrin2010 [14]

The question is incomplete, here is the complete question:

Recall that m(t) = m.(1/2)^t/h for radioactive decay, where h is the half-life. Suppose that a 500 g sample of phosphorus-32 decays to 356 g over 7 days. Calculate the half life of the sample.

<u>Answer:</u> The half life of the sample of phosphorus-32 is 14.28days^{-1}

<u>Step-by-step explanation:</u>

The equation used to calculate the half life of the sample is given as:

m(t)=m_o(1/2)^{t/h}

where,

m(t) =  amount of sample after time 't' = 356 g

m_o = initial amount of the sample = 500 g

t = time period = 7 days

h = half life of the sample = ?

Putting values in above equation, we get:

356=500\times (\frac{1}{2})^{7/h}\\\\h=14.28days^{-1}

Hence, the half life of the sample of phosphorus-32 is 14.28days^{-1}

7 0
4 years ago
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