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3 years ago

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)
s = 8 km

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.

Answer: The half life of the sample of phosphorus-32 is $14.28days^{-1}$

Step-by-step explanation:

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