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maksim [4K]
4 years ago
15

A ball is launched into the air from a height of 12 feet at time t = 0. The function that models this situation is h(t) = -2t2 +

10t + 12, where t is measured in seconds and h is the height in feet.
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What is the height of the ball after 2 seconds?
---
When will the ball hit the ground?
Mathematics
2 answers:
marshall27 [118]4 years ago
6 0

Answer:

Step-by-step explanation:

50

Natalka [10]4 years ago
6 0
NfnzmmMknx is Cual. It will gto until then
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vova2212 [387]

Answer:

5 units

Step-by-step explanation:

create a point in (1,1) and a right angled triangle will be formed, you can then use distance formula and the distance from the first point to (1,1) will be 3 units and from second point to (1,1) will be 4 units and verify using pythagoras theorem

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3 years ago
How do I rewrite with no zero or negative exponents?
oksian1 [2.3K]
Do it with positive exponets
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3 years ago
Cells use the hydrolysis of adenosine triphosphate, abbreviated as ATP, as a source of energy. Symbolically, this reaction can b
nataly862011 [7]

Answer:

Step-by-step explanation:

From the given information:

ΔG° = -30.5 kJ/mol

By applying the following equation to calculate the value of K.

ΔG° =-RT㏑K

making ㏑ K  the subject of the formula:

\mathtt{ In  \ K} = \dfrac{\Delta G^0}{-RT}

where;

Temperature at 25° C = (25 + 273)K

= 298K

R = 8.3145 J/mol.K (gas cosntant)

\mathtt{ In  \ K} = \dfrac{-30.5 \times 10^{3}\ J /mol} {-(8.3145 \ J/mol. K \times 298 \ K}

\mathtt{ In  \ K} = \dfrac{-30.5 \times 10^{3}\ J /mol} {-2477.721 J/mol }

㏑K = 12.309

K = e^{12.309}

K = 221682.17

K = 2.22 × 10⁵

b) The reaction for the metabolism of glucose is given as:

C_6H_{12} O_6 + 6O_{2(g)} \to  + 6CO_{2(g)} + 6H_2O_{(l)}

From the above expression, let calculate the Gibbs free energy by using the formula:

\Delta G^0_{rx n }= \Delta G^0_{product}- \Delta G^0_{reactant}

\Delta G^0_{rx n }= [6 \times \Delta G^0_{f}(CO_2) + 6 \times \Delta G^0_{f}(H_2O)] - [1 \times \Delta G^0_{f}(C_6H_{12}O_6) + 6 \times \Delta G^0_{f}(O_2)]

At standard conditions;

The values of corresponding compounds are substituted into the equation above:

Thus,

\Delta G^0_{rx n }= [6 \times (-394) + 6 \times (-237)] - [1 \times (-911) + 6 \times (0)] \ kJ/mol

\Delta G^0_{rx n }= [-2364-1422] - [-911+0] \ kJ/mol

\Delta G^0_{rx n }= -3786 +911 \ kJ/mol

\Delta G^0_{rx n }= -2875 \ kJ/mol

\Delta G^0_{rx n }= -2875000 \ J/mol

Now, the no of ATP molecules generated = \dfrac{\Delta G^0 \text{of metabolism for glucose}}{\Delta G^0 \text{of hydrolysis  for ATP}}

= (-2875000 J/mol ) / -30500 J/mol

= 94.26

≅ 94 ATP molecules generated

3 0
3 years ago
Kris has a box of 8 crayons. Sylvia's box has 6 times as many crayon's as Kri's box. How many crayons are in Sylvia's box?
svp [43]
Sylvia would have 48 crayons
5 0
4 years ago
Read 2 more answers
Solve 2x + y = 7 for y<br> Y=?
Vladimir [108]

Answer:

The Answer is: y = -2x + 7

Step-by-step explanation:

2x + y = 7

Subtract 2x from both sides:

y = -2x + 7

7 0
3 years ago
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