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3 years ago

Pls give me the correct answer

Mathematics

1 answer:

skelet666 [1.2K]3 years ago

6 0

Answer:

-21, -16, -14, -12, -11.

and

p≤−11

Hope this helps!

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Please help! functions operations. explain please

Lina20 [59]

Answer:

$f^{-1}(x)=x^2-3$

Step-by-step explanation:

Given $f(x)=\sqrt{x+3}$

Let $y=\sqrt{x+3}$

Interchange x and y

$x=\sqrt{y+3}$

Solve for y, by first squaring both sides

$x^2=y+3$

$x^2-3=y$

Hence

$f^{-1}(x)=x^2-3$

The last choice is correct

5 0

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

What number is divisible by both 85 and 51

Vitek1552 [10]

17
85/17 is 5 and 51/17 is 3

7 0

3 years ago

Read 2 more answers

1 1/5 divided by 3/10

MissTica

Answer: <em><u>4.</u></em> This is because 1 and 1/5 = 1.2 and 3/10 = 0.3 and 1.2 divided by 0.3 = 4. That's because 1.2 can be split into 4 equal groups and each group has 0.3.

4 0

3 years ago

Need a correct answer please

bezimeni [28]

Answer:

102°

Step-by-step explanation:

∠PQR = ∠PQS + ∠SQR

12x - 6 = 16 + 9x + 17

3x = 27

x = 9

∠PQR = 102°

6 0

3 years ago