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3 years ago

15.

Mathematics

1 answer:

slamgirl [31]3 years ago

8 0

Answer:

Step-by-step explanation:

❌Option A is incorrect

Rationale: from the graph, 1 hr on the x-axis doesn't correspond to $100 on the y-axis

❌Option B is incorrect

Rationale: from the graph, 2 hrs on the x-axis doesn't correspond to $200 on the y-axis. It can't also be the unit rate because unit rate would be the amount of dollars per 1 hr.

✅Option C is CORRECT

Rationale: from the graph, 8 hrs on the x-axis corresponds to $500 on the y-axis

❌Option D is incorrect

Rationale: from the graph, 4 hr on the x-axis doesn't correspond to $300 on the y-axis

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Hey guys i need help

77julia77 [94]

Answer:

9 yards

Step-by-step explanation:

from the maple-> end, it is 28 2/6 yards. that means we add the segments that are already there to find how much of 28 2/6 is there between the maple and oak.

11 3/6 + 7 5/6 = 19 1/3

28 2/6 - 19 1/3 = 9

7 0

2 years ago

City Park: You are desinigng a marble planter for a city park. You want the length of the planter to be sic times the height, an

Alex Ar [27]

Answer: 6 feet x 3 feet

Step-by-step explanation:

Let the height be given by= x

The length is= 6x -2 (1) from both sides= 6x-2

The width is= 3x-2(1) from both sides= 3x-2

The total volume= length * width * height

4=(6x-2)*(3x-2)*x

Solving we get,

x=1 and other factor is not the valid option.

So the outer dimensions should be 6 feet x 3 feet

8 0

3 years ago

Simplify.

Ket [755]

First off, nicely done on the exponents. You'd be surprised on how bad some look ,:D

a)
$(3xy^{3})(2x^{2}y)\\ (3*2)(x*x^{2})(y^{3}*y)\\\\ 6x^{3}y^{4}$

b)
$2y(2y+3y^{2})-y^{2}(3-y)\\ (4y^{2}+6y^{3}) + (-3y^{2}+y^{3})\\ (6y^{3}+y^{3})+(4y^{2}-3y^{2})\\ 7y^{3}+y^{2}\\\\ y^{2}(7y+1)$

8 0

3 years ago

Calculate the pH of a buffer solution made by mixing 300 mL of 0.2 M acetic acid, CH3COOH, and 200 mL of 0.3 M of its salt sodiu

Lesechka [4]

Answer:

Approximately $4.75$ .

Step-by-step explanation:

Remark: this approach make use of the fact that in the original solution, the concentration of $\rm CH_3COOH$ and $\rm CH_3COO^{-}$ are equal.

${\rm CH_3COOH} \rightleftharpoons {\rm CH_3COO^{-}} + {\rm H^{+}}$

Since $\rm CH_3COONa$ is a salt soluble in water. Once in water, it would readily ionize to give $\rm CH_3COO^{-}$ and $\rm Na^{+}$ ions.

Assume that the $\rm CH_3COOH$ and $\rm CH_3COO^{-}$ ions in this solution did not disintegrate at all. The solution would contain:

$0.3\; \rm L \times 0.2\; \rm mol \cdot L^{-1} = 0.06\; \rm mol$ of $\rm CH_3COOH$ , and

$0.06\; \rm mol$ of $\rm CH_3COO^{-}$ from $0.2\; \rm L \times 0.3\; \rm mol \cdot L^{-1} = 0.06\; \rm mol$ of $\rm CH_3COONa$ .

Accordingly, the concentration of $\rm CH_3COOH$ and $\rm CH_3COO^{-}$ would be:

$\begin{aligned} & c({\rm CH_3COOH}) \\ &= \frac{n({\rm CH_3COOH})}{V} \\ &= \frac{0.06\; \rm mol}{0.5\; \rm L} = 0.12\; \rm mol \cdot L^{-1} \end{aligned}$ .

$\begin{aligned} & c({\rm CH_3COO^{-}}) \\ &= \frac{n({\rm CH_3COO^{-}})}{V} \\ &= \frac{0.06\; \rm mol}{0.5\; \rm L} = 0.12\; \rm mol \cdot L^{-1} \end{aligned}$ .

In other words, in this buffer solution, the initial concentration of the weak acid $\rm CH_3COOH$ is the same as that of its conjugate base, $\rm CH_3COO^{-}$ .

Hence, once in equilibrium, the $\rm pH$ of this buffer solution would be the same as the ${\rm pK}_{a}$ of $\rm CH_3COOH$ .

Calculate the ${\rm pK}_{a}$ of $\rm CH_3COOH$ from its ${\rm K}_{a}$ :

$\begin{aligned} & {\rm pH}(\text{solution}) \\ &= {\rm pK}_{a} \\ &= -\log_{10}({\rm K}_{a}) \\ &= -\log_{10} (1.76 \times 10^{-5}) \\ &\approx 4.75\end{aligned}$ .

7 0

2 years ago

Set up and evaluate the optimization problems. (Enter your answers as comma-separated lists.) Find two positive integers such th

alexira [117]

Answer and Step-by-step explanation:

Let x and y be two positive integers and their sum is 14:

X + y = 14

And the sum of square of this number is:

f = x2 + y2

= x2+ (14 – x)2

Differentiate with respect to x, we get:

F’(x) = [ x2 + (14 – x)2]’ = 0

2x + 2(14-x)(-1) = 0

2x +( 28 – 2x)(-1) = 0

2x – 28 +2x = 0

2x + 2x = 28

4x = 28

X = 7

Hence, y = 14 – x = 14 -7 = 7

Now taking second derivative test:

F”(x) > 0

For x = y = 7,f reaches its maximum value:

(7)2 + (7)2 = 49 + 49

= 98

F at endpoints x Є [ 0, 14]

F(0) = 02 + (14 – 0)2

= 196

F(14) = (14)2 + (14 – 14)2

= 196

Hence the sum of squares of these numbers is minimum when x = y = 7

And maximum when numbers are 0 and 14.

6 0

2 years ago