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2 years ago

A Pentagon has angle measure of 101°,103°,115°and 113° what must the fifth angle measure? The fifth angle measures

Mathematics

1 answer:

eimsori [14]2 years ago

8 0

Add all angles.
101+103+115+113=432
then subtract 432 from the total angle of a pentagon which is 5

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Whitney wants to buy a sweater for $36. If the sales tax rate is 8.25%, what will be the amount of sales tax on Whitney’s purcha

Sedaia [141]

Answer:297.00

Step-by-step explanation:

There's 2 total decimal places in both numbers.

Ignore the decimal places and complete the multiplication as if operating on two integers.

3 6

× 8 2 5

+ 1 8 0

+ 7 2

+ 2 8 8

= 2 9 7 0 0Rewrite the product with 2 total decimal places.

Answer = 297.00

Therefore:

36 × 8.25 = 297.00

6 0

2 years ago

(2+2)-(45+67)=x solve for x

Minchanka [31]

Answer:

108

HOPE THIS HELPS

-Todo<3

Step-by-step explanation:

4-112=x

108=x

4 0

2 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

Fill in the sign that completes the equation below<br><br> 7,300=730 10

Zepler [3.9K]

It is a multiplication sign. *
I hope I helped.

7 0

2 years ago

Read 2 more answers

Scientist are studying a bacteria sample. The function f(x)=360(1.07)^x gives the number of bacteria in the sample at the end of

Lapatulllka [165]

Options :

A. The initial number of bacteria is 7.

B. The initial of bacteria decreases at a rate of 93% each day.

C. The number of bacteria increases at a rate of 7% each day.

D. The number of bacteria at the end of one day is 360.

Answer:

C. The number of bacteria increases at a rate of 7% each day.

Step-by-step explanation:

Given the function :

f(x)=360(1.07)^x ; Number of bacteria in sample at the end of x days :

The function above represents an exponential growth function :

With the general form ; Ab^x

Where A = initial amount ;

b = growth rate

x = time

For the function :

A = initial amount of bacteria = 360

b = growth rate = (1 + r) = 1.07

If ; (1 + r) = 1.07 ; we can solve for r to obtain the daily growth rate ;

1 + r = 1.07

r = 1.07 - 1

r = 0.07

r as a percentage ;

0.07 * 100% = 7%

7 0

2 years ago