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11111nata11111 [884]
2 years ago
14

Write an essay comparing the two arguments made about whether dodgyball should be played in school. Use textual evidence from th

e article to support your answer. TITLE- A Dodgy call AUTHOR Tum Jones Introduction Hook- Bridge-​
Mathematics
1 answer:
sergij07 [2.7K]2 years ago
3 0

Answer:

(This is not math)

Step-by-step explanation:

I think that dodgeball should be allowed un our school because it gets our legs moving and lets us extrasice and practice our swiftness

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10.<br> 25<br> xº<br> 40<br> Solve for x. Round to the nearest tenth
xxTIMURxx [149]

Answer:

51.32

Step-by-step explanation:

∠B = arcsin(b·sin(A)a)

= 0.67513 rad = 38.682° = 38°40'56"

∠C = 180° - A - B = 0.89566 rad = 51.318° = 51°19'4"

c = a·sin(C)sin(A) = 31.22499

3 0
3 years ago
If 100 cm id 1 m than how many cm are there are 13m
postnew [5]
The question requires you skills in doing multiplication or division. Ratio and proportion would also do here.
The topic here is about conversion.
You are given that 100 cm = 1m
Then how about 13 m?

100 cm : 1 m =  X : 13m
100 cm * 13m = 1m * x
1300 = 1m * X
X = 1300 cm

So 13m is equal to 1,300 cm 
3 0
3 years ago
Read 2 more answers
Each investment matures in 3 years. The interest compounds annually.
Sveta_85 [38]

bearing in mind that 4¾ is simply 4.75.

\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$600\\ r=rate\to 5\%\to \frac{5}{100}\dotfill &0.05\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &3 \end{cases} \\\\\\ A=600\left(1+\frac{0.05}{1}\right)^{1\cdot 3}\implies A=600(1.05)^3\implies A=694.575 \\\\[-0.35em] ~\dotfill

\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$750\\ r=rate\to 4.75\%\to \frac{4.75}{100}\dotfill &0.0475\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &3 \end{cases} \\\\\\ A=750\left(1+\frac{0.0475}{1}\right)^{1\cdot 3}\implies A=750(1.0475)^3\implies A\approx 862.032

well, the interest for each is simply A - P

695.575 - 600 = 95.575.

862.032 - 750 = 112.032.

4 0
3 years ago
Am I correct? If not please explain and which one!? THANK SO MUCH ❤️
Rzqust [24]

Answer:

No, your answer is incorrect.

simplify the question that would be the last one D.

Step-by-step explanation:

if this helped you please mark brainliest <3

6 0
2 years ago
In which section of the number line is √17
Murljashka [212]

\bf \sqrt{17}\approx 4.123 \\\\[-0.35em] ~\dotfill\\\\ \boxed{4}\rule[0.35em]{2em}{0.25pt}\stackrel{\sqrt{17}}{4.123}\rule[0.35em]{10em}{0.25pt}\boxed{5}\qquad \leftarrow \textit{between 4 and 5}

3 0
3 years ago
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