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4 years ago

Is 0.575 as a rational

Mathematics

2 answers:

earnstyle [38]4 years ago

7 0

It is a rational number because it can be expressed as a fraction as 23/40

inessss [21]4 years ago

4 0

The answer would be 23/41

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8 + 4 5 b = 9 10 b The solution is b =

andre [41]

Answer:

b=0.0092

Step-by-step explanation:

8 + 4 5b = 9 10b

-45b -45b

----------------------------

8=865b

divide by 865

8/865=865b/865

0.0092=b

b=0.0092

3 0

4 years ago

Fill in the blanks 6p^2=

den301095 [7]

Answer:

6p^2=36p^2

Step-by-step explanation:

3 0

3 years ago

Kylie draws a triangle on a coordinate grid. She then performs the following

zmey [24]

B PLEASE GIVE BRAINLIEST

7 0

3 years ago

In quadrilateral ABCD what is m

sineoko [7]

Answer:

B.99°

Step-by-step explanation:

105°+ 66°+90°+ angle c=360°

angle c= 360° - 261

=99°

8 0

3 years ago

Use the binomial theorem to expand the following binomial expressions.

andre [41]

Answer:

Remember, the expansion of $(x+y)^n$ is $(x+y)^n=\sum_{k=0}^n \binom{n}{k}x^{n-k}y^k$ , where $\binom{n}{k}=\frac{n!}{(n-k)!k!}$ .

$(1+\sqrt{2})^5=\sum_{k=0}^5 \binom{5}{k}1^{5-k}\sqrt{2}^k=\sum_{k=0}^5 \binom{5}{k}2^{\frac{k}{2}}\\=\binom{5}{0}2^{\frac{0}{2}}+\binom{5}{1}2^{\frac{1}{2}}+\binom{5}{2}2^{\frac{2}{2}}+\binom{5}{3}2^{\frac{3}{2}}+\binom{5}{4}2^{\frac{4}{2}}+\binom{5}{5}2^{\frac{5}{2}}\\=1+5\sqrt{2}+10*2+10*2^{\frac{3}{2}}+5*4+1*2^{\frac{5}{2}}\\=41+5\sqrt{2}+10*2^{\frac{3}{2}}+2^{\frac{5}{2}}$

$(1+i)^9=\sum_{k=0}^9 \binom{9}{k}1^{9-k}i^k=\sum_{k=0}^9 \binom{9}{k}i^k\\=\binom{9}{0}i^0+\binom{9}{1}i^1+\binom{9}{2}i^2+\binom{9}{3}i^3+\binom{9}{4}i^4+\binom{9}{5}i^5+\binom{9}{6}i^6+\\+\binom{9}{7}i^7+\binom{9}{8}i^8+\binom{9}{9}i^9\\=1+9i-36-84i+126+126i-+84-36i+9+i\\=16+16i$

$(1-\pi)^5=\sum_{k=0}^5 \binom{5}{k}1^{9-k}(-\pi)^k=\sum_{k=0}^5 \binom{5}{k}(-\pi)^k\\=\binom{5}{0}(-\pi)^0+\binom{5}{1}(-\pi)^1+\binom{5}{2}(-\pi)^2+\binom{5}{3}(-\pi)^3+\binom{5}{4}(-\pi)^4+\binom{5}{5}(-\pi)^5\\=1-5+10\pi^2-10\pi^3+5\pi^4-\pi^5$

$(\sqrt{2}+i)^6=\sum_{k=0}^6 \binom{6}{k}\sqrt{2}^{6-k}i^k\\=\binom{6}{0}\sqrt{2}^{6}i^0+\binom{6}{1}\sqrt{2}^{5}i+\binom{6}{2}\sqrt{2}^{4}i^2+\binom{6}{3}\sqrt{2}^{3}i^3+\binom{6}{4}\sqrt{2}^{2}i^4+\binom{6}{5}\sqrt{2}i^5+\binom{6}{6}\sqrt{2}^{0}i^6$

$(2-i)^6=\sum_{k=0}^6 \binom{6}{k}2^{6-k}(-i)^k\\=\binom{6}{0}2^{6}(-i)^0+\binom{6}{1}2^{5}(-i)^1+\binom{6}{2}2^{4}(-i)^2+\binom{6}{3}2^{3}(-i)^3+\binom{6}{4}2^{2}(-i)^4+\binom{6}{5}2^{1}(-i)^5+\binom{6}{k}2^{0}(-i)^6\\=1-32i+\binom{6}{2}16i^2-\binom{6}{3}8i^3+\binom{6}{4}4i^4-\binom{6}{5}2i^5+\binom{6}{k}i^6$

5 0

4 years ago