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2 years ago

Find the highest common factor of 360 and 900

Mathematics

1 answer:

yuradex [85]2 years ago

7 0

Answer:

180

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Solve. - 1/3b = 9 A. -3 B. 3 C. - 27 D. 27

ruslelena [56]

$-\dfrac{1}{3}b=9\ \ \ |multiply\ both\ sides\ by\ (-3)\\\\-3\cdot\left(-\dfrac{1}{3}b\right)=-3\cdot9\\\\\huge\boxed{b=-27}\leftarrow\boxed{C}$

3 0

3 years ago

Read 2 more answers

Simplify the expression (5 3i) (4-4i) a. 8i b. 9-i c. 9 i d. 9-7i

poizon [28]

The expression will be d. 9-7i

8 0

3 years ago

In the figure, two poles are 25 m and 35 m high. A cable 23 m long Joins the tops of the poles. Find the distance between the po

sukhopar [10]

Answer: 20.71 m

Step-by-step explanation:

Given

the length of the poles are 25m and 35 m

The length of the cable is 23 m

Suppose the distance between them is x

From the figure, apply Pythagoras theorem

$\Rightarrow 23^=x^2+10^2\\\Rightarrow 529-100=x^2\\\Rightarrow x^2=429\\\\\Rightarrow x=\sqrt{429}=20.71\ m$

Distance between Poles is 20.71 m

6 0

3 years ago

If there are 19 questions and each question has 4 answers, what is the probability she will answer at least one question correct

gregori [183]

Answer:

0.9958

Step-by-step explanation:

P(being correct) = 1/4 = 0.25

Hence, p = 0.25

n = 19

P(x ≥ 1) = p(x = 1) + p(x = 2) +... + p(x = 19)

Using the binomial probability formula :

P(x =x) = nCx * p^x * (1 - p)^(n - x)

However, to save computation time, we could use a calculator :

Using a calculator,

P(x ≥ 1) = 0.99577

P(x ≥ 1) = 0.9958

7 0

2 years ago

If f (n)(0) = (n + 1)! for n = 0, 1, 2, , find the taylor series at a=0 for f.

Pie

Given that $f^{(n)}(0)=(n+1)!$ , we have for $f(x)$ the Taylor series expansion about 0 as

$f(x)=\displaystyle\sum_{n=0}^\infty\frac{(n+1)!}{n!}x^n=\sum_{n=0}^\infty(n+1)x^n$

Replace $n+1$ with $n$ , so that the series is equivalent to

$f(x)=\displaystyle\sum_{n=1}^\infty nx^{n-1}$

and notice that

$\displaystyle\frac{\mathrm d}{\mathrm dx}\sum_{n=0}^\infty x^n=\sum_{n=1}^\infty nx^{n-1}$

Recall that for $|x|$ , we have

$\displaystyle\sum_{n=0}^\infty x^n=\frac1{1-x}$

which means

$f(x)=\displaystyle\sum_{n=1}^\infty nx^{n-1}=\frac{\mathrm d}{\mathrm dx}\frac1{1-x}$
$\implies f(x)=\dfrac1{(1-x)^2}$

5 0

3 years ago