Ksjjsksjskskdjslslfjsklskskskskdkd

Mathematics

2 answers:

andrezito [222]2 years ago

7 0

Answer:

bgtbdjnshdbgfdhjsknbfgrgeybxhnsjinxbcybuxhnjinxuycrcbhnsjdfgdhjnrbgybhbvcxesexdfghunjtrxezsxdcfvgbhunjhgyfrxeszsxdcfvgbhnjihgyftdrxeszsexdfhujihgfrdesedrfgyhujhgfrdeswedrtyujijuhygfrdswedfghujhgtfdsdfghjhgtfrdeszasxdcfyuxsedrcfvbjiyvcvibvcxcfvmbvcxcdfvgbnminbvcxesdcfvbnubvtrcexertyubvcxsdcfvgybhunyvtfrcddcfvgbhunijnbvccdrftvybunbvcxcvbunijbvccfvgbhnjbvfcdxsezexdcfvgbhunhbvcrexrctvybunibvcxdcfgbhnjbvtcrxdcfvbhnjinbvcxdcfvbninbvcxeecvbnijbvcxecvbnjnbvcxcvbnjvchy7tfdcfvgbhuytrdtvyhuihu8yfdfcgvyhu8y7tdfcvhjuigfdgvhgfc gvgfvcb vnbmgbdvx 6f vgv ghy7u . bjnebrtg bhvcxu8shebfjcdwhdfcb njxshdeybcjxswhedqrbfcjmsqnhxjswdrnwdfrvrfjcdsmiunytrexjkijuhgfdxdcfvghbjbgvcxzxdcfghbgvcxzxdcfghjnhbgvfcdxxcvgbhjnnhbvcxcfghjkm ncxrtyhn bxdrtgyhng dfgthn r4tgb 5tgb gt5gb bhyujm u7yhb cwsdxcftgyh

Step-by-step explanation:

Paraphin [41]2 years ago

6 0

Answer:

Step-by-step explanation:

Jkjkjkjkj idk what you want answered so jwkksnfiwe

You might be interested in

The circle below is centred at point (2,-1) and has a radius of length 3 what is its equation ?

NeTakaya

Answer:

Option C

Step-by-step explanation:

The standard form of equation of a cirle is:

(x-h)^2+ (y-k)^2=r^2

In the given question as the point is given and the radius of circle is given:

So,

(h,k)=(2,-1)

and

r=3

Here,

h=2

k= -1

Putting the values of h,k and r in standard form

(x-2)^2+ (y-(-1))^2=(3)^2

(x-2)^2+ (y+1)^2=(3)^2

So the equation of circle is:

(x-2)^2+ (y+1)^2=9(3)^2

Option C is the correct answer ..

7 0

2 years ago

Please help!!

yarga [219]

Because this is a fourth degree that will be tricky to factor as it is, we will do a u substitution. Let $x^2=u$ . We can now rewrite that polynomial in terms of u: $u^2-7u-8=0$ . Filling into the quadratic formula we have $u=\frac{7+/-\sqrt{49-4(1)(-8)}}{2}$ . Simplifying down $u=\frac{7+/-\sqrt{81}}{2}$ and $u=\frac{7+9}{2}$ or $u=\frac{7-9}{2}$ . That means that u = 8 or u = -1. But don't forget that we let $u=x^2$ , so we have to put x-squared back in for u. That gives us $x^2=8$ which simplifies down to $+/-2\sqrt{2}$ . That also gives us $x^2=-1$ . When we take the square root of -1, we have to use the fact that -1 = i^2, so we sub that in to get $x=+/-i$ . All in all, your solutions are as follows: $2\sqrt{2},-2\sqrt{2},i,-i$ which is choice b.