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Levart [38]
3 years ago
11

What is the sq root of 99 over the sq root of 11 ???

Mathematics
2 answers:
Evgen [1.6K]3 years ago
4 0
9.94987437/<span>3.31662479 or 3
Hope this helps!
</span>
Step2247 [10]3 years ago
4 0
The answer is 3. =D

Your welcome.
You might be interested in
The answer choices are -3, 3/2, 3, and -12/5.<br> thanks in advance!! :)
BabaBlast [244]

Answer:

<em>(C). {3} </em>

Step-by-step explanation:

\frac{x}{2(x+1)} = \frac{-2x}{4(x+1)} + \frac{2x-3}{x+1} ( x ≠ - 1 )

\frac{x}{2(x+1)} - \frac{2x-3}{x+1} + \frac{x}{2(x+1)} = 0

\frac{2x-3}{x+1} - \frac{x}{(x+1)} = 0

\frac{x-3}{x+1} = 0 ⇒ <em>x = 3</em>

8 0
3 years ago
Can you pls help i need to submit it soon ..i'll mark you brainlist
garik1379 [7]

Here is the answer with explanation

8 0
2 years ago
Read 2 more answers
Please help me..............................
myrzilka [38]

Answer:

y = 12

Step-by-step explanation:

∵ m∠ABC = 40

∵ BD ray is the bisector of ∠ABC

∴ m∠ABD = m∠CBD = 20

In 2 Δ BAD and BCD:

∵ m∠BAD = m∠BCD = 90°

∵ m∠ABD = m∠CBD = 20°

∵ BD is a common side in the two triangles

∴ ΔABD congruent to ΔCBD⇒(AAS)

∴ AD = CD

∴ 3y + 6 = 5y - 18

∴ 5y - 3y = 6 + 18

∴ 2y = 24

∴ y = 24/2 = 12

6 0
3 years ago
Consider the parabola r​(t)equalsleft angle at squared plus 1 comma t right angle​, for minusinfinityless thantless thaninfinity
kodGreya [7K]

Given:-   r(t)=< at^2+1,t>  ; -\infty < t< \infty , where a is any positive real number.

Consider the helix parabolic equation :  

                                              r(t)=< at^2+1,t>

now, take the derivatives we get;

                                            r{}'(t)=

As, we know that two vectors are orthogonal if their dot product is zero.

Here,  r(t) and r{}'(t)  are orthogonal i.e,   r\cdot r{}'=0

Therefore, we have ,

                                  < at^2+1,t>\cdot < 2at,1>=0

< at^2+1,t>\cdot < 2at,1>=

                                              =2a^2t^3+2at+t

2a^2t^3+2at+t=0

take t common in above equation we get,

t\cdot \left (2a^2t^2+2a+1\right )=0

⇒t=0 or 2a^2t^2+2a+1=0

To find the solution for t;

take 2a^2t^2+2a+1=0

The numberD = b^2 -4ac determined from the coefficients of the equation ax^2 + bx + c = 0.

The determinant D=0-4(2a^2)(2a+1)=-8a^2\cdot(2a+1)

Since, for any positive value of a determinant is negative.

Therefore, there is no solution.

The only solution, we have t=0.

Hence, we have only one points on the parabola  r(t)=< at^2+1,t> i.e <1,0>




                                               




6 0
3 years ago
HELP!!! <br> write the following in algebraic expressions:
Nadusha1986 [10]

Step-by-step explanation:

Here is the answer for the unkown n(k+m)

5 0
3 years ago
Read 2 more answers
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