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Natali [406]
2 years ago
10

Compute the length of sides AB, AC.

Mathematics
1 answer:
Paha777 [63]2 years ago
8 0
AB=24 AC=24. The triangle is SSS Which means all sides are congruent so just setting AB = AC then finding x. Then plugging it in u get your answer
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1/3 of students at a school are boys. If there are 600 students at the school, how many are girls?
Temka [501]

Answer:

200 students are boys and rest of 400 are girls

Step-by-step explanation:

Given that:

Total students = 300

1/3 students of total are boys

so.

Boys = 1/3(600)

Boys = 200

For finding number of girls:

Girls = total - boys

Girls = 600 - 200

Girls = 400

i hope it will help you!

8 0
4 years ago
Read 2 more answers
7. Find f(9) when f(x) = 12 (17 – 3x) *​
Radda [10]

Answer:

-120

Step-by-step explanation:

6 0
3 years ago
PLZ help will give brainliest and 60 points asap
vovangra [49]

Answer:

A. -2 + 12 -2 ^ 3 + 2 ^ 0 * 3

= 10 - 2 ^ 3 + 2 ^ 0 * 3

= 10 - 8 + 1

= 3

B. 9 ^ (1/2)(25) - 8 + 2 + 3 ^ 2

= 9 ^ (1/2)(25) - 6 + 9

= 3 ^ 25 + 3

= 3 ^ 25 + 3

= 847288609446

C. 1/2 * 3/5 * (2 ^ 3 - 6) + 1/5

= 1/2 * 3/5 * 2 + 1/5

= 3/5 + 1/5

= 4/5

3 0
3 years ago
Read 2 more answers
PLS I NEED HELP ON THIS QUESTION..
Sonbull [250]

Answer:

it's literraly right there boy

Step-by-step explanation:

bc it's right there

7 0
3 years ago
Read 2 more answers
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
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