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3 years ago

Which set of side lengths will form a right triangle 8,15,17 7,24,26 6,8,11 10,24,25

1 answer:

6 0

Let’s try the pythagorean formula.

1) 8^2+15^2=17^2
289=289 Yes

2) 7^2+24^2=26^2
625 ≠ 676 No

3) 6^2+8^2=11^2
100 ≠ 121 No

4) 10^2+24^2=25^2
676 ≠ 625 No

Therefor, 8,15,17 is your answer

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For all of the questions <br>with method <br>thankyou

Dmitry_Shevchenko [17]

Answer:

(i) 7/10

(ii) 3/10

(iii) 1/5

(iv) Rs 40,000

Step-by-step explanation:

The fraction of the salary spent on food = 1/2

The fraction of the salary spent on rented house fee = 1/5

(i) The fraction spent for both food and rental fee = (1/2) + (1/5) = (5 + 2)/10 = 7/10

(ii) The remainder (rest) of the salary = 1 - 7/10 = 3/10

The fraction of the remainder spent for children's education = 1/3

The fraction of the total salary spent for the children's education = (1/3) × (3/10) = 1/10

(iii) The remaining portion deposited in the bank = 1 - (1/10 + 7/10)) = 2/10 = 1/5

(iv) The amount equal to portion of 1/5 of his salary deposited in the bank is Rs 8000

Let <em>x</em> represent his whole salary, we have;

(1/5) × x = Rs 8,000

x = 5 × Rs 8,000 = Rs 40,000

His whole salary is Rs 40,000.

3 0

3 years ago

Find the maximum and minimum values attained by f(x, y, z) = 5xyz on the unit ball x2 + y2 + z2 ≤ 1.

Allushta [10]

Check for critical points within the unit ball by solving for when the first-order partial derivatives vanish:
$f_x=5yz=0\implies y=0\text{ or }z=0$
$f_y=5xz=0\implies x=0\text{ or }z=0$
$f_z=5xy=0\implies x=0\text{ or }y=0$

Taken together, we find that (0, 0, 0) appears to be the only critical point on $f$ within the ball. At this point, we have $f(0,0,0)=0$ .

Now let's use the method of Lagrange multipliers to look for critical points on the boundary. We have the Lagrangian

$L(x,y,z,\lambda)=5xyz+\lambda(x^2+y^2+z^2-1)$

with partial derivatives (set to 0)

$L_x=5yz+2\lambda x=0$
$L_y=5xz+2\lambda y=0$
$L_z=5xy+2\lambda z=0$
$L_\lambda=x^2+y^2+z^2-1=0$

We then observe that

$xL_x+yL_y+zL_z=0\implies15xyz+2\lambda=0\implies\lambda=-\dfrac{15xyz}2$

So, ignoring the critical point we've already found at (0, 0, 0),

$5yz+2\left(-\dfrac{15xyz}2\right)x=0\implies5yz(1-3x^2)=0\implies x=\pm\dfrac1{\sqrt3}$
$5xz+2\left(-\dfrac{15xyz}2\right)y=0\implies5xz(1-3y^2)=0\implies y=\pm\dfrac1{\sqrt3}$
$5xy+2\left(-\dfrac{15xyz}2\right)z=0\implies5xy(1-3z^2)=0\implies z=\pm\dfrac1{\sqrt3}$

So ultimately, we have 9 critical points - 1 at the origin (0, 0, 0), and 8 at the various combinations of $\left(\pm\dfrac1{\sqrt3},\pm\dfrac1{\sqrt3},\pm\dfrac1{\sqrt3}\right)$ , at which points we get a value of either of $\pm\dfrac5{\sqrt3}$ , with the maximum being the positive value and the minimum being the negative one.

5 0

3 years ago

3 (x - 3) + x (r + 4)

docker41 [41]

Answer:

Step-by-step explanation:

3(x-3)+x(r+4)

3x-9+xr+4x

7x+9+xr

7 0

3 years ago

Pls check picture! Pls give me answer pls! ASAP Right answer. if you dont know dont answer pls!!!! Right Answer for brainliest!

lyudmila [28]

Answer: The answer is 6 and 4.

Step-by-step explanation:

6 0

3 years ago

Divide £945 in the ratio 2:5

dybincka [34]

The answer is 675. 2270. 657

5 0

3 years ago