All categories

3 years ago

Question in picture #21 only

Mathematics

2 answers:

inn [45]3 years ago

6 0

Answer:

The side is 8

Step-by-step explanation:

Retracing in Pythagorean Theorem

17^2 - 15^2 = 64

$\sqrt{64}=8$

krek1111 [17]3 years ago

3 0

Number 21 is the first answer a.) 8

You might be interested in

Robbin is using the recipe to make chicken noodle soup. He plans to make 6 batches of the soup. He has 2/3 teaspoon of black pep

weqwewe [10]

That is 4 teaspoons in all

5 0

3 years ago

11) - Simplify: -(16 - 5x) A) 5x=16 5x + 16 -5x - 16 - 5x+16

MatroZZZ [7]

A 5x=16 is the answer

7 0

2 years ago

Read 2 more answers

Darnell bought 6 packs of baseball card. There were 6 cards in each pack Darnell put all his card in card book. He put 9 cards o

irga5000 [103]

Answer:

Darnell used 4 pages.

Step-by-step explanation:

We know Darnell has 36 cards, with this given we can do the simple division of 36 / 9 to get the answer of 4.

Pls give brainliest!

7 0

2 years ago

Read 2 more answers

Can you create 2 dividing decimal questions where the answer would be 14?

Alexxandr [17]

Answer:

2.24/0.16

736.434 × 0.019011

Step-by-step explanation:

The above questions have a solution of 14.

5 0

3 years ago

Find the extreme values of f subject to both constraints f(x,y) = 2x^2+3y^2-4x-5, x^2 + y^2 <=16

Softa [21]

$f(x,y)=2x^2+3y^2-4x-5$
$f_x=4x-4=0\implies x=1$
$f_y=6y=0\implies y=0$

$f(x,y)$ has only one critical point at $(x,y)=(1,0)$ . The function has Hessian

$\mathbf H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}4&0\\0&6\end{bmatrix}$

which is positive definite for all $(x,y)$ , which means $f(x,y)$ attains a minimum at the critical point with a value of $f(1,0)=-7$ .

To find the extrema (if any) along the boundary, parameterize it by $x=4\cos t$ and $y=4\sin t$ , with $0\le t$ . On the boundary, we have

$f(x(t),y(t))=F(t)=2(4\cos t)^2+3(4\sin t)^2-4(4\cos t)-5=32\cos^2t+48\sin^2t-16\cos t-5$
$F(t)=35-16\cos t-8\cos2t$

Find the critical points along the boundary:

$F'(t)=16\sin t+16\sin2t=16\sin t+32\sin t\cos t=16\sin t(1+2\cos t)=0$
$\implies t=0,\dfrac{2\pi}3,\pi,\dfrac{4\pi}3$

Respectively, plugging these values into $F(t)$ gives 11, 47, 43, and 47. We omit the first and third, as we can see the absolute extrema occur when $F(t)=47$ .

Now, solve for $x,y$ for both cases:

$t=\dfrac{2\pi}3\implies\begin{cases}x=4\cos t=-2\\y=4\sin t=2\sqrt3\end{cases}$

$t=\dfrac{4\pi}3\implies\begin{cases}x=4\cos t=-2\\y=4\sin t=-2\sqrt3\end{cases}$

so $f(x,y)$ has two absolute maxima at $(x,y)=(-2,\pm2\sqrt3)$ with the same value of 47.

5 0

2 years ago