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3 years ago

Plzzz help me well mark brainliest if correct!!...

Mathematics

2 answers:

Temka [501]3 years ago

3 0

Answer:

Step-by-step explanation:

because I said so

Snezhnost [94]3 years ago

3 0

Answer: C, Third grade girls

Step-by-step explanation:

The tallest bar graph indicates the most tickets sold, which is C

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Please help me asap I will give brainliest or whatever you want

dangina [55]

Answer/Step-by-step explanation:

Part A:

$Key:\left[1 Adult = 4 Student]$

$Formula: Divide$ $by$ $4$

12 Student = 3 Adult

24 Student = 6 Adult

40 Student = 10 Adult

Part B:

33 Student

Hence, divide 33 by 4 = 8 with a remainder of 1.

Therefore, 8 Adult and for the remainder 1 student either one Adult takes 5 Student or Needed 9 Adult.

7 0

2 years ago

Let the (x; y) coordinates represent locations on the ground. The height h of

grigory [225]

The critical points of h(x,y) occur wherever its partial derivatives $h_x$ and $h_y$ vanish simultaneously. We have

$h_x = 8-4y-8x = 0 \implies y=2-2x \\\\ h_y = 10-4x-12y^2 = 0 \implies 2x+6y^2=5$

Substitute y in the second equation and solve for x, then for y :

$2x+6(2-2x)^2=5 \\\\ 24x^2-46x+19=0 \\\\ \implies x=\dfrac{23\pm\sqrt{73}}{24}\text{ and }y=\dfrac{1\mp\sqrt{73}}{12}$

This is to say there are two critical points,

$(x,y)=\left(\dfrac{23+\sqrt{73}}{24},\dfrac{1-\sqrt{73}}{12}\right)\text{ and }(x,y)=\left(\dfrac{23-\sqrt{73}}{24},\dfrac{1+\sqrt{73}}{12}\right)$

To classify these critical points, we carry out the second partial derivative test. h(x,y) has Hessian

$H(x,y) = \begin{bmatrix}h_{xx}&h_{xy}\\h_{yx}&h_{yy}\end{bmatrix} = \begin{bmatrix}-8&-4\\-4&-24y\end{bmatrix}$

whose determinant is $192y-16$ . Now,

• if the Hessian determinant is negative at a given critical point, then you have a saddle point

• if both the determinant and $h_{xx}$ are positive at the point, then it's a local minimum

• if the determinant is positive and $h_{xx}$ is negative, then it's a local maximum

• otherwise the test fails

We have

$\det\left(H\left(\dfrac{23+\sqrt{73}}{24},\dfrac{1-\sqrt{73}}{12}\right)\right) = -16\sqrt{73} < 0$

while

$\det\left(H\left(\dfrac{23-\sqrt{73}}{24},\dfrac{1+\sqrt{73}}{12}\right)\right) = 16\sqrt{73}>0 \\\\ \text{ and } \\\\ h_{xx}\left(\dfrac{23+\sqrt{73}}{24},\dfrac{1-\sqrt{73}}{12}\right)=-8 < 0$

So, we end up with

$h\left(\dfrac{23+\sqrt{73}}{24},\dfrac{1-\sqrt{73}}{12}\right)=-\dfrac{4247+37\sqrt{73}}{72} \text{ (saddle point)}\\\\\text{ and }\\\\h\left(\dfrac{23-\sqrt{73}}{24},\dfrac{1+\sqrt{73}}{12}\right)=-\dfrac{4247-37\sqrt{73}}{72} \text{ (local max)}$

7 0

3 years ago

I need an answer please! The question says Find the value of x that would make the following expression equivalent to sixteen ti

aev [14]

Square root of 6 times 16 will give us 39.19

The square root of 48 times 32 [which gives us 1536] is also equal to 39.19

The square root of 1536 is equal to 39.19

Hence, x is equal to 32

Hope this helped! :D

8 0

3 years ago

Another question for fun..... b+182=293

AURORKA [14]

The answer should be b=111

4 0

4 years ago

Read 2 more answers

logs are stacked in a pile on the bottom row and 15 on the top row . there are 10 rows in all with each row having one more log

WITCHER [35]

Since there is a common difference of one, this is an arithmetic sequence of the form a(n)=a+d(n-1) which in this case is:

a(n)=15+1(n-1)

All arithmetic sequences have a sum which is equal to the average of the first and last terms time the number of terms, mathematically:

s(n)=(2an+dn^2-dn)/2 (that is just the result of ((a+a+d(n-1))(n/2)) )

Since a=15 and d=1 this becomes:

s(n)=(30n+n^2-n)/2

s(n)=(n^2+29n)/2 so

s(10)=(100+290)/2

s(10)=195 logs

8 0

4 years ago

Read 2 more answers