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3 years ago

PLZZ HELP ME OUT THIS ACTUALLY DUE TODAY AND GOING BREAK DOWN I NEED SOMEONE HELP PLEASE!!!!

Mathematics

2 answers:

mylen [45]3 years ago

6 0

Answer:

Step-by-step explanation:

Hey there!

A. 3x5 is 15! However since one of the number is a negetive, we must add a negetive sign to the 15, making it -15!

Remember this: A negetive number times a negetive number is a posititve number, but a negetive number times a positive number is a negetive!

B. 5+-5 is equal to 0, so 3x0 is 0!

C. No matter which way you mutliply it, whether it's -5x3 or 3x-5, the answer will always be -15! (-5)x(3+-3) is: 3+ -3 is 0, so -5 times 0 is the same thing as 5x0, which is still 0!

D. Like I said before, a negetive times a negetive is a positive, so -3x-5 is the same as saying 3x5, which is 15!

I hope this explanation helps you! Please mark me Brainliest!

maks197457 [2]3 years ago

3 0

Answer:

Step-by-step explanation:

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What is the value of x? enter your answer in the box.

Mamont248 [21]

Answer:

if 9x8 =72

then 56/8= 7

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3 years ago

Find the absolute maximum and absolute minimum values of the function f(x, y) = x 2 + y 2 − x 2 y + 7 on the set d = {(x, y) : |

dsp73

Looks like $f(x,y)=x^2+y^2-x^2y+7$ .

$f_x=2x-2xy=0\implies2x(1-y)=0\implies x=0\text{ or }y=1$

$f_y=2y-x^2=0\implies2y=x^2$

If $x=0$ , then $y=0$ - critical point at (0, 0).
If $y=1$ , then $x=\pm\sqrt2$ - two critical points at $(-\sqrt2,1)$ and $(\sqrt2,1)$

The latter two critical points occur outside of $D$ since $|\pm\sqrt2|>1$ so we ignore those points.

The Hessian matrix for this function is

$H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}2-2y&-2x\\-2x&2\end{bmatrix}$

The value of its determinant at (0, 0) is $\det H(0,0)=4>0$ , which means a minimum occurs at the point, and we have $f(0,0)=7$ .

Now consider each boundary:

If $x=1$ , then

$f(1,y)=8-y+y^2=\left(y-\dfrac12\right)^2+\dfrac{31}4$

which has 3 extreme values over the interval $-1\le y\le1$ of 31/4 = 7.75 at the point (1, 1/2); 8 at (1, 1); and 10 at (1, -1).

If $x=-1$ , then

$f(-1,y)=8-y+y^2$

and we get the same extrema as in the previous case: 8 at (-1, 1), and 10 at (-1, -1).

If $y=1$ , then

$f(x,1)=8$

which doesn't tell us about anything we don't already know (namely that 8 is an extreme value).

If $y=-1$ , then

$f(x,-1)=2x^2+8$

which has 3 extreme values, but the previous cases already include them.

Hence $f(x,y)$ has absolute maxima of 10 at the points (1, -1) and (-1, -1) and an absolute minimum of 0 at (0, 0).

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3 years ago

For each sequence write an explicit formula 96,48,24,12,6

STALIN [3.7K]

$a_1=96;\ a_2=48;\ a_3=24;\ a_4=12;\ a_5=6\\\\a_1=a_2:2\to96:2=48\\a_2=a_3:2\to48:2=24\\a_3=a_2:2\to24:2=12\\a_4=a_3:2\to12:2=6\\\\therefore\\\\a_n=96:2^{n-1}=\frac {96}{2^{n-1}}\\\\check:\\a_1=\frac{96}{2^{1-1}}=\frac{96}{1^0}=96\\a_2=\frac{96}{2^{2-1}}=\frac{96}{2^1}=\frac{96}{2}=48\\a_3=\frac{96}{2^{3-1}}=\frac{96}{2^2}=\frac{96}{4}=24\\\vdots$

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3 years ago

Bronson builds a rectangular deck for his friend. the width of the deck is 29 feet. the perimeter of the deck must be at least 1

olga_2 [115]

<span>Answer: a) Greater than or equal to 38 feet b)Since there is no restriction on the perimeter there exists many possible values for the length of the deck. Explanation: Given that the width of the deck is 29 ft. and the perimeter of the deck is at least 134 ft. 134 = 2(length + width) 134 = (2 x length) + (2 x 29) 134 = (2 x length) + 58 2 x length = 134 - 58 2 x length = 76 length = 76 / 2 length = 38 ft Thus, the inequality will be: a)Length â‰Ą 38 ft b)since there is no restriction on the perimeter there exists many possible values for length of deck.</span>

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4 years ago

How many ways can 10 runners finish first, second, and third in a race?

olga nikolaevna [1]

Let's call the 10 runners A B C D E F G H I J
ABC
ABD
ABE
ABF
ABG
ABH
ABI
ABJ
ACB
ACD
ACE
ACF
ACG
ACH
ACI
ACJ
ADB
ADC
ADE
ADF
ADG
ADH
ADI
ADJ
AEB
AEC
AED
AEF
AEG
AEH
AEI
AEJ
AFB
AFC
AFD
AFE
AFG
AFH
AFI
AFJ
AGB
AGC
AGD
AGE
AGF
AGH
AGI
AGJ
AHB
AHC
AHD
AHE
AHF
AHG
AHI
AHJ
AIB
AIC
AID
AIE
AIF
AIG
AIH
AIJ
AJB
AJC
AJD
AJE
AJF
AJG
AJH
AJI
The above is 60 different ways, but now B C D E F G H I J have to come first, and the proses starts again, until we have covered them all. I'm not going to write them all out but in theory, there is 540 different ways. 9*60=540.

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3 years ago