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LenKa [72]
3 years ago
9

A distributive property for 7 times 63

Mathematics
2 answers:
djverab [1.8K]3 years ago
8 0
The answer to this question is 431
SVETLANKA909090 [29]3 years ago
6 0
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Hello can anyone help me ? here the question
lesantik [10]

Answer:

A. 1/5k - 2/3j and -2/3j +1/5k

Step-by-step explanation:

A. 1/5k - 2/3j and -2/3j +1/5k

B. 1/5k - 2/3j and -1/5k +2/3j

There is a change in the signs of each term

1/5k changed to -1/5k

-2/3j changed to +2/3j

Not equivalent

C. 1/5k - 2/3j and 1/5j - 2/3k

There is a change in the variables

1/5k changed to 1/5j

-2/3j changed to -2/3k

D. 1/5k - 2/3j and 2/3j - 1/5k

The is a change in the signs of each term

1/5k changed to -1/5k

-2/3j changed to +2/3j

The only equivalent expression is

A. 1/5k - 2/3j and -2/3j +1/5k

3 0
3 years ago
Write and simplify the integral that gives the arc length of the following curve on the given interval b. If necessary, use tech
LiRa [457]

Answer:

L = 4.103

Step-by-step explanation:

we have length of curve

L = \int\limits^b_a {\sqrt{(f'(x))²+1} } \, dx

where f(x) = d/dx(3*in(x)) = 3/x

substituting for f(x), we have L = \int\limits^5_2 {\sqrt{(3/x)²+1} } \, dx

(since the limit is 2≤ x ≤5)

solving,  L = \int\limits^5_2 {\sqrt{9/x²+1} } \, dx

Simplifying this integral, we have

L = 4.10321

8 0
4 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).

3 0
3 years ago
If m is 7 and x is 7 so what is m​
olchik [2.2K]
Can u just post the whole question or elaborate further
6 0
3 years ago
Read 2 more answers
Please help me out with this problem!!!!!!
guapka [62]

Answer:

b)

Step-by-step explanation:

The line with positive slope is 3x-5y≥15 and the line with negative slope is 2x+3y≤12.

This is because such inequalities represent half planes which are divided by the corresponding lines.

Put (0,0) in both lines to check if the origin lies in the shaded half plane or not.

4 0
4 years ago
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