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svetoff [14.1K]
3 years ago
11

HELP ME PLEASE I DONT UNDERSTAND

Mathematics
1 answer:
yulyashka [42]3 years ago
6 0

Answer:

Step-by-step explanation:

acid=16.5 % of 348

=0.165×348

≈57.4 ml

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Find the limits of integration ly, uy, lx, ux, lz, uz (some of which will involve variables x,y,z) so that ∫uz lz∫uxlx∫uyly????y
SSSSS [86.1K]

Answer:

X from 0 to 21

Y from 0 to 7

Z from 0 to 3

Step-by-step explanation:

Since we are being asked by the integration limits in first octant (positive x, positive y and positive z) we need to know where does the plane intersect this axes. For this we have:

for x=0 and y=0

7z=21

z=3

for x=0 and z=0

3y=21

y=7

for z=0 and y=0

x=21

This means that the integration limits are:

X from 0 to 21

Y from 0 to 7

Z from 0 to 3

8 0
4 years ago
What is the solution for -10 < x - 9?
storchak [24]

Answer:

x > -1

Step-by-step explanation:

-10 < x - 9 add 9 to both sides of the inequation

9 + (-10) < x - 9 + 9 minus 9 will eliminate the positive 9 on the right side of the inequation therefore the answer is:

-1 < x  and x > -1

7 0
2 years ago
GIVING BRAINLIEST AND CREDIT ;)))))) DONT LEAVE ME HANGING
melamori03 [73]

Answer:

Type the equations given into desmos and it will give you the answer


8 0
3 years ago
Need help really pllzzzz.....
Dominik [7]
G(x)=-2x-4
f(x)=15+0.75x

the common solution Q is their intersection-> calculate x by setting them equal:
-2x-4=15+0.75x
-2x-(3/4)x=19
-(11/4)x=19
x=-4*19/11
x=-76/11

calculate y:
f(-76/11)=15+(3/4)*(-76/11)
=15+(-3*19/11)
=15+(-57/11)
=15+-5-(2/11)
=10-(2/11)
=108/11

so the solution is a) (-76/11, 108/11)
4 0
3 years ago
Find the average squared distance between the points of r{(x,y): 0x, 0y} and the point (,)
ivolga24 [154]

The average squared distance between the points is their variance

The average squared distance is 8/3

<h3>How to determine the average squared distance?</h3>

The given parameters are:

R={(x,y): 0<=x<=2, 0<=y<=2}

The point = (2,2).

f(x,y) = (x-2)² + (y-2)²

The squared distance is calculated as:

D² = f(x,y) = (x-2)² + (y-2)²

Where the area (A) is:

A = xy

Substitute the maximum values of x and y in A = xy

A = 2 * 2

A = 4

The minimum values of x and y are 0.

So, the limits for the integrals are 0 to 2

The integral becomes

D\² = \int \int (x-2)\² + (y-2)\² dx dy

Expand

D\² = \int \int x\² -4x + 4 + (y-2)\² dx dy

Evaluate the inner integral with respect to x from 0 to 2

D\² = \int \frac 13x^3 -2x^2 + 4x + x(y-2)\² |\limits^2_0 dy

Expand

D\² = \int [\frac 13(2)^3 -2(2)^2 + 4(2) + (2)(y-2)\²] dy

Simplify

D\² = \int \frac 83 + (2)(y-2)\² dy

Expand

D\² = \int \frac 83 + 2y^2-8y + 8 \ dy

Evaluate the integral with respect to y from 0 to 2

D\² = [\frac 83y + \frac 23y^3- 4y^2 + 8y ]|\limits^2_0

Expand

D\² = \frac 83*2 + \frac 23*2^3- 4*2^2 + 8*2

D² = 32/3

The average squared distance (AD²) is calculated as:

AD² = D²/A

So, we have:

AD² = 32/3 \div 4

Evaluate the quotient

AD² = 32/12

Simplify

AD² = 8/3

Hence, the average squared distance is 8/3

Read more about average distance or variance at:

brainly.com/question/15858152

8 0
3 years ago
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