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3 years ago

Can someone help me with this one

Mathematics

2 answers:

erma4kov [3.2K]3 years ago

7 0

The answe is 4..you can see the pattern if you add 4 in the x

sergey [27]3 years ago

6 0

The answer is 4. If you add 4 to any of the x values, it equals the y value next to it

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Each side of a quilt square measures approximately 4.25 inches. If there are about 2.54 centimeters in 1 inch, how long is each

IrinaK [193]

Answer: approximately 10.8 centimeters

Step-by-step explanation:

We have a square, where each side measures approx. 4.25 in

Now we know that 1in ≈ 2.54 cm

Then, in 4.25 in, we have 4.25 times 1 inch, so we have 4.25 times the length of 2.54 cm

So the approximate measure of the sides in centimeters is:

4.25*(2.54)cm = 10.8 cm

So we have that each side measures approximately 10.8 centimeters

5 0

3 years ago

What is the length of the segment connecting the points (−9, 3) and (−9, −2)?

Stels [109]

It is 5 units

because -9,3 is up and -9,-2 is down
5

6 0

3 years ago

What is 2/5 ×5/6 in simplist form

yanalaym [24]

Answer:

1/3

0.3 (repeating)

3 0

3 years ago

Read 2 more answers

Can somebody help me please

Romashka-Z-Leto [24]

Answer:

23) y=3x-7, 3x-y=7

24) y=-6x+9, 6x-y=9

25) y=1/4x+11, x-4y=-44

26) y=-1/2x+4, x+2y=8

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Use the Divergence Theorem to evaluate the following integral S F · N dS and find the outward flux of F through the surface of t

Xelga [282]

Answer:

Result;

$\int\limits\int\limits_S { \textbf{F}} \, \cdot \textbf{N} d {S} = 32\pi$

Step-by-step explanation:

Where:

F(x, y, z) = 2(x·i +y·j +z·k) and

S: z = 0, z = 4 -x² - y²

For the solid region between the paraboloid

z = 4 - x² - y²

div F

For S: z = 0, z = 4 -x² - y²

We have the equation of a parabola

To verify the result for F(x, y, z) = 2(x·i +y·j +z·k)

We have for the surface S₁ the outward normal is N₁ = -k and the outward normal for surface S₂ is N₂ given by

$N_2 = \frac{2x \textbf{i} +2y \textbf{j} + \textbf{k}}{\sqrt{4x^2+4y^2+1} }$

Solving we have;

$\int\limits\int\limits_S { \textbf{F}} \, \cdot \textbf{N} d {S} = \int\limits\int\limits_{S1} { \textbf{F}} \, \cdot \textbf{N}_1 d {S} + \int\limits\int\limits_{S2} { \textbf{F}} \, \cdot \textbf{N}_2 d {S}$

Plugging the values for N₁ and N₂, we have

$= \int\limits\int\limits_{S1} { \textbf{F}} \, \cdot \textbf{(-k)}d {S} + \int\limits\int\limits_{S2} { \textbf{F}} \, \cdot \frac{2x \textbf{i} +2y \textbf{j} + \textbf{k}}{\sqrt{4x^2+4y^2+1} } d {S}$

Where:

F(x, y, z) = 2(xi +yj +zk) we have

$= -\int\limits\int\limits_{S1} 2z \ dA + \int\limits\int\limits_{S2} 4x^2+4y^2+2z \ dA$

$= -\int\limits^2_{-2} \int\limits^{\sqrt{4-y^2}} _{-\sqrt{4-y^2}} 2z \ dA + \int\limits^2_{-2} \int\limits^{\sqrt{4-y^2}} _{-\sqrt{4-y^2}} 4x^2+4y^2+2z \ dA$

$= \int\limits^2_{-2} \int\limits^{\sqrt{4-y^2}} _{-\sqrt{4-y^2}} 4x^2+4y^2 \ dxdy$

$= \int\limits^2_{-2} \frac{(16y^2 +32)\sqrt{-(y^2-4)} }{3} dy$

= 32π.

6 0

4 years ago