All categories

3 years ago

6 7/8 + 5 3/4 = what

Mathematics

1 answer:

Shtirlitz [24]3 years ago

6 0

Answer:

$12\frac{5}{8}$

Is your answer, have a good day

You might be interested in

Please help 20 points

12345 [234]

Answer:

x =± (√ 317/ 2 )+( 15 /2)

6 0

3 years ago

He was a newcomer in the land, a chechaquo, and this was his first winter. The trouble with him was that he was without imaginat

kolbaska11 [484]

Answer:

TBH I'm kind of confused on what you are being taught

Step-by-step explanation:

?????????

3 0

3 years ago

What is the answer to this question? (pls help)

prohojiy [21]

Answer:

Step-by-step explanation:

if you divide the money by the ounce, D is the cheapest with 0.24 per ounce

4 0

3 years ago

Read 2 more answers

Carla bought 2 pounds of red beads three-quarter pounds of green beads and 10 ounces of string at the craft store how much do Ka

skad [1K]

70 ounces OR 4.375 pounds OR 4 pounds and 6 ounces
[To get this answer, first make pounds into ounces. There are 16 ounces in a pound. Multiply 2 (pounds) and 16 (ounces in a pound) to get 32 ounces of red beads. Next, do the same to the 1 3/4 pounds of green beads: 1 3/4 (pounds) times 16 (ounces in a pound) make 28 ounces. Last add 32 (oz) + 28 (oz) + 10 (oz) together to get the total weight of her supplies: 70 oz OR 4.375 lbs OR 6 oz. It's up to you which answer (they are the same but in different forms) you want to use.]

Hope I helped! Have a nice day! :)

8 0

3 years ago

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