All categories

2 years ago

I'm dumb so like I don't understand (╥ω╥`)

Mathematics

2 answers:

olya-2409 [2.1K]2 years ago

7 0

i think you need 27 grams of fiber a day

kobusy [5.1K]2 years ago

5 0

A. you should eat 25.92 grams of fiber daily.
b. you would eat 2.376 grams of fiber.

You might be interested in

determine os valores de x e y sabendo que os numeros x,6 e 27 sao diramente proporcionais aos numeros 2, y e 18 nessa ordem

bazaltina [42]

Answer:

x=3 e y=4

Step-by-step explanation:

x/2=6/y=27/18

...

x/2=27/18

18.x=2.(27)

18x=54

x=54/18

x= 3

...

6/y=27/18

27y=6.(18)

27y=108

y=108/27

y=4

3 0

3 years ago

James defines a line segment as "a portion of a line." His statement is not precise enough because he should specify that

vekshin1

Answer: D. the segment has two endpoints.

Step-by-step explanation:

A line segment refers to a part of a line which is bounded by two distinct end points. The line segment is part of the line which helps in the connection of two points that are considered to be the endpoints.

A line segment is the distance between the two points and which one measure.

They can be used to form a side of a polygon because they have a defined length.

8 0

2 years ago

Make d the subject of the formula h=<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7B3%7D" id="TexFormula1" title="\frac{d}{3}

lakkis [162]

First you would subtract 2 to get d/3 by itself. On the left side you would have h-2 and one the right d/3 (h-2=d/3). Then you divide by 3 to get d by itself. So the answer is h/3-2/3=d

4 0

2 years ago

Help . will mark best answer

ZanzabumX [31]

answer: infinitely many :)

5 0

2 years ago

Solve the equation <img src="https://tex.z-dn.net/?f=%2835%20x%5E%7B4%7D%20y%2B14%20x%5E%7B5%7D%20y-2%20y%5E%7B3%7D-4x%20y%5E%7B

jeka94

$\underbrace{35x^4y+14x^5y-2y^3-4xy^3}_M\,\mathrm dx+\underbrace{7x^5-6xy^2}_N\,\mathrm dy=0$

$M_y=35x^4+14x^5-6y^2-12xy^2$
$N_x=35x^4-6y^2$

$\dfrac{N_x-M_y}N=\dfrac{-14x^5+12xy^2}{7x^5-6xy^2}=-2$

This suggests an integrating factor depending on $x$ only is possible, and given by

$\mu(x)=\exp\left(-\displaystyle\int\frac{N_x-M_y}N\,\mathrm dx\right)=e^{2x}$

Distributing across the ODE, we end up with

$\underbrace{(35x^4y+14x^5y-2y^3-4xy^3)e^{2x}}_{M^*}\,\mathrm dx+\underbrace{(7x^5-6xy^2)e^{2x}}_{N^*}\,\mathrm dy=0$

The equation is now exact, with

${M^*}_y={N^*}_x=(35x^4+14x^5-6y^2-12xy^2)e^{2x}$

Now we find the solution:

$F_x=M^*$
$F=\displaystyle\int(35x^4+14x^5-2y^3-4xy^3)e^{2x}\,\mathrm dx$
$F=(7x^5y-2xy^3)e^{2x}+f(y)$

$F_y=N^*$
$(7x^5-6xy^2)e^{2x}+f'(y)=(7x^5-6xy^2)e^{2x}$
$f'(y)=0$
$\implies f(y)=C$

The general solution is then

$F(x,y)=(7x^5y-2xy^3)e^{2x}=C$

7 0

3 years ago