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Arturiano [62]
4 years ago
9

Cells lining the gut need to take in glucose but at a certain time, the concentration of extracellular glucose is lower than the

concentration already stored in cells. What process is needed here?
Simple diffusion
Active transport
Facilitated diffusion
Mathematics
1 answer:
Artist 52 [7]4 years ago
7 0

Answer:

hm

Step-by-step explanation:

yes i think so :)

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Choose the correct simplification of the expression (−2x + 4y)(3x − 7y).
Rudik [331]
-2x • 3x = -6x2 (the 2 is an exponent)
4y • -7y = -28y2 (the 2 is an exponent)
-2x • -7y = 14xy
4y • 3x = 12xy
14xy + 12xy = 26xy

—> -6x2(expo) - 28y2(expo) + 26xy

Check the answer again just to be sure, it’s been a while since I’ve had algebra!
3 0
3 years ago
Find the value of X in the isosceles triangle shown below.
BigorU [14]

Answer:

D) Square root of 48

Step-by-step explanation:

We have to split the triangle in half and then multiply that by two for the answer. Then we use a²+b²=c² because the two triangles are right triangles, and we know b=4 and c=8, so a²+(4²)=8², or a²+16=64

a² would have to equal 48. a=√48.

7 0
4 years ago
12 ÷ 2 over 5 I JUST NEED HELP OKAY PEOPLE! WHAT DO OYOU NOT UNDER STAND IF YOU HELP ME I WILL MARK YOU BRI......
blagie [28]

Step-by-step explanation:

12 ÷ 2 = 6 over 5 would be 6/5 but if over 5 means to divide then 1.2 is ur answer.

3 0
3 years ago
A student incorrectly says the volume of the regular hexagonal​ prism, to the nearest cubic​ centimeter, is 1,619 cm3. What is t
Greeley [361]

Answer:

17.5

Step-by-step explanation:

3 0
3 years ago
Find a polynomial P(x) with real coefficients having a degree 4, leading coefficient 4, and zeros 3-i and 4i.
Alisiya [41]

now, this polynomial has roots of 3-i and 4i, namely 3 - i and 0 + 4i.

let's bear in mind that a complex root never comes all by her lonesome, her sibling is always with her, the conjugate, so if 3 - i is there, 3 + i is also coming along, likewise if 0 + 4i is there, her sibling 0 - 4i is also there.

\bf \begin{cases} x=3-i\implies &x-3+i=0\\ x=3+i\implies &x-3-i=0\\ x=4i\implies &x-4i=0\\ x=-4i\implies &x+4i=0 \end{cases}\\\\[-0.35em] ~\dotfill\\\\ (x-3+i)(x-3-i)(x-4i)(x+4i)=\stackrel{y}{0} \\[2em] \underset{\textit{difference of squares}}{[(x-3)+i][(x-3)-i]}\underset{\textit{difference of squares}}{[x-4i][x+4i]}=0

\bf [(x-3)^2-i^2][x^2-(4i)^2]=y\implies [(x-3)^2-(-1)][x^2-(4^2i^2)]=0 \\[2em] [(x-3)^2-(-1)][x^2-(16(-1))]=0\implies [(x-3)^2+1][x^2+16]=0 \\[2em] [(x^2-6x+9)+1][x^2+16]=y\implies (x^2-6x+10)(x^2+16)=0 \\\\\\ x^4-6x^3+10x^2+16x^2-96x+160=0 \\\\\\ x^4-6x^3+26x^2-96x+160=0 \\\\\\ \stackrel{\textit{multiplying both sides by 4}}{4(x^4-6x^3+26x^2-96x+160)=4(0)} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill 4x^4-24x^3+104x^2-384x+640=y~\hfill

3 0
4 years ago
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