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2 years ago

Please explain how this is incorrect.

Mathematics

1 answer:

MatroZZZ [7]2 years ago

5 0

Step-by-step explanation:

The spores always x3 after a month so going a month back they ÷3 instead so 900÷3 is 300 spores on the first month

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1. Evaluate each expression without using a calculator.<br> (a) <img src="https://tex.z-dn.net/?f=%28-3%29%5E%7B4%7D" id="TexFor

Rufina [12.5K]

The result of the exponential expression is 81

<h3>Indices expressions</h3>

Indices are written in term of exponents

According to the expoenent law of indices

a^4 = a * a * a * a

Given the expression

(-3)^4

This means the product of -3 in four places. Mathematically

(-3)^4 = -3 * -3 * -3 * -3
(-3)^4 = 9 * 9
(-3)^4 = 81

Hence the result of the exponential expression is 81

Learn more on exponents here: brainly.com/question/11975096

5 0

2 years ago

PLS HELP ASAP I AM ALMOSF DONE PLS

vitfil [10]

Answer:

im saying d is the answer

Step-by-step explanation:

7 0

3 years ago

A system of linear equations is graphed on this coordinate grid.

Serjik [45]

I think it is B (1,5) because that is where it intersects

4 0

3 years ago

The slope of the line tangent to the curve y^2 + (xy+1)^3 = 0 at (2, -1) is ...?

Lina20 [59]

$\frac{d}{dx}(y^2 + (xy+1)^3 = 0)
\\
\\\frac{d}{dx}y^2 + \frac{d}{dx}(xy+1)^3 = \frac{d}{dx}0
\\
\\\frac{dy}{dx}2y + \frac{d}{dx}(xy+1)\ *3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(\frac{d}{dx}(xy)+\frac{d}{dx}1\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(\frac{d}{dx}xy+x\frac{d}{dx}y+\frac{dx}{dx}\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(\frac{dx}{dx}y+x\frac{dy}{dx}+\frac{dx}{dx}0\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(y+x\frac{dy}{dx}\right)\ * 3(xy+1)^2 = 0$
$\frac{dy}{dx}2y + \left(y+x\frac{dy}{dx}\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(3y+3x\frac{dy}{dx}\right) (xy+1)^2 = 0
\\
\\2y\ y' + \left(3y+3x\ y'\right) (xy+1)^2 = 0
\\
\\2y\ y' + \left(3y+3x\ y'\right) ((xy)^2+2xy+1) = 0
\\
\\2y\ y' + \left(3y+3x\ y'\right) ((xy)^2+2xy+1) = 0
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\\2y\ y' + 3x^2y^3 +6xy^2+3y+(3x y' +6x^2y y'+3x^2y y') = 0
\\
\\3x^2y^3 +6xy^2+3y+(3x y' +6x^2y y'+3x^2y y'+2yy' ) = 0
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\\y'(3x +6x^2y+3x^2y+2y ) = -3x^2y^3 -6xy^2-3y

$
$y' = \frac{-3x^2y^3 -6xy^2-3y}{(3x +6x^2y+3x^2y+2y )};\ x=2,y=-1
\\
\\y' = \frac{-3(2)^2(-1)^3 -6(2)(-1)^2-3(-1)}{(3(2) +6(2)^2(-1)+3(2)^2(-1)+2(-1) )}
\\
\\y' = \frac{-3(4)(-1) -6(2)(1)+3}{(6 +6(4)(-1)+3(4)(-1)-2 )}
\\
\\y' = \frac{12 -12+3}{(6 -24-12-2 )}
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\\y' = \frac{3}{( -32 )}$

7 0

3 years ago

Find the value of xx, yy, and zz in the parallelogram below.

Scrat [10]

Answer: x = 92, y = -119

Step-by-step explanation:

Opposite sides of a parallelogram are equal.

x + 6 = 98

x = 92

127 = -y + 8

-y = 119

y = -119

5 0

2 years ago

Read 2 more answers