64 is decreased to 48

Mathematics

1 answer:

choli [55]3 years ago

8 0

Answer:

33.28, I think this is it

Step-by-step explanation:

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How to solve 4(x+1)-(2x-4)=6 using distributive property PLS HELP! THANK U!! giving BRAINLIEST

Pie

Answer: $x=\Large\boxed{-1}$

Step-by-step explanation:

Given equation

$4(x+1)-(2x-4)=7$

Expand the parenthesis and apply the distributive property

$4\times x+4\times1-2x+4=6$

$4x+4-2x+4=6$

Combine like terms

$4x-2x+4+4=6$

$2x+8=6$

Subtract 8 on both sides

$2x+8-8=6-8$

$2x=-2$

Divide 2 on both sides

$2x\div2=-2\div2$

$x=\Large\boxed{-1}$

Hope this helps!! :)

Please let me know if you have any questions

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Read 2 more answers

For what values of x does the curve y^2x^3-15x^2=4 have horizontal tangent lines

enyata [817]

We have the following curve:

$y^2x^3-15x^2=4$

To find the tangent lines to the curve we need to use the concept of derivative, but we can’t solve this problem for $y$ , thus, let's apply implicit differentiation, so:

$\frac{d}{dx}(y^2x^3-15x^2=4) \\ \\
\frac{d}{dx}(y^2x^3)-\frac{d}{dx}(15x^2)=\frac{d}{dx}(4) \\ \\
(y^{2})'x^3+(x^3)'y^2-15(x^2)'=0 \\ \\ 2yy'x^3+3x^2y^2-30x=0 \\ \\
y'=\frac{dy}{dx}=\frac{30x-3x^2y^2}{2x^3y}=\frac{x(30-3xy^2)}{2x^3y} \\ \\
y'=\frac{30-3xy^2}{2x^2y}$

Therefore the horizontal lines occurs when $y'=0$ , then:

$\frac{30-3xy^2}{2x^2y}=0 \\ \\ That \ is, \
when: \\ \\ 30-3xy^2=0 \\ \\ \therefore xy^2=10 \rightarrow y^2=\frac{10}{x}$

If we substitute this in the original equation we have:

$\frac{10}{x}x^3-15x^2=4 \\ \\ \therefore
10x^2-15x^2=4 \\ \\ \therefore x^2=-\frac{4}{5}$

This is an absurd result because it is impossible for a squared number to get a negative number. So the conclusion is that there is no any value of $x$ in which the curve has horizontal tangent lines.