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2 years ago

Solve the inequality: 4k-(5+3k)>2

Mathematics

2 answers:

Elanso [62]2 years ago

5 0

Answer:

k > 7

Step-by-step explanation:

Given the inequality statement: 4k - (5 + 3k) > 2

Distribute -1 into (5 + 3k):

4k -5 - 3k > 2

Next, add like terms:

k - 5 > 2

Add 5 to both sides of the inequality:

k - 5 + 5 > 2 + 5

k > 7

Please mark my answers as the Brainliest if you find my explanations helpful :)

Alik [6]2 years ago

4 0

Answer:

$k>7$

Step-by-step explanation:

$4k-\left(5+3k\right)>2$

$-\left(5+3k\right)$

$-5-3k$

$4k-5-3k$

$4k-3k=k$

$k-5$

$k-5>2$

Add 5 to both sides

$k-5+5>2+5$

$k>7$

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A country's population in 1992 was 72 million. In 1998 it was 76 million. Estimate the population in 2012 using the exponential

stepan [7]

We assume the annual percentage increase is the same. We will work out the constant k using the formula

We have
P = 76
A = 72
t = 6 years (from 1992 to 1998)

Substitute into the formula
$76=72e^{6k}$
$\frac{76}{72}=e^{8k}$
Take log both sides give
$ln( \frac{76}{72})=ln(e^{6k})$
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We can now work out the number of population by the end of 2012 using the formula

We have
A = 72 million
k = 0.009
t = 10 years (1992 to 2012)

$P=72e^{0.009*10}$ $P=78.78$ ≈79 million

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3 years ago

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VladimirAG [237]

Answer:

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Step-by-step explanation:

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3 years ago

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otez555 [7]

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3 years ago

I need help with #11

Troyanec [42]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\

\begin{array}{rllll} 
% left side templates
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}
\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\end{array}$

$\bf \begin{array}{llll}


\bullet \textit{ vertical shift by }{{ D}}\\
\qquad if\ {{ D}}\textit{ is negative, downwards}\\\\
\qquad if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{ B}}}
\end{array}$

now, with that template in mind, let's take a peek at this function

$\bf \begin{array}{lllcclll}
y=&2(&1x&-2)^2&-4\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}\\\\
-----------------------------\\\\
A\cdot B=2\impliedby \textit{shrunk by a factor of 2, of half-size}\\\\
\cfrac{C}{B}= \cfrac{-2}{1}\implies -2\impliedby \textit{horizontal right shift of 2 units}\\\\
D=-4\impliedby \textit{vertical down shift of 4 units}$

so, the graph of y=2(x-2)²-4, is really the same graph of y=x², BUT, narrower, and moved about horizontally and vertically

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3 years ago

Read 2 more answers

The figure is the net for a rectangular prism.

oksano4ka [1.4K]

Answer:

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