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3 years ago

Choose the linear inequality that describes the graph. The gray area represents the shaded region.

Mathematics

1 answer:

jek_recluse [69]3 years ago

6 0

Answer:

2x + 3y ≥ 5

Step-by-step explanation:

See the graph attached.

The bold straight line passes through the points (1,1) and (4,-1).

Therefore, the equation of the straight line will be

$\frac{y + 1}{- 1 - 1} = \frac{x - 4}{4 - 1}$

⇒ 3(y + 1) = - 2(x - 4)

⇒ 3y + 3 = - 2x + 8

⇒ 2x + 3y = 5 ............. (1)

Now, the shaded region i.e. the solution to the inequality does not include the origin(0,0).

So, putting x = 0 and y = 0 in the equation (1) we get, 0 < 5

Therefore, the inequality equation is 2x + 3y ≥ 5 (Answer)

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Answer:

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Step-by-step explanation:

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2 years ago

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3 years ago

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3 years ago

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3 years ago

Anyone know the answer to this algebra problem?

ikadub [295]

Answer: $\bold{a=1\qquad b=\dfrac{1}{16}\qquad c=\dfrac{1}{64}\qquad d=1\qquad e=\dfrac{4}{9}\qquad f=\dfrac{16}{81}}$

Step-by-step explanation:

$\begin{array}{c|l}\underline{\quad x\quad}&\underline{\quad 4^{-x}\qquad \qquad}\\-1&4^{-(-1)}=4^1=4\\\\0&4^{-(0)}=4^0=1\\\\2&4^{-(2)}=\dfrac{1}{4^2}=\dfrac{1}{16}\\\\4&4^{-(4)}=\dfrac{1}{4^4}=\dfrac{1}{64}\end{array}$

$\begin{array}{c|l}\underline{\quad \bigg x \quad}&\underline{\quad \bigg(\dfrac{2}{3}\bigg)^x\qquad \qquad}\\\\-1&\bigg(\dfrac{2}{3}\bigg)^{-1}=\dfrac{3}{2}\\\\0&\bigg(\dfrac{2}{3}\bigg)^{0}=1\\\\2&\bigg(\dfrac{2}{3}\bigg)^{2}=\dfrac{2^2}{3^2}=\dfrac{4}{9}\\\\4&\bigg(\dfrac{2}{3}\bigg)^{4}=\dfrac{2^4}{3^4}=\dfrac{16}{81}\end{array}$

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3 years ago