Solve the following linear inequality: -3 ≤ 2x + 5 /3 < 6

1 answer:

5 0

$-3\leq\dfrac{2x+5}{3} < 6\ \ \ \ \ |\cdot3\\\\-9\leq2x+5 < 18\ \ \ \ |-5\\\\-14\leq2x < 13\ \ \ \ \ |:2\\\\-7\leq x < \dfrac{13}{2}\to x\in\left[-7,\ \dfrac{13}{2}\right)$

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Inessa [10]

Answer:

Step-by-step explanation:

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3 years ago

Find the slope of the line whose equation is 7x - 2y = 4. 7/2 -2 4/7

Nookie1986 [14]

To find the slope, we must put the equation in y = mx + b form, and m represents ur slope.

7x - 2y = 4....subtract 7x from both sides
-2y = -7x + 4 ...now divide both sides by -2
y = (-7/-2)x + (4/-2)
y = 7/2x - 2....now we have it in y = mx + b form, and since m is ur slope, ur slope in this equation is 7/2.

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3 years ago

What is mBC ?<br> A. 28<br> B. 56<br> C. 70<br> D. 140

const2013 [10]

56 is the right answer trust me

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Read 2 more answers

The simplified expression

Romashka-Z-Leto [24]

Answer:

$5x^2 y^2$

Step-by-step explanation:

We need to use the properties shown below to solve this:

1. $\sqrt[n]{x^a} =x^{\frac{a}{n}}$

2. $\sqrt{x}\sqrt{x} =x$

3. $\sqrt{x} \sqrt{y}=\sqrt{x*y}$

Area of a triangle is given by 1/2 * base * height, so we do that and simplify:

$A=\frac{1}{2}(\sqrt{5x^3} )(2\sqrt{5xy^4} )\\A=\frac{1}{2}(5x^3)^{\frac{1}{2}}*2*(5xy^4)^{\frac{1}{2}}\\A=\sqrt{5}x^{\frac{3}{2}}*\sqrt{5}\sqrt{x} } y^2\\A=\sqrt{5} \sqrt{5}x^{\frac{3}{2}} x^{\frac{1}{2}}y^2\\A=5*x^2y^2\\A=5x^2 y^2$