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Rudiy27
3 years ago
8

Determine the relationship between the point (1, –5) and the given system of inequalities.

Mathematics
2 answers:
zimovet [89]3 years ago
8 0
Algebraically, the point (1, -5) satisfies the first inequality, but it does not satisfy the second inequality because -5 is not greater than -5. Graphically, the point (1, -5) lies in the shaded area of the first inequality but lies on the dashed line of the second inequality, which is not inclusive. Therefore (1, -5) is not a solution to the given system of inequalities.
Darya [45]3 years ago
4 0
Y < = 3x + 2...subbing in (1,-5)
-5 < = 3(1) + 2
-5 < = 3 + 2
-5 < = 5 (true)

y > -2x - 3...subbing in (1,-5)
-5 > -2(1) - 3
-5 > -2 - 3
-5 > -5 (false)

response : Algebraically, the point (1,-5) satisfies the first inequality, but does not satisfy the second inequality because -5 is not greater then -5.Graphically, the point (1,-5) lies in the shaded area of the first inequality but lies on the dashed line of the second inequality, which is not inclusive. Therefore, (1,5) is not a solution to the given system of inequalities.

lol...ur sample response is the correct answer for this inequality...exactly


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What is -(5x+6)-x=-6-6x
Veronika [31]

Answer: All real numbers are solutions.

Step-by-step explanation:

<u>Let's solve your equation step-by-step.</u>

−(5x+6)−x=−6−6x

<u>Step 1: Simplify both sides of the equation.</u>

−(5x+6)−x=−6−6x

−5x+−6+−x=−6+−6x(Distribute)

(−5x+−x)+(−6)=−6x−6(Combine Like Terms)

−6x+−6=−6x−6

−6x−6=−6x−6

<u>Step 2: Add 6x to both sides.</u>

−6x−6+6x=−6x−6+6x

−6=−6

<u>Step 3: Add 6 to both sides.</u>

−6+6=−6+6

0=0

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3 years ago
A plank is 6.7 feet long. Arachna wants to cut the plank into 0.8-foot pieces. how many 0.8-foot pieces can she get?
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Step-by-step explanation:

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4. Find the vertex, focus, and directrix of the parabola. Sketch its graph, showing the focus and the directrix.
valentina_108 [34]

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focus - (0, 3/4)

directrix - 3/4

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8 0
2 years ago
What is the distance between the points (3, 7) and (15, 16) on a coordinate
AysviL [449]

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6 0
3 years ago
Please solve, answer choices included.
qaws [65]
4. To solve this problem, we divide the two expressions step by step:

\frac{x+2}{x-1}* \frac{x^{2}+4x-5 }{x+4}
Here we have inverted the second term since division is just multiplying the inverse of the term.

\frac{x+2}{x-1}* \frac{(x+5)(x-1)}{x+4}
In this step we factor out the quadratic equation.


\frac{x+2}{1}* \frac{(x+5)}{x+4}
Then, we cancel out the like term which is x-1.

We then solve for the final combined expression:
\frac{(x+2)(x+5)}{(x+4)}

For the restrictions, we just need to prevent the denominators of the two original terms to reach zero since this would make the expression undefined:

x-1\neq0
x+5\neq0
x+4\neq0

Therefore, x should not be equal to 1, -5, or -4.

Comparing these to the choices, we can tell the correct answer.

ANSWER: \frac{(x+2)(x+5)}{(x+4)}; x\neq1,-4,-5

5. To get the ratio of the volume of the candle to its surface area, we simply divide the two terms with the volume on the numerator and the surface area on the denominator:

\frac{ \frac{1}{3} \pi  r^{2}h }{ \pi  r^{2}+ \pi r \sqrt{ r^{2}  +h^{2} }  }

We can simplify this expression by factoring out the denominator and cancelling like terms.

\frac{ \frac{1}{3} \pi r^{2}h }{ \pi r(r+ \sqrt{ r^{2} +h^{2} } )}
\frac{ rh }{ 3(r+ \sqrt{ r^{2} +h^{2} } )}
\frac{ rh }{ 3r+ 3\sqrt{ r^{2} +h^{2} } }

We then rationalize the denominator:

\frac{rh}{3r+3 \sqrt{ r^{2} + h^{2} }}  * \frac{3r-3 \sqrt{ r^{2} + h^{2} }}{3r-3 \sqrt{ r^{2} + h^{2} }}
\frac{rh(3r-3 \sqrt{ r^{2} + h^{2} })}{(3r)^{2}-(3 \sqrt{ r^{2} + h^{2} })^{2}}}=\frac{3 r^{2}h -3rh \sqrt{ r^{2} + h^{2} }}{9r^{2} -9 (r^{2} + h^{2} )}=\frac{3rh(r -\sqrt{ r^{2} + h^{2} })}{9[r^{2} -(r^{2} + h^{2} )]}=\frac{rh(r -\sqrt{ r^{2} + h^{2} })}{3[r^{2} -(r^{2} + h^{2} )]}

Since the height is equal to the length of the radius, we can replace h with r and further simplify the expression:

\frac{r*r(r -\sqrt{ r^{2} + r^{2} })}{3[r^{2} -(r^{2} + r^{2} )]}=\frac{ r^{2} (r -\sqrt{2 r^{2} })}{3[r^{2} -(2r^{2} )]}=\frac{ r^{2} (r -r\sqrt{2 })}{-3r^{2} }=\frac{r -r\sqrt{2 }}{-3 }=\frac{r(1 -\sqrt{2 })}{-3 }

By examining the choices, we can see one option similar to the answer.

ANSWER: \frac{r(1 -\sqrt{2 })}{-3 }
8 0
3 years ago
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