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siniylev [52]
3 years ago
7

Which values from the set (-6,-4,-3,-1,0,2) satisfy this inequality? - 1/2 x + 3 > 5

Mathematics
1 answer:
andrew-mc [135]3 years ago
4 0

Answer:

-6

Step-by-step explanation:

Manipulating the initial inequality, we get x < -4

Out of the given ones, only -6 is STRICTLY smaller than -4, thus the only one that satisfies the inequality.

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2 tbsp. of peanut butter have 0.25 of total carbohydrate a day what fraction is that?
sergij07 [2.7K]

0.25 is 1/4 if that's what youre asking, i'm not sure,

6 0
2 years ago
Law of radicals
andrey2020 [161]

Answer:

1)  \sqrt{x^7}=x^{\frac{7}{2}

2)  \sqrt[3]{y^5}=y^{\frac{5}{3}

3) \sqrt[5]{a^{12}}=a^{\frac{12}{5} }

4) \sqrt[4]{z^{9}}=z^\frac{9}{4}

Step-by-step explanation:

1) \sqrt{x^7}

We know that \sqrt{x}=x^{\frac{1}{2}

So, \sqrt{x^7}=x^{\frac{7}{2}

2) \sqrt[3]{y^5}

We know that \sqrt[3]{x}=x^{\frac{1}{3}

So, \sqrt[3]{y^5}=y^{\frac{5}{3}

3) \sqrt[5]{a^{12}}

We know that \sqrt[5]{x}=x^{\frac{1}{5}

So, \sqrt[5]{a^{12}}=a^{\frac{12}{5} }

4) \sqrt[4]{z^{9}}

We know that \sqrt[4]{x}=x^{\frac{1}{4}

So, \sqrt[4]{z^{9}}=z^\frac{9}{4}

6 0
3 years ago
Which ratio is equivalent to the ratio 30:48?
lord [1]

Answer: 5:8

Step-by-step explanation:

By reducing both sides equally, the ending answer is 5:8

(There can be many answers to your question but the one i told you is the most reduced one)

7 0
3 years ago
Read 2 more answers
Derive the equation of the parabola with a focus at (3,1) and a directrix of y = 5
serg [7]
So hmmm  check the picture below

so... the vertex is "p" distance from the focus and the directrix, thus, the vertex is really half-way between both

in this case, 2 units up from the focus or 2 units down from the directrix, and thus it lands at 3,3

now, the "p" distance is 2, however, the directrix is up, the focus point is below it, the parabola opens towards the focus point, thus, the parabola is opening downwards, and the squared variable is the "x"

because the parabola opens downwards, "p" is negative, and thus, -2

now, let's plug all those fellows in then

\bf \begin{array}{llll}&#10;(x-{{ h}})^2=4{{ p}}(y-{{ k}})\\&#10;\end{array}&#10;\qquad &#10;\begin{array}{llll}&#10;vertex\ ({{ h}},{{ k}})\\&#10;{{ p}}=\textit{distance from vertex to }\\&#10;\qquad \textit{ focus or directrix}&#10;\end{array}\\\\&#10;-----------------------------\\\\&#10;&#10;\begin{cases}&#10;h=3\\&#10;k=3\\&#10;p=-2&#10;\end{cases}\implies (x-3)^2=4(-2)(y-3)\implies (x-3)^2=-8(y-3)&#10;\\\\\\&#10;-\cfrac{(x-3)^2}{8}=y-3\implies \boxed{-\cfrac{1}{8}(x-3)^2+3=y}

5 0
3 years ago
What is the solution to this inequality x-7&gt;8
Veronika [31]

Answer:

Choice a: x>5

Step-by-step explanation:

All you have to do on this one is add the 7 on both sides.

x-7>8\\x>5

So the answer is anything greater than 5.

7 0
2 years ago
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