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Vlad1618 [11]
3 years ago
9

Which inequality is true when

Mathematics
1 answer:
Mars2501 [29]3 years ago
7 0

Answer:

D

Step-by-step explanation:

If we use the calculator will find out that x= √151/2 ≈ 6.1441

D. 6 < x < 7

x is less then 7 and greater than 6

Without the calculator you can think that 151 is almost 144 that is 12² so

we can approximate √151 to be a bit more then 12

√151/2 = a bit more then 12 /2 = 6 and a bit more

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Example:The missing value is x than Sin 7 / 18
Darya [45]

Answer:

44.9

Step-by-step explanation:

12/17 = PQ/PR

(sin^-1)(12/17) = 44.9

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3 years ago
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Find the circumference of circle P in terms of pi , P = 5ft
ra1l [238]
If by P=5ft you mean the radius is 5 then c = 2pi r or c = 10pi or approx 31.4 ft
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3 years ago
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PLEASE HELP ME OUT!!!
igor_vitrenko [27]
Well, first, the equation that you are looking for is Y = MX + B. To find B, look at the first dot on the Y-axis. It starts at positive 3, so, you know your equation will end with +3. Next, to find MX, start at the either dot, and find the Rise/Run. In this particular equation, the rise is 1, and, the run is -2, because it's going backwards (negative line). Meaning, the line's equation would be, f( x )= -1/2x + 3.
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3 years ago
The length of each side of an equilateral triangle is increased by 20%, resulting in triangle ABC. If the length of each side of
kompoz [17]

Answer: Area of ΔABC is 2.25x the area of ΔDEF.

Step-by-step explanation: Because equilateral triangle has 3 equal sides, area is calculated as

A=\frac{\sqrt{3} }{4} a^{2}

with a as side of the triangle.

Triangle ABC is 20% bigger than the original, which means its side (a₁) measures, compared to the original:

a₁ = 1.2a

Then, its area is

A_{1}=\frac{\sqrt{3} }{4}(1.2a)^{2}

A_{1}=\frac{\sqrt{3} }{4}1.44a^{2}

Triangle DEF is 20% smaller than the original, which means its side is:

a₂ = 0.8a

So, area is

A_{2}=\frac{\sqrt{3} }{4} (0.8a)^{2}

A_{2}=\frac{\sqrt{3} }{4} 0.64a^{2}

Now, comparing areas:

\frac{A_{1}}{A_{2}}= (\frac{\sqrt{3}.1.44a^{2} }{4})(\frac{4}{\sqrt{3}.0.64a^{2} } )

\frac{A_{1}}{A_{2}} = 2.25

<u>The area of ΔABC is </u><u>2.25x</u><u> greater than the area of ΔDEF.</u>

7 0
3 years ago
P³ = 1/8 please help me if anybody can .
ivanzaharov [21]

Answer:

p= \dfrac{1}{2}

Step-by-step explanation:

Given equation:

p^3=\dfrac{1}{8}

Cube root both sides:

\implies \sqrt[3]{p^3}= \sqrt[3]{\dfrac{1}{8}}

\implies p= \sqrt[3]{\dfrac{1}{8}}

\textsf{Apply exponent rule} \quad \sqrt[n]{a}=a^{\frac{1}{n}}:

\implies p= \left(\dfrac{1}{8}\right)^{\frac{1}{3}}

\textsf{Apply exponent rule} \quad \left(\dfrac{a}{b}\right)^c=\dfrac{a^c}{b^c}:

\implies p= \dfrac{1^{\frac{1}{3}}}{8^{\frac{1}{3}}}

\textsf{Apply exponent rule} \quad 1^a=1:

\implies p= \dfrac{1}{8^{\frac{1}{3}}}

Rewrite 8 as 2³:

\implies p= \dfrac{1}{(2^3)^{\frac{1}{3}}}

\textsf{Apply exponent rule} \quad (a^b)^c=a^{bc}:

\implies p= \dfrac{1}{2^{(3 \cdot \frac{1}{3})}}

Simplify:

\implies p= \dfrac{1}{2^{\frac{3}{3}}}

\implies p= \dfrac{1}{2^{1}}

\implies p= \dfrac{1}{2}

3 0
1 year ago
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