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Kay [80]
3 years ago
9

which expression is equivalent to 2m(3/2m+1) +3 (5/3m-2) ? A .3m^2+5m-1 B. 3/4m^2 +23/9m-6 C. 3m^2+7m-6 D. 3/4m^2+5/9m-1 please

explain with steps
Mathematics
2 answers:
vlabodo [156]3 years ago
8 0

Answer:

b

Step-by-step explanation:

horrorfan [7]3 years ago
5 0

Answer:

66

Step-by-step explanation:

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Write the standard equation of the circle with the given center is (0,0) and radius is 2.5
BlackZzzverrR [31]

Answer:

x² + y² = 6.25

Step-by-step explanation:

The standard equation of a circle is ( x – h)² + ( y - k)² = r² where the center is (h, k) and r is the radius.

From your given information that the center is (0, 0) then both h and k are 0. Then r is 2.5, so

( x – 0)² + ( y - 0)² = (2.5)²

x² + y² = (2.5)²

x² + y² = 6.25

7 0
3 years ago
Please help please help please help please help please help please help please help please help
Burka [1]
I Believe it’s the first one
5 0
3 years ago
How do I write each number as a power of the given base? <br><br>example: 49; base 7
VLD [36.1K]
7^n=49
we know that 7*7=49 so thus 7^2=49
7^n=7^2
therefor n=2
it is written as 7^2
7 0
3 years ago
Let O be an angle in quadrant III such that cos 0 = -2/5 Find the exact values of csco and tan 0.​
vivado [14]

well, we know that θ is in the III Quadrant, where the sine is negative and the cosine is negative as well, or if you wish, where "x" as well as "y" are both negative, now, the hypotenuse or radius of the circle is just a distance amount, so is never negative, so in the equation of cos(θ) = - (2/5), the negative must be the adjacent side, thus

cos(\theta)=\cfrac{\stackrel{adjacent}{-2}}{\underset{hypotenuse}{5}}\qquad \textit{let's find the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{5^2 - (-2)^2}=b\implies \pm\sqrt{25-4}\implies \pm\sqrt{21}=b\implies \stackrel{III~Quadrant}{-\sqrt{21}=b}

\dotfill\\\\ csc(\theta)\implies \cfrac{\stackrel{hypotenuse}{5}}{\underset{opposite}{-\sqrt{21}}}\implies \stackrel{\textit{rationalizing the denominator}}{-\cfrac{5}{\sqrt{21}}\cdot \cfrac{\sqrt{21}}{\sqrt{21}}\implies -\cfrac{5\sqrt{21}}{21}} \\\\\\ tan(\theta)=\cfrac{\stackrel{opposite}{-\sqrt{21}}}{\underset{adjacent}{-2}}\implies tan(\theta)=\cfrac{\sqrt{21}}{2}

4 0
2 years ago
How many significant digits are there in the product of 0.005 and 25.725
Furkat [3]
A is the correct answe
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3 years ago
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