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2 years ago

Two times what equals twenty-two

Mathematics

2 answers:

forsale [732]2 years ago

8 0

The are is eleven because 2×11= 22

saul85 [17]2 years ago

6 0

2 multiplied by 11 is 22. so the answer is 11

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-1/2 +2=-x+7 how do you figure this out

Feliz [49]

Answer:

x = 5.5

Step-by-step explanation:

Solve for x by simplifying both sides of the equation, then isolating the variable.

$-\frac{1}{2} +2=-x+7$

$-\frac{1}{2} +2= 1.5$

$1.5=-x+7$

-7 -7

$-5.5 = -x$

Divide both sides by -1.

$5.5 = x$

3 0

2 years ago

The sum of the areas of two circles is 80 pie square meters. Find the length of a radius of each circle if one of them is twice

Svetach [21]

Answer:

The radius of the circles are $4\ m$ and $8\ m$

Step-by-step explanation:

Let

x-----> the radius of larger circle

y----> the radius of smaller circle

we know that

$\pi x^{2} +\pi y^{2}=80\pi$

$x^{2} +y^{2}=80$ -----> equation A

Remember that

$x=2y$ -----> equation B

substitute equation B in equation A and solve for y

$(2y)^{2} +y^{2}=80$

$4y^{2} +y^{2}=80$

$5y^{2}=80$

$y^{2}=16$

$y=4\ m$

Find the value of x

$x=2y$

$x=2(4)=8\ m$

therefore

The radius of the circles are $4\ m$ and $8\ m$

4 0

2 years ago

Find (f/g) (x) and state the domain restriction. f (x) = 6x - 3 and g(x) = 12x^2 - 6x

Rufina [12.5K]

9514 1404 393

Answer:

(f/g)(x) = 1/(2x)

x ≠ 1/2

Step-by-step explanation:

$(f/g)(x)=\dfrac{f(x)}{g(x)}=\dfrac{6x-3}{12x^2-6x}=\dfrac{6x-3}{2x(6x-3)}\\\\\boxed{(f/g)(x)=\dfrac{1}{2x}}\quad x\ne\dfrac{1}{2}$

The domain restriction x ≠ 1/2 comes from the requirement to prevent the denominator factor 6x-3 from being zero.

8 0

2 years ago

If A+B+C=<img src="https://tex.z-dn.net/?f=%5Cpi" id="TexFormula1" title="\pi" alt="\pi" align="absmiddle" class="latex-formula"

seraphim [82]

Answer:

$a + b + c = \pi \\ = > c= \pi - a - b \\ < = > \tan(c) = \tan(\pi - a - b) = -\tan(a + b)$

Step-by-step explanation:

we have:

$\tan(a) + \tan(b) + \tan(c) \\ = \tan(a) + \tan(b) - \tan(a + b) \\ = \tan( a) + \tan(b) - \frac{ \tan(a) + \tan(b) }{1 - \tan(a) \tan(b) } \\ = \frac{ ( \tan(a) + \tan(b) ) \tan(a) \tan(b) }{ \tan(a) \tan(b) - 1 } (1)$

we also have:

$\tan(a) \tan(b) \tan(c) \\ = - \tan(a) \tan(b) \tan(a + b) \\ = \frac{ -(\tan( a ) + \tan(b) ) \tan(a) \tan(b) }{1 - \tan(a) \tan(b) } \\ = \frac{( \tan(a) + \tan(b)) \tan(a) \tan(b) }{ \tan(a) \tan(b) - 1 } (2)$

from (1)(2) => proven

5 0

2 years ago

Which expression is equivalent to 2(a+3) when a equals 3 2(3) 5(3) 2(3)+3 2(3)+6

sammy [17]

2(3+3) or 2(3)+6, so the answer is the third option. It’s just using the distributive property.

8 0

3 years ago

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