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UkoKoshka [18]
3 years ago
11

If c=1/3 and b=3 solve 3c divided by (2b to the 2nd power

Mathematics
1 answer:
valentina_108 [34]3 years ago
3 0

Answer:

4 1/2

Step-by-step explanation:

3(1/3) divided by 2(3)2nd power

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balandron [24]

Answer: The first step would be to multiply the first equation by 3 and the second by 2 so you can eliminate x.

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Find 4 × 0.001.<br><br> A) 0.0004 <br> B) 0.004 <br> C) 0.04 <br> D) 0.4
maksim [4K]

Answer:

B) 0.004

Step-by-step explanation:

0.001

x      4

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0.004

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A rectangular prism and a square pyramid were joined to form a composite figure.
azamat

Answer:

B. 333 in.2

Step-by-step explanation:

The area of the base is 81^2

lateral is 45 x 4 = 180^2  (9x5x4)

180^2 add the 72 pyramid = 252^2 + base of 81^2 = 333^2

Composite figueres i do like this as shown to you a few seconds ago.

The triangle shows us just the height

4 inches

We can see that height is smaller central isosceles height across the center base point.

We also can remember to use the length 9inches but divide by 2 and get each triangle area this way.

4 x 1/2 base = 4x 1/2 4.5 = 4 x 2.25 = 9^2 each right side triangle

9 x 8 = 72^2

we add the areas 72+ 81+lateral 180 = 333 inches^2

8 0
3 years ago
The 6% state income tax on a $42.00 salary​
prisoha [69]

Answer:

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Step-by-step explanation:

5 0
3 years ago
PLEASE HELP I NEED THIS ASAP<br> Find the vertex of y=x^2-7x+4
Goshia [24]

Answer:

\mathrm{Minimum}\space\left(\frac{7}{2},\:-\frac{33}{4}\right)

Step-by-step explanation:

y=x^2-7x+4\\\mathrm{The\:vertex\:of\:an\:up-down\:facing\:parabola\:of\:the\:form}\:y=ax^2+bx+c\:\mathrm{is}\:x_v=-\frac{b}{2a}\\\mathrm{The\:parabola\:params\:are:}\\a=1,\:b=-7,\:c=4\\x_v=-\frac{b}{2a}\\x_v=-\frac{\left(-7\right)}{2\cdot \:1}\\\mathrm{Simplify}\\x_v=\frac{7}{2}\\\mathrm{Plug\:in}\:\:x_v=\frac{7}{2}\:\mathrm{to\:find\:the}\:y_v\:\mathrm{value}\\y_v=\left(\frac{7}{2}\right)^2-7\cdot \frac{7}{2}+4\\

\mathrm{Simplify\:}\left(\frac{7}{2}\right)^2-7\cdot \frac{7}{2}+4:\quad -\frac{33}{4}\\y_v=-\frac{33}{4}\\Therefore\:the\:parabola\:vertex\:is\\\left(\frac{7}{2},\:-\frac{33}{4}\right)\\\mathrm{If}\:a0,\:\mathrm{then\:the\:vertex\:is\:a\:minimum\:value}\\a=1\\\mathrm{Minimum}\space\left(\frac{7}{2},\:-\frac{33}{4}\right)

6 0
3 years ago
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