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Pie
2 years ago
11

Subtract 6 from z, then multiply 3 by the result

Mathematics
2 answers:
Helen [10]2 years ago
6 0

Answer:

3(z-6)

Step-by-step explanation:

First Subtract 6 from z,

=> z - 6

then multiply by 3,

=>3(z-6)

elena-s [515]2 years ago
6 0

Answer:

\Huge{\bf  \boxed{\bf 3 \times (z - 6)}}

Step-by-step explanation:

Topic- Algebra

<u>Subtract 6 from z will be rewritten as,</u>

\sf \longmapsto \: z - 6

On this algebraic expression, There is one term and a number.We can't simplify it.<em>(If we could Simplify the given expression then also we can't as there is a condition)</em>

<u>Then , We have asked to multiply the result (z-6) by 3.</u>

That is,

\sf \longmapsto3(z - 6)

But,the condition is that we should not simplify any part of the given expression.So we can't simplify this expression.

\sf \longmapsto3 \times (z - 6)

\rule{175pt}{2pt}

I hope this helps!

Please let me know if you have any questions.I am joyous to help!

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PLEASE HELP!! Using the SAS Congruence Theorem
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Answer:

SAS

Step-by-step explanation:

JK is congruent to LM   (given)

L is midpoint of JN (given)

JL is congruent to LN (definition of midpoint)

JK is parallel to LM (given)

Angle LJK is congruent to Angle NLM (corresponding angles)

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3 years ago
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Steve likes to entertain friends at parties with "wire tricks." Suppose he takes a piece of wire 60 inches long and cuts it into
Alex_Xolod [135]

Answer:

a) the length of the wire for the circle = (\frac{60\pi }{\pi+4}) in

b)the length of the wire for the square = (\frac{240}{\pi+4}) in

c) the smallest possible area = 126.02 in² into two decimal places

Step-by-step explanation:

If one piece of wire for the square is y; and another piece of wire for circle is (60-y).

Then; we can say; let the side of the square be b

so 4(b)=y

         b=\frac{y}{4}

Area of the square which is L² can now be said to be;

A_S=(\frac{y}{4})^2 = \frac{y^2}{16}

On the otherhand; let the radius (r) of the  circle be;

2πr = 60-y

r = \frac{60-y}{2\pi }

Area of the circle which is πr² can now be;

A_C= \pi (\frac{60-y}{2\pi } )^2

     =( \frac{60-y}{4\pi } )^2

Total Area (A);

A = A_S+A_C

   = \frac{y^2}{16} +(\frac{60-y}{4\pi } )^2

For the smallest possible area; \frac{dA}{dy}=0

∴ \frac{2y}{16}+\frac{2(60-y)(-1)}{4\pi}=0

If we divide through with (2) and each entity move to the opposite side; we have:

\frac{y}{18}=\frac{(60-y)}{2\pi}

By cross multiplying; we have:

2πy = 480 - 8y

collect like terms

(2π + 8) y = 480

which can be reduced to (π + 4)y = 240 by dividing through with 2

y= \frac{240}{\pi+4}

∴ since y= \frac{240}{\pi+4}, we can determine for the length of the circle ;

60-y can now be;

= 60-\frac{240}{\pi+4}

= \frac{(\pi+4)*60-240}{\pi+40}

= \frac{60\pi+240-240}{\pi+4}

= (\frac{60\pi}{\pi+4})in

also, the length of wire for the square  (y) ; y= (\frac{240}{\pi+4})in

The smallest possible area (A) = \frac{1}{16} (\frac{240}{\pi+4})^2+(\frac{60\pi}{\pi+y})^2(\frac{1}{4\pi})

= 126.0223095 in²

≅ 126.02 in² ( to two decimal places)

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Step-by-step explanation:

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Answer:

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Answer:

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