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3 years ago

7

om was using wire of the following thicknesses .33 mm, .275 mm, .25 mm, and .3 mm for some electrical work. Order the wire from

thickest to thinnest. A. .275, .25, .3, .33 B. .33, .3, .275, .25 C. .3, .33, .25, .275 D. .3, .33, .275, .25

1 answer:

bogdanovich [222]3 years ago

8 0

B. .33 .3 .275 .25 yup, just gotta get over 20 characters

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At 3 P.M., the temperature is 12°F. It then began dropping 2°F every hour for 7 hours. What was the temperature after 7 hours?

fiasKO [112]

All you need to do to answer this question is multiply 2 (the dropping temperature) by 7 (time in hours) which is 14. Subtract 14 from 12 and your answer is -2 degrees Fahrenheit

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2 years ago

What is the next term of the arithmetic sequence? -2,-4,-6

hichkok12 [17]

Answer:

- 8

Step-by-step explanation:

The common difference d between consecutive terms is

d = $a_{2}$ - $a_{1}$ , that is

d = - 4 - (- 2) = - 4 + 2 = - 2

To obtain the next term subtract - 2 from - 6

next term = - 6 - 2 = - 8

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2 years ago

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use the matrix tool to solve the system of equations enter the answer as an ordered pair. 8x+5y=9 -x+y=7

Dimas [21]

Answer:

x = -44/13

y = -65/13

Step-by-step explanation:

Using matrix form means using the crammers rule

The matrix form of the expression is written as;

$\left[\begin{array}{ccc}8&5\\-1&1\\\end{array}\right] \left[\begin{array}{ccc}x\\y\\\end{array}\right] = \left[\begin{array}{ccc}9\\7\\\end{array}\right]$

AX = B

taking the determinant of A;

|A| = 8(1) - 5(-1)

|A| = 8 + 5

|A| = 13

After replacing the first row with the column matrix;

$A_x =\left[\begin{array}{ccc}9&5\\7&-1\\\end{array}\right]$

|Ax| = 9(-1)-5(7)

||Ax| = -9 - 35

|Ax| = -44

x = |Ax|/|A|

x = -44/13

similarly for y

$A_x =\left[\begin{array}{ccc}8&9\\-1&7\\\end{array}\right]$

|Ay| = 8(7)+9

|Ay| = 56+9

|Ay| = 65

y = |Ay|/|A|

y = -65/13

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2 years ago

Simplified product ?

mel-nik [20]

Answer:

Last choice is correct.

Step-by-step explanation:

$\left(\sqrt{10x^4}-x\sqrt{5x^2}\right)\left(2\sqrt{15x^4}+\sqrt{3x^3}\right)$

$\left(x^2\sqrt{10}-x\cdot x\sqrt{5}\right)\left(2\cdot x^2\sqrt{15}+x\sqrt{3x}\right)$

$\left(x^2\sqrt{10}-x^2\sqrt{5}\right)\left(2x^2\sqrt{15}+x\sqrt{3x}\right)$

$x^2\sqrt{10}\left(2x^2\sqrt{15}+x\sqrt{3x}\right)-x^2\sqrt{5}\left(2x^2\sqrt{15}+x\sqrt{3x}\right)$

$2x^4\sqrt{150}+x^3\sqrt{30x}-2\sqrt{75}x^4-x^3\sqrt{15x}$

$2x^4\cdot5\sqrt{6}+x^3\sqrt{30x}-2\cdot5\sqrt{3}x^4-x^3\sqrt{15x}$

$10x^4\sqrt{6}+x^3\sqrt{30x}-10\sqrt{3}x^4-x^3\sqrt{15x}$

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Hence final answer is $10x^4\sqrt{6}+x^3\sqrt{30x}-10x^4\sqrt{3}-x^3\sqrt{15x}$

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2 years ago

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horsena [70]

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2 years ago

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