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vampirchik [111]
3 years ago
9

Which of the following is the absolute value of 2+4i? A 2√2 B 2√3 C √6 D 2√5

Mathematics
1 answer:
Alika [10]3 years ago
4 0

2/5 because it is basically looking up at google.

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1. If I get a loan of $30,000 and I promise to pay in 10 equal instalments of $3,000 each semester, how much interest in total d
Gekata [30.6K]

Answer:

yes because when you multiply 3000 by 10 that gives you a total of 30000 that you paid off

7 0
3 years ago
Which is the equation of a line with a slope of -1/3 and a Y intercept at (0,1)
Len [333]

Answer:

y = -1/3x + 1

Step-by-step explanation:

substitute the given values in y = mx + b where m is the slope and b is the y-intercept

y = -1/3x + b

at point (0, 1)

1 = -1/3 × 0 + b

1 = 0 + b

b = 1

y = -1/3 + 1

3 0
3 years ago
Read 2 more answers
If Frank can produce either 20 pizzas or 40 cheesecakes per day, the opportunity cost per
navik [9.2K]
20p = 40ck
1/2 pizzas to cheesecakes.
4 0
3 years ago
What is the rule used to reflect the triangle ABC to its image?
Debora [2.8K]

Answer:

The rule used to reflect Δ ABC to its image is Reflect over y = x ⇒ B

Step-by-step explanation:

  • If the point (x, y) reflected across the x-axis , then its image is (x, -y)
  • If the point (x, y) reflected across the y-axis , then its image is (-x, y)
  • If the point (x, y) reflected across the line y = x , then its image is (y, x)
  • If the point (x, y) reflected across the line y = -x , then its image is (-y, -x)

From the given figure

∵ The coordinates of point A are (-4.5, 6)

∵ The coordinates of point A' are (6, -4.5)

→ The coordinates are switched ⇒ 3rd rule

∴ Point A is reflected over the line y = x

∴ Δ ABC is reflected over the line y = x

∴ The rule used to reflect Δ ABC to its image is Reflect over y = x

<em>Note: Point B' on the graph should be C' and point C' should be B' (correct it)</em>

3 0
2 years ago
For a resistor in a direct current circuit that does not vary its resistance, the power that a resistor must dissipate is direct
Mrrafil [7]

THe problem is basically telling us: P=kV^2

where P is the power disappated and V^2 is our voltage squared.

\frac{1}{16}=k*14^2\implies\\ \frac{1}{16*196}=k \implies \\ \frac{1}{3136}=k

So, for the second example to find the power we simply have to plug k and our voltage back in, so:P=\frac{14^2*3^2}{14^2*6} \implies \\ P=\frac{9}{6}= \frac{3}{2}

6 0
3 years ago
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