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Reil [10]
3 years ago
5

The pic is the question! Please help

Mathematics
2 answers:
bulgar [2K]3 years ago
4 0

Answer:

93 is your correct answer

Step-by-step explanation:

kirza4 [7]3 years ago
3 0

Answer:

93

Step-by-step explanation:

150×62%

150×62/100

15×62/10

=930/10

=93

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Point a is located at 4,-7. The point is reflected in the x-axis it's image is located at
Vilka [71]
The answer should be (4,7)
7 0
3 years ago
A line passes through (1, 8) and is perpendicular to the graph of y = 2x+1. What equation
Genrish500 [490]

Answer:

y = -1/2x + 17/2

Explanation:

y = 2x + 1 is in slope intercept form

in the equation y = mx+b, m is the slope, and in the equation m =2

the slope of a perpendicular line is the negative reciprocal of the other slope.

slope of perpendicular= -1/m = -1/2

y = -1/2x + b

now find b by substituting in (1,8) into the partial equation

8 = -1/2 + b

b = 8+ 1/2

b = 17/2

please give thanks :) hope this helps

5 0
3 years ago
Need help please thanks
Ivahew [28]

Answer:

A

Step-by-step explanation:

5 0
2 years ago
A transformer increases the input voltage from 12V to 12,000V. What is the ratio of the number of wire turns on the primary coil
erma4kov [3.2K]

Answer:

\frac{n_p}{n_s}=\frac{1}{1000}

Step-by-step explanation:

Given voltage of primary coil is V_p=12V

And voltage of secondary coil is V_s=12,000V

A transformer is a device that contains primary and secondary coil. These coils are mounted on soft iron. The change in voltage depends upon the number of turns of primary coil (n_p) and secondary coil (n_s).

Following equation represents the relation between them.

\frac{V_s}{V_p} =\frac{n_s}{n_p}\\\\\frac{12,000}{12}=\frac{n_s}{n_p}\\\\1000=\frac{n_s}{n_p}

Then, ratio of the number of wire turns on the primary coil to the number of turns on the secondary coil is

\frac{n_p}{n_s}=\frac{1}{1000}

5 0
3 years ago
Use synthetic division to solve the following: 2x^4 + 11x^3 + 13x^2 + 2x – 8 ÷ x +4​
const2013 [10]

Answer:

x=-5

Step-by-step explanation:

5 0
2 years ago
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