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goldenfox [79]
3 years ago
15

Complete the equation for the horizontal line that passes through (5 -7)

Mathematics
1 answer:
professor190 [17]3 years ago
7 0

Answer:

The equation of the horizontal line that passes through (5 , -7) is y = -7

Step-by-step explanation:

The equation of any in the slope-intercept form is y = mx + c, where

1. m is the slope of the line

2. c is the y-intercept

3. The slope of a horizontal line is zero

4. The vertical line has no slope

If the line is horizontal that means all the y-coordinates of the points lie

on the line are equal

Ex: If a horizontal line passes through point (2 , -1), then the equation

     of the line is y = -1

If the line is vertical that means all the x-coordinates of the points lie

on the line are equal

Ex: If a vertical line passes through point (3 , 2), then the equation

     of the line is x = 3

∵ The horizontal line passes through point (5 , -7)

∴ The equation of the line is y = -7

The equation of the horizontal line that passes through (5 , -7) is

y = -7

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Two angles are supplementary. One angle is 16° less than three times the other. Find the measures of the angles
Ivan

Answer:

Step-by-step explanation:

x and y are supplementary angles, so their sum is 180°.

x+y = 180°

x is 16° less than three times y

x = 3y - 16°

(3y-16°) + y = 180°

4y = 196°

y = 49°

x = 3y-16° = 131°

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3 years ago
Please solve the following problem
kirza4 [7]

Answer:

x = 27

Step-by-step explanation:

Given

\frac{x}{9} = 3

Multiply both sides by 9 to clear the fraction

x = 9 × 3 = 27

5 0
3 years ago
Given that 8 bolts and 6 nuts weigh 138grams and 3 bolts and 5 nuts weigh 71 grams. Find the weight of;(a)one bolt and one nut (
Viktor [21]

Answer:

(a) Weight of a bolt = 12 grms and weight of a nut is 7 gms. Weight of one bolt and one nut = 12 + 7 = 19 gms

(b) 69 gms

(c) 209 gms

Step-by-step explanation:

This problem relates to solution of 2 unknowns using simultaneous equations.

Let B be the weight of a single bolt and N be the weight of a single nut

Since 8 bolts and 6 nuts weigh 138 grams we get one of the equations as

8B + 6N = 138    (1)

The second equation in the unknowns is

3B + 5N = 71  (2)

To solve, we eliminate one of the unknowns by making its coefficients the same and subtracting one from the other

Multiplying (1) by 5 ===>   40B + 30N = 690   (3)

Multiplying (2) by 6 ===>  18B + 30N = 426    (4)

(3) - (4) eliminates the N variables and yields

22B = 264   ==> B = 264/22 ==> B = 12

So the weight of a single bolt is 12 grams

We can find the weight of a single nut by substituting this value of B into any of the equations (1), (2), (3) or (4) and solving for N. Let's use equation (2)

3(12) + 5N = 71 ==>  36 + 5N = 71 ==> 5N = 71-36 = 35 ==>  N= 7

So the weight of a single nut is 7 grams

Weight of one bolt and one nut is the sum of the above individual weights = 12 + 7 = 19 gms

To solve (b) and (c) we could set up two other equations and plug in values for B and N

(b) 4B + 3N = 4.12 + 3.7 = 69
However, an alternate way is to perceive that 4B + 3N is exactly half of 8B + 6N so the value of that must be 138/2 = 69

(c) If we add both equations (1) and (2), we get

11B + 11N = 138 + 71 = 209 which is the equation for the total weight of 11 bolts and 11 nuts

5 0
1 year ago
If f(a)=4a-5 what is f(3)-f(1.25)
ladessa [460]

Answer:

f(a)=4a-5

1:    a=  -5  / f−4

f(3)-f(1.25)

2:   1.75f

Step-by-step explanation:

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3 years ago
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