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castortr0y [4]
3 years ago
12

PLEASE HELPPP pleaseee

Mathematics
1 answer:
erastovalidia [21]3 years ago
8 0
Ok so do you know how to do it?
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Linda gets paid $80 per day. Unfortunately Linda went home sick after working 1/3 of the day on Friday. At her company they don’
Ludmilka [50]

Answer:

26.67

Step-by-step explanation:

Take the total amount per day and multiply by the fraction of the day that she worked

80 *1/3 =26.66666

Rounding to the nearest cent

26.67

7 0
2 years ago
Mandy begins cycling west at 30 miles per hour at 11a.M. If Liz leaves from same point 20 minutes later bicycling west at 36 mph
RoseWind [281]

The answer is 1:00 PM


8 0
3 years ago
Help me with this it's confusing :(
fredd [130]
For the equation it would be y=my+b
3 0
3 years ago
Write in standard form the equation of a line with a slope of -3/5 and a y intercept of -3
Vlada [557]

Answer:

5y+3x=-9

Step-by-step explanation:

Let us start by the general form of the standard equation, Ax+By=C. One way we can solve this problem is by finding the <u>slope-intercept form</u> of this equation, y=mx+b, and converting it into the standard equation. In the slope-intercept form, m represents the slope, b represents the y intercept.

From this problem, we are given both the slope and the y intercept. We know have the equation:

y=-\frac{3}{5}x -3

Great! Now let us rearrange the terms so that the y and x terms are on one side of the equation.

y+\frac{3}{5} x=-3

This seems right, but a standard equation must have coefficient values that are real numbers. So, A and B must be real numbers. We can do this by multiplying the entire equation by 5 and ridding the denominator of the A term.

5y+3x=-9

<em>I hope this helps! Please let me know if you have any questions :)</em>

7 0
3 years ago
"A line that goes through (4, 7) and (-2, 1)"
Olenka [21]
Y= x + 3
1. Find the gradient(slope)

1 - 7 / - 2 - 4
=1

2. Find the y. Intercept ( c )
Y = x + c
Replace by any of the two coordinates given (4 , 7 )
7 = 4 + c
C= 3

Equation => Y = x + 3
6 0
3 years ago
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