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Vadim26 [7]
3 years ago
8

Write the equation of a line that passes through the point (-5,-3) and has a slope of -3/5 [SLOPE INTERCEPT FORM y=mx+b]

Mathematics
1 answer:
EastWind [94]3 years ago
5 0

Answer:

y= -3/5x -6

Step-by-step explanation:

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The softball team stopped for ice cream after a big game. five 2-scoop cones cost the same as three sundaes. If they bought nine
Savatey [412]
<span>Let the cost of 2-scoop cones be x, and the cost of sundaes be y.
The first statement tells us that 5x = 3y
The second statement tells us that 9x + 6y = 19.95
Then we need to calculate the value of y (the cost of a sundae).
From our first equation, y = 5x/3. Substituting this into the second equation:
9x + 6(5x/3) = 19.95
9x + 10x = 19.95
19x = 19.95
x = 1.05
Substituting back into the first equation: 5(1.05) = 3y
5.25 = 3y
y = 1.75
Therefore the cost of a sundae is $1.75.
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3 0
3 years ago
HW 7.4/7.5 help mhhhh
Hatshy [7]

Answer:

1. LI=18

2. x=10

3. Yes (Y)

4. Yes (Y)

5. x=15

6. x=18

7. x=8

8. x=6

   y=6.5

Step-by-step explanation:

1. LI/JL=KH/JK

Replacing the given values:

LI/6=21/7

Dividing on the right side of the equation:

LI/6=3

Solving for LI: Multiplying both sides of the equation by 6:

6(LI/6)=6(3)

LI=18


2.TV/VS=RU/US

Replacing the given values:

x/17.5=8/14

Simplifying the fraction on the right side of the equation: Dividing numerator and denominator by 2:

x/17.5=(8/2) / (14/2)

x/17.5=4/7

Solving for x: Multiplying both sides of the equation by 17.5:

17.5(x/17.5)=17.5(4/7)=(17.5/1)(4/7)

Multiplying:

x=(17.5 x 4) / (1 x 7)

x=70/7

Dividing:

x=10


3. BC is parallel to DE if AD/DB is equal to AE/EC:

AD/DB=15/12=(15/3) / (12/3)→AD/DB=5/4

AE/EC=10/8=(10/2) / (8/2)→AE/EC=5/4

Like AD/DB=5/4=AE/EC → BC is parallel to DE


4. BC is parallel to DE if AD/DB is equal to AE/EC:

AD/DB= 2DB / DB→AD/DB=2

AE/EC=30 / (AC-AE)=30 / (45-30)=30/15→AE/EC=2

Like AD/DB=2=AE/EC → BC is parallel to DE


5. If JH is a midsegment of triangle KLM:

x=(1/2)(30)

x=15


6. If JH is a midsegment of triangle KLM:

x=2(9)

x=18


7. If JH is a midsegment of triangle KLM: x=8


8. 2x+1=x+7

Solving for x. Grouping the x's on the left side of the equation: Subtracting x both sides of the equation:

2x+1-x=x+7-x

Subtracting:

x+1=7

Subtracting 1 both sides of the equation:

x+1-1=7-1

Subtracting:

x=6

2x+1=2(6)+1=12+1→2x+1=13

x+7=6+7→x+7=13


(3y-8)/(y+5)=(2x+1)/(x+7)

(3y-8)/(y+5)=13/13

(3y-8)/(y+5)=1

3y-8=y+5

Solving for y. Grouping the y's on the left side of the equation: Subtracting y both sides of the equation:

3y-8-y=y+5-y

Subtracting:

2y-8=5

Adding 8 both sides of the equation:

2y-8+8=5+8

Subtracting:

2y=13

Dividing both sides of the equation by 2:

2y/2=13/2

Dividing:

y=6.5



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