Solve : 11 (9-v)=0 a) -9 b) 9 c) 99 d)-99

Answers

Answer 1

Answer: B. V=9Multiply it out, then subtract 99 from both sides, then divide by 11

Answer 2

Answer: the answer would be 9

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Miriam is going to the county fair. Each ride ticket costs $2 and admission costs $7. Miriam has $25 to spend at the fair.

Answers

Answer:

Step-by-step explanation:

Given data

Ride ticket= $2

Admission= $7

Amount= $25

We want to find the greatest number of rides

let the number of rides be x

The expression to describe how she spends the amount is given as

y=7+2x

for y= 25, let us find x

25= 7+2x

25-7= 2x

18=3x

x= 18/2

x= 9

Hence the number of rides is 9

Which unit would you use to measure the weight of a car?
O mg
O kg
O g
O km

Answers

Answer:

The weight of an object is defined as the force of gravity on the object and may be calculated as the mass times the acceleration of gravity, w = mg. Since the weight is a force, its SI unit is the newton.

hope that helps love!

1. Juan bought a pair of shoes costing P923.00 and a tie costing P275.00. He gave the seller P2, 000.00. How much change did he receive?

Answers

$2000.00-923.00-275.00=802.00$

For today’s lunch, a school cafeteria’s budget allows it to purchase at most 60 cans of beans and 45 cans of corn. 1 can of beans feeds 5 students, and 1 can of corn feeds 6 students. Each student will have beans or corn, but not both, and there will be a maximum of 420 students at lunch. If a can of beans cost $2.00 and a can of corn cost $3.00, what is the maximum amount of money required to feed all of the students either beans or corn?A: $180

B: $195

C: $210

D: $225

Answers

Answer:

Option B.

Step-by-step explanation:

Let x be the number of cans of beans and y be the number of cans of corn.

Cafeteria’s budget allows it to purchase at most 60 cans of beans and 45 cans of corn.

$x\leq 60$

$y\leq 45$

1 can of beans feeds 5 students, and 1 can of corn feeds 6 students. Each student will have beans or corn, but not both, and there will be a maximum of 420 students at lunch.

$5x+6y\leq 420$

Can of beans cost $2.00 and a can of corn cost $3.00.

Objective function, Z=2x+3y

The required linear programming problem is

Objective function, Z=2x+3y

Subject to the constraints

$x\leq 60$

$y\leq 45$

$5x+6y\leq 420$

$x\geq 0, y\geq 0$ (Only 1st quadrant)

Draw the graph of these constraints as shown below.

The verities of common shaded region are (0,45), (30,45), (60,20), (60,0), (0,0).

Points Z=2x+3y

(0,0) 0

(0,45) 135

(30,45) 195

(60,20) 180

(60,0) 120

The maximum amount of money required to feed all of the students either beans or corn is $195.

Number of cans of beans = 30

Number of cans of corn = 45

Therefore, the correct option is B.

I would go with B, one hundred ninety five dollars. Hope that this would help you..

A store uses the expression –2p + 50 to model the number of backpacks it sells per day, where the price, p, can be anywhere from $9 to $15. Which price gives the store the maximum amount of revenue, and what is the maximum revenue? (Revenue = price mc023-1.jpg number of backpacks.) $9.00 per backpack gives the maximum revenue; the maximum revenue is $32.00. $12.00 per backpack gives the maximum revenue; the maximum revenue is $312.00. $12.50 per backpack gives the maximum revenue; the maximum revenue is $312.50. $15.00 per backpack gives the maximum revenue; the maximum revenue is $20.00.

Answers

Revenue = price * number of backpacks
number of backpacks = -2p + 50

p = 9 ; -2(9) + 50 = -18 + 50 = 32
revenue = 9 * 32 = 288

p = 12 ; -2(12) + 50 = -24 + 50 = 26
revenue = 12 * 26 = 312

p = 12.50 ; -2(12.50) + 50 = -25 + 50 = 25
revenue = 12.50 * 25 = 312.50 GIVES THE MAXIMUM REVENUE

p = 15 ; -2(15) + 50 = -30 + 50 = 20
revenue = 15 * 20 = 300

Answer:

Its C on Edge

Step-by-step explanation:

$12.50 per and maximum of $312.50

Took the Test and got it right.

2x + 3y = 15
x + y = 6