Find an equation for the line that passes through the points (-3,-6) and (3,3)

A grocer has two kinds of candies: one sells for 90 cents per pound and the other sells for 40 cents per pound. How many pounds

Stels [109]

Answer:

45 lbs of 90 cent candies and 5 lbs of 40 cent candies

Step-by-step explanation:

5 0

3 years ago

Write an algebraic equation twice a number increased by 10 is 50

salantis [7]

You need to represent it as 2x + 10 = 50

4 0

3 years ago

Amy needs to mail a gift card to a friend. She uses 47-cent stamps and 6-cent stamps to pay $2.42 in postage. How many of each s

AVprozaik [17]

Answer:

Answer:Amy used 4 41-cent stamps and 8 6-cent stamps.

Step-by-step explanation:

Let x represent the number of 41-cent stamps that Amy used. Let y represent the number of 6-cent stamps that Amy used.

41 cents = 41/100 = $0.41

6 cents = 6/100 = $0.06

She uses 41-cent stamps and 6-cent stamps to pay $2.12 in postage. It means that

0.41x + 0.06y = 2.12

Multiplying through by 100, it becomes

41x + 6y = 212

6y = 212 -41x

We would test for corresponding values of x and y that satisfies the equation and they must be whole numbers.

If x = 3,

6y = 212 - 41 × 3 = 89

y = 89/6 = 14.8333

If x = 4,

6y = 212 - 41 × 4 = 48

y = 48/6 = 8

Answer:Amy used 4 41-cent stamps and 8 6-cent stamps.

Step-by-step explanation:

Let x represent the number of 41-cent stamps that Amy used. Let y represent the number of 6-cent stamps that Amy used.

41 cents = 41/100 = $0.41

6 cents = 6/100 = $0.06

She uses 41-cent stamps and 6-cent stamps to pay $2.12 in postage. It means that

0.41x + 0.06y = 2.12

Multiplying through by 100, it becomes

41x + 6y = 212

6y = 212 -41x

We would test for corresponding values of x and y that satisfies the equation and they must be whole numbers.

If x = 3,

6y = 212 - 41 × 3 = 89

y = 89/6 = 14.8333

If x = 4,

6y = 212 - 41 × 4 = 48

y = 48/6 = 8

8 0

3 years ago

Read 2 more answers

4m/5-2m/3=1 solve the equation

katen-ka-za [31]

Answer:

m = 7.5

Step-by-step explanation:

4m/5 - 2m/3 =1

We want to get rid of the fractions so multiply each side by 5*3 or 15

15 *(4m/5 - 2m/3 ) =1*15

15 *(4m/5) -15( 2m/3 ) =1*15

Canceling like factors

3* 4m -5*2m = 15

Multiply

12m - 10m = 15

2m = 15

Divide each side by 2

2m/2 = 15/2

m = 7.5

8 0

3 years ago

Determine as a linear relation in x, y, z the plane given by the vector function F(u, v) = a + u b + v c when a = 2 i − 2 j + k,

Ostrovityanka [42]

Answer:

2x - y - 3z = 0

Step-by-step explanation:

Since the set

{i, j} = {(1,0), (0,1)}

is a base in $\mathbb{R}^2$

and F is linear, then

would be a base of the plane generated by F.

F(1,0) = a+b = (2i-2j+k)+(i+2j+k) = 3i+2k

F(0,1) = a+c = (2i-2j+k)+(2i+j+2k) = 4i-j+3k

Now, we just have to find the equation of the plane that contains the vectors 3i+2k and 4i-j+3k

We need a normal vector which is the cross product of 3i+2k and 4i-j+3k

(3i+2k)X(4i-j+3k) = 2i-j-3k

The equation of the plane whose normal vector is 2i-j-3k and contains the point (3,0,2) (the end of the vector F(1,0)) is given by

2(x-3) -1(y-0) -3(z-2) = 0

or what is the same

2x - y - 3z = 0

3 0

4 years ago