Determine the solution to a−5=3 that makes the equation true.

Josh has a rewards card for a movie theater. - He receives 15 points for becoming a rewards card holder. - He earns 3.5 points f

jeka94

Answer:

Option A. 3.5x + 15 ≥ 55

Step-by-step explanation:

Points Rosh received for becoming a member = 15

Points Rosh received for visiting the moving theater = 3.5

Total points needed for a free movie ticket = 55

For each visit he receives 3.5 points, so for x visits he will receive 3.5x points.

Total points = Points received on becoming a member + Points received on x visits

So,

Total Points = 15 + 3.5x

These points must be at least 55 for a free movie ticket. At least means equal to or greater than. So the total points must be equal to or greater than 55 in order to earn a free movie ticket. This can be expressed as:

3.5x + 15 ≥ 55

So, option A gives the correct answer

8 0

3 years ago

Read 2 more answers

What is the correct answer???

LUCKY_DIMON [66]

Answer:

C

Step-by-step explanation:

4 0

4 years ago

21 Maria sold small boxes of candy for $1 and

Anna35 [415]

Answer:

There where 10 more Small boxes sold than Large boxes.

Step-by-step explanation:

This is a typical question that can be solved by obtaining a system of two linear equations and solving for each variable.

In principle Linear Equations are algebraic expressions denoting a relationship between a Dependent variable $y$ and an Independent variable $x$ . In a system of Two Linear equations we have two equations of the same variable sets (thus Two Independent variables) so in this case both $y$ and $x$ will be variable terms.

Now with respect to the question and the given information, here our two Variable terms will be the small and the large boxes.

Given Information:

Small Boxes (lets call them $s$ ) cost $1 per box

Large Boxes (lets call them $l$ ) cost $4 per box
Total Number of Boxes sold is 30
Total Profit from sold Boxes is $60

Thus from the above we can obtain one equation denoting the Total Number of Boxes sold and one equation denoting the Total Profit from sold boxes, respectively, as follow:

$s+l=30$ Eqn(1): Total Number of Boxes

$1s+4l=60$ Eqn(2): Total Profit from sold boxes

Now we have a system of two linear equations which we can solve and find the number of small and large boxes, $s$ and $l$ respectively.

From Eqn(1) we see that

$s=30-l$ Eqn(3).

Plugging Eqn(3) in Eqn(2) we can solve for $l$ as:

$1(30-l)+4l=60$

$30-l+4l=60$ Factored out bracket

$3l=60-30$ Gather similar terms on each side and simplify

$3l=30$

$l=\frac{30}{3}$

$l=10$

Plugging in the value for $l=10$ in Eqn(3) we have

$s=30-10\\s=20$

So we know that they were 10 Large Boxes and 20 Small Boxes sold, thus to answer our question, there where 10 more Small boxes sold than Large boxes.

5 0

4 years ago

Seth found two new ways to get to school. Which route is faster for Seth, when school is 17 miles away?

nignag [31]

Answer:answer is D relative house

Step-by-step explanation:

6 0

3 years ago

Clayton Road is 2.35 miles long. Wood Pike Road is 1.7 miles long. Draw a quick picture to find the combined length of Clayton R

marysya [2.9K]

Answer:

4.05 miles.

Step-by-step explanation:

See the attached diagram. (not in scale)

Let AB is the Clayton Road and its length is 2.35 miles and BC is the Wood Pike Road and it is 1.7 miles long.

So, the combined length of Clayton Road and Wood Pike Road is AC (= AB + BC)

So, the length of AC will be (2.35 +1.7) miles = 4.05 miles. (Answer)

8 0

3 years ago