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1 year ago

What is the difference between d and 7?

Mathematics

1 answer:

Goryan [66]1 year ago

3 0

Answer:

d - 7

Step-by-step explanation:

The word "difference" tells us that we will be using subtraction. Also, subtraction is not the same if you reverse the terms like, x - 2 and 2 - x are not the same. In your question it say "d and 7" so that's the order we use in the algebraic expression.

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Gavin drank 3/4 of a liter of orange juice from from a container, which was 3/8 of the orange that was originally in the contain

kow [346]

Answer:

2 liters of orange juice

Step-by-step explanation:

We know that Gaby drank 3/4 liters of juice

3/4 liters is the same as saying 0.75 liters

3/4 = 0.75

if 0.75 is 3/8 of the container we just have to divide it by 3 and multiply it by 8 and so we will get the liters of the container

(0.75/3) * 8 =

0.25 * 8 =

then initially in the container there were 2 liters of orange juice

3 0

3 years ago

Seventy four trees were planted in the garden. Forty nine more bushes were planted than trees in the garden. How many bushes wer

kap26 [50]

No. of trees=74
49 more bushes were palnted
hence
74+49=123 bushes

3 0

3 years ago

Plz help me with it

vovangra [49]

$\sqrt{2x + 7} + 1 = \sqrt{2x + 14 } \\ x = 1$

5 0

3 years ago

Read 2 more answers

4. Parking fees at IIUM are RM 5.00 for IIUM students and RM 7.50 for non-IIUM students. At the

professor190 [17]

Answer:

(a) 780 students and 960 non-students

(b) No. The maximum revenue is RM9000 from 1200 non-students.

(c). Revenue is maximum of RM9000 at 1200 non-students, decreasing by RM2.50 per student to a minimum of RM6000 at 1200 students

Step-by-step explanation:

Let x = IIUM students and

and y = non-IIUM students

You have two conditions

(a) x + y = total vehicles parked

(b) 5.00x + 7.50y = total gross receipts

(a) Wednesday

From your table,

x + y = 1740

5.00x + 7.70y = RM11 100

Solve the simultaneous equations

$\begin{array}{rrcrl}(1) & x + y & = &1740&\\(2) & 5.00x + 7.50y & = & 11 100\\(3)& 5.00x + 5.00y & = & 8700 & \text{Multiplied (1) by 5}\\&2.50 y & = &2400 &\text{Subtracted (3) from (2)}\\(4)&y& = &\mathbf{960} &\text{Divided each side by 2.50}\\& x +960& = &1740& \text{Substituted (4) into (1)}\\& x& = &\mathbf{780}& \\\end{array}\\\text{There are $\large \boxed{\textbf{780 students and 960 non-students}}$}$

(b) Can 1200 vehicles bring in RM10000?

No. Even if all the cars were from non-students, the most you could get is

1200 × 7.50 = RM9000

(c) Possible combinations for 1200 vehicles

Revenue = 5.00x + 7.50y = 5.00x + 7.50(1200 -x) = 5.00x + 9000 - 7.50x =

Revenue = 9000 - 2.50x

The maximum revenue of RM9000 occurs when there are no student cars and 1200 non-student cars.

For each student car that enters and displaces a non-student, the revenue drops by RM2.50.

Finally. when there are 1200 student cars and no non-students, the revenue has dropped to a minimum of RM6000.

3 0

3 years ago

Evaluate the following

fgiga [73]

12C5 = 792 5P4 = 120

6 0

3 years ago