It considers the function f: R — R >, f (x) = p x + q, where p
and q are real numbers.

a) find the real numbers p and q knowing that

f (1) = 1 and
f (2) =-1

b) for p = -2 and q = 3, graph the function using
xOy coordinates system.

Free mathematics solved problems and tutorials for elementary, middle or high school

It considers the function f: R — R >, f (x) = p x + q, where p
and q are real numbers.

a) find the real numbers p and q knowing that

f (1) = 1 and
f (2) =-1

b) for p = -2 and q = 3, graph the function using
xOy coordinates system.

The
figure is represented schematically an Aquarium in the form of a right prism
with square floor, which features 8 dm floor side and the lateral edge of Aquarium of 5 dm. The faces
of the Aquarium are made of glass. The Aquarium floor is made from a different
material. Aquarium is not covered. Into Aquarium is located water up to a
height of 4dm (glass thickness neglect).

a) Calculate how many litres of water
are in the Aquarium,

b) Calculate how many square metres of
glass are needed to into 100 pcs. of Aquarium, having the dimensions given in
the hypotesis,

c) Show that, at any time, the distance
between two fish is less than or equal to 12 dm.

The triangle ABC has the measure of 90 degrees of the angle A, the bisector of the angle ABC intersects in point D the side AC and DE is the altitude of the triangle BDC. Find that:

a) AE is perpendicular to BD ;

b) AE is parallel to FC, where F is the intersecting point of AB and DE:

{F} = AB ∩ DE.

In the figure bellow
it is represented the outline of a rectangular room with an area of 48 square meters.
We note ABCD this rectangle.

It is known
that the width is three quarters (¾ ) of the length of the room.
Inside the room
is a fireplace (a stove), represented in the figure by the square MNPD with the
one meter side.

In the room it
is installed hardwood floor except shaded area MNPD where is the stove. a) find the length of the room;

b) knowing that losses represent 10% of the floor area that will be covered with parquet flooring, show that it is enough to buy 51.7 square meter of needed parquet flooring,

c) the parquet flooring is sold in packs each containing 2,5 square meters of parquet flooring. The price of each box with parquet flooring is 43 $. Find the minimum amount of money required to purchase needed parquet flooring.

Find the natural number that simultaneously satisfy the
conditions:

a) divided by 4 the remainig is 3

b) divided by 10, the remaning is 1;

c) divided by 12, the remaining is 3;

d) the sum of quotients of the three divisions from points
a), b) and c) is higher by 16 than one third of the number.

In this issue we use the Euclidean division.

This theorem says the dividend (D) is equal to the product
of the divisor (Q) and the quotient (C) added the remaining (R)

D = Q ∙ C + R

If we denote the number with the letter D and apply the above theorem
we can write conditions:

a) D = 4 ∙ C1 + 3

b) D = 10 ∙ C2 + 1 ∙

c) D = 12 ∙ C3 + 3

For the last condition we write: C1 + C2 + C3 = D / 3 + 16

a) D = 4 ∙
C1 + 3

D
- 3 = 4 ∙ C1

C1 = (D-3) / 4

b) D = 10 ∙ C2 + 1 ∙

D - 1 = 10 ∙ C2

C2 = (D-1) / 10

c) D = 12 ∙ C3 + 3

D - 3 = 12 ∙ C3

C3 = (D-3) / 12

Replacing expressions obtained for C1, C2 and C3 to the
condition d) we obtain:

(D-3)
/ 4 + (D - 1) / 10 + (D - 3) / 12= D / 3 + 16

The common denominator of these fractions:

4 = 2∙2

10 = 2 ∙ 5

12 = 2∙2 ∙ 3

3 = 3∙1

The common denominator = 2∙2 ∙ 3 ∙ 5 = 60

Multiply each fraction reach 60: first by 15, the second by
6, third by 5, fourth by 20 and the free term (whole number) by 60:

15 (D-3) + 6 (D-1) + 5 (D-3) =20 D + 16 ∙ 60

15 D - 45 +6 D -6 +5 D – 15 = 20 D + 960

26 D - 20 D = 960 + 45 + 6 + 15

6 D = 1026

D = 171

Solution: The number required is 171

The problem requires us to find a two digit number ab, where a is the tens digit and b is the units’ digit.

Figures may have values from 1 to 9 (for a=0 result 0^{4} + 0^{2 }= 5 · b => b=0, and the number ab=0 is not a two-digit number).

From a^{4} + a^{2 }= 5 · b. We note that 5b is a multiple of 5.

a^{4} + a^{2 }= a^{2 }(a^{2 }+1) must be multiple of 5.

a^{2} can’t be 5 (a is a natural number)

a^{2 }+1 = 5;

a^{2} = 5-1 = 4;

a = 2^{}

4*5 = 5 *b

b = 4

Other case, for the next multiple of 5

a^{2 }+1=10

a^{2}=9

a=3

a^{2}(a^{2}+1) = 9*10
= 90 = 5*b =>b = 90/5 = 18 >9, impossible, b is a digit, b< 9

The only number ab that is solution to the problem is 24.

The problem gives us an
equality between the numbers x, y, z and u.

7x+5y – 2z – 2u = 0 (the
hypothesis)

If you look at this
equality you can see that the relation could be processed moving two terms on
the right side of the equality.

-2z and -2u moved to
the right side with the changed sign become:
2z and 2u.

7x + 5y = 2z + 2u. We
can remove the common factor 2.

7x + 5y = 2 (z + u).

From this form of the
last relationship we note the following:

to the right we have a bracket
multiplied by 2, which means that this product is an even number.

If on the right side of
the equality it is an even number, then we also have an even number on the left
of equal sign.

So, 7x + 5y is an even
number.

Here the sum is even.

How do we get an even
number in the sum? if we add two even numbers or two odd numbers!

How 7 and 5 are odd, it follows that x and y
are either both even or both odd (in other words: they have the same parity).

We conclude that the
amount x + y is even (if we add two even numbers the result is an even number,
or if we add two odd numbers the result is an even number also).

The problem asks us the
last digit of the number A

A = (11x + 9y) *(z + u
- x).

.

Number A is a product
of two brackets. We see that z is assembled with u and this amount (z + u) is
in the hypothesis multiplied by 2: 7x
+ 5y = 2 (z + u).

.

To be able to use the
hypothesis equality we multiply by 2 the number A and we replace 2(z + u) by
7x+5y. And we get:

2 A = (11x + 9 y) * 2 *
(z + u - x) = (11x + 9 y) (2 (x + u) - 2 x) = (11x + 9 y) (7 x + 5 y - 2 x) =
(11x + 9 y) (5x + 5y) = (11x + 9 y) * 5 (x + y)

We have demonstrated
that x and y have the same parity, and x + y is an even number, x + y is
divisible by 2.

For parenthesis (11x + 9y)
we note:

If x and y are both even
11x + 9y is an even number, so it is divisible by 2.

If x and y are both odd
11x is odd, 9y is also odd and the sum of two odd numbers is always an even
number, so 11x + 9y is divisible by 2.

It follows that number
**2A** is divisible by the number 2 from the bracket 11x + 9y, once it is divisible
by 2 by the even number x + y and it is also divisible by 5.

Number **A** will be
divisible by 2 and 5.

Therefore, number A is divisible
by 2 and 5, so it is divisible by 10.

From the criterion of divisibility
by 10 (any number that has the last digit “0” is divisible by 10) the last digit
of the number A is „0”.

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