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3 years ago

The sum of two numbers is 58. If their difference is 28, find both numbers

Mathematics

2 answers:

Alina [70]3 years ago

7 0

Answer:

the numbers are 43 and 15

Step-by-step explanation:

Let the numbers are x and y

x+y=58

x-y=28

from that two equations,

2x=86

x=43

substitute x value in any equation we will get ,

y=15

juin [17]3 years ago

4 0

Answer:

x = 43 and y = 28

Step-by-step explanation:

Consider two number x and y such that x + y = 58 and x - y = 28

from equation (2) x = y + 28

now, we substitute x by y + 28 in equation (1) we get:

x + y = 58 means y + 28 + y = 58 the. 2y + 28 = 58 then y = 30/2 = 15

since, x = y + 28 then x = 15 + 28 = 43

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David is saving $250 to buy a bicycle. He saved $8 each week for 12 weeks. He wants to buy the bicycle in 7 more weeks. How much

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Step-by-step explanation:

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3 years ago

He vertices of square pqrs are p -4,0 q 4,3 r 7,-5 and s -1,-18.Show that the diagonals of square pqrs are congruent perpendicul

Anit [1.1K]

Answer:

Step-by-step explanation:

The vertices of the square given are P(-4, 0), Q(4, 3), R(7, -5) and, S(-1, -18)

For this diagonal to be right angle the slope of the diagonal must be m1=-1/m2

So let find the slope of diagonal 1

The two points are P and R

P(-4, 0), R(7, -5)

Slope is given as

m1=∆y/∆x

m1=(y2-y1)/(x2-x1)

m1=-5-0/7--4

m1=-5/7+4

m1=-5/11

Slope of the second diagonal

Which is Q and S

Q(4, 3), S(-1, -18)

m2=∆y/∆x

m2=(y2-y1)/(x2-x1)

m2=(-18-3)/(-1-4)

m2=-21/-5

m2=21/5

So, slope of diagonal 1 is not equal to slope two

This shows that the diagonal of the square are not diagonal.

But the diagonal of a square should be perpendicular, this shows that this is not a square, let prove that with distance between two points

Given two points

(x1,y1) and (x2,y2)

Distance between the two points is

D=√(y2-y1)²+(x2-x1)²

For line PQ

P(-4, 0), Q(4, 3)

PQ=√(3-0)²+(4--4)²

PQ=√(3)²+(4+4)²

PQ=√9+8²

PQ=√9+64

PQ=√73

Also let fine RS

R(7, -5) and, S(-1, -18)

RS=√(-18--5)+(-1-7)

RS=√(-18+5)²+(-1-7)²

RS=√(-13)²+(-8)²

RS=√169+64

RS=√233

Since RS is not equal to PQ then this is not a square, a square is suppose to have equal sides

But I suspect one of the vertices is wrong, vertices S it should have been (-1,-8) and not (-1,-18)

So using S(-1,-8)

Let apply this to the slope

Q(4, 3), S(-1, -8)

m2=∆y/∆x

m2=(y2-y1)/(x2-x1)

m2=(-8-3)/(-1-4)

m2=-11/-5

m2=11/5

Now,

Let find the negative reciprocal of m2

Reciprocal of m2 is 5/11

Then negative of it is -5/11

Which is equal to m1

Then, the square diagonal is perpendicular

6 0

4 years ago

Find y when x = 3 of the equation for y=2x+12

ser-zykov [4K]

Answer:

y = 18

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Step-by-step explanation:

Step 1: Define

y = 2x + 12

x = 3

Step 2: Evaluate

Substitute in x: y = 2(3) + 12
Multiply: y = 6 + 12
Add: y = 18

3 0

3 years ago

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Help with this math question

tresset_1 [31]

The given inequality is:

$|-8x+24| \leq 16$

This inequality can be divided in two parts as:

a) $-16 \leq -8x +24$
b) $-8x + 24 \leq 16$

Solving part a:

$-16 \leq -8x+24 \\ \\ 
-40 \leq -8x \\ \\ 
5 \geq x \\ 
or \\ 
x \leq 5$

Solving part b:

$-8x+24 \leq 16 \\ \\ 
-8x \leq -8 \\ \\ 
x \geq 1$

Therefore, the solution to the given inequality is $x \leq 5$ and $x \geq 1$ . Combining both the ranges we get the solution: $1 \leq x \leq 5$ .

In interval notation, this solution can be expressed as [1,5]

3 0

3 years ago

Read 2 more answers