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sineoko [7]
3 years ago
8

0.36x4 +0.47x3 -0.13x2-0.01x3 + 0.19x2 + x0.44x2+0.33x20.31xThe sum is​

Mathematics
2 answers:
blagie [28]3 years ago
7 0
0.31x3 would be the answer
Elena-2011 [213]3 years ago
5 0

Answer:

0.31x3 is your correct answer

You might be interested in
2/x + 3/y = 13 ; 5/x - 4/y = -2​
Tomtit [17]

Given that

(2/x)+(3/y) = 13--------(1)

(5/x)-(4/y) = -2 -------(2)

Put 1/x = a and 1/y = b then

2a + 3b = 13 ----------(3)

On multiplying with 5 then

10a +15 b = 65 -------(4)

and

5a -4b= -2 ----------(5)

On multiplying with 2 then

10 a - 8b = -4 -------(6)

On Subtracting (6) from (4) then

10a + 15b = 65

10a - 8b = -4

(-)

_____________

0 + 23 b = 69

______________

⇛ 23b = 69

⇛ b = 69/23

⇛ b =3

On Substituting the value of b in (5)

5a -4b= -2

⇛ 5a -4(3) = -2

⇛ 5a -12 = -2

⇛ 5a = -2+12

⇛ 5a = 10

⇛ a = 10/5

⇛ a = 2

Now we have

a = 2

⇛1/x = 2

⇛ x = 1/2

and

b = 3

⇛1/y = 3

⇛ y = 1/3

<u>Answer :-</u>The solution for the given problem is (1/2,1/3)

<u>Check</u>: If x = 1/2 and y = 1/3 then

LHS = (2/x)+(3/y)

= 2/(1/2)+3/(1/3)

= (2×2)+(3×3)

= 4+9

= 13

= RHS

LHS=RHS is true

and

LHS=(5/x)-(4/y)

⇛ 5/(1/2)- 4/(1/3)

⇛(5×2)-(4×3)

⇛ 10-12

⇛ -2

⇛RHS

LHS = RHS is true

5 0
2 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
3 years ago
Help evaluating the question for 14,16
Olenka [21]
Number 14 is 61 and Number 16 is 36..,.



5 0
3 years ago
Read 2 more answers
The diagram below shows a shaded parallelogram drawn inside a rectangle. Using Pythagoras, find the hypotenuse of triangle A and
Julli [10]

Answer: 5.9 cm for both triangles.

Step-by-step explanation:

Hi, since the situation forms 2 right triangles we have to apply the Pythagorean Theorem:  

c^2 = a^2 + b^2  

Where c is the hypotenuse of a triangle (the longest side of the triangle) and a and b are the other sides.  

Replacing with the values given:  

c^2 = 3^2 + 5^2  

c^2 = 9+25

c^2 = 34

c = √34

c = 5.9 cm

Since both triangles are identical ( same side lengths) the hypotenuse is the same for both, 5.9 cm.

Feel free to ask for more if needed or if you did not understand something.  

5 0
3 years ago
Subtract 6 from me then multiply by 2 if you subtract 40 then divide by 4 you get 8 what number I am
stellarik [79]

Answer:

42

Step-by-step explanation:

42-6 = 36

36*2 = 72

72-40 = 32

32/4 = 8

8 0
2 years ago
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