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Nataly [62]
3 years ago
12

5.40(10)+2.50(x)=3.00(10+x) Solve

Mathematics
1 answer:
iris [78.8K]3 years ago
3 0

Answer:

x=48

Step-by-step explanation:

5.40(10)+2.50(x)=3.00(10+x)\\\\54.0+2.50x=3.00(10)+3.00(x)\\\\54+2.5x=30.0+3.00x\\\\54+2.5x=30+3x\\\\54+2.5x-2.5x=30+3x-2.5x\\\\54+0=30+(3-2.5)x\\\\54=30+0.5x\\\\54-30=30+0.5x-30\\\\24=0.5x\\\\10(24)=10(0.5x)\\\\240=5x\\\\\frac{1}{5} (240)=\frac{1}{5} (5x)\\\\48=x\\\\x=48

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Math sequences number patterns
grandymaker [24]

Answer:

see explanation

Step-by-step explanation:

(a)

Note the squared value in column 3 which is the square of 1 more than the row number, that is

row 2 → (2 + 1)² →3²

row 3 → (3 + 1)² → 4²

Find the square root of 676 = 26 → (25 + 1)² = 26²

Hence the row number is 25

(b)

The pattern in column 1 is [ row number × (row number + 2 ) + 1 ]

row number is n then n(n + 2) + 1 = n² + 2n + 1

4 0
3 years ago
Complete all show all your work <br> 40÷5 (2+11)=
Ulleksa [173]
40÷5 (2+11)
8 (13)
104
8 0
3 years ago
Read 2 more answers
Need help finding x pls help
neonofarm [45]

Answer:

C: x = 11

Step-by-step explanation:

if 28 / 4 = 7, then you divide 40 / 4 to get what WH's side is.

if the side is 10 the x has to be 11 because 11 - 1 = 10

3 0
3 years ago
PLS HELP ME GUYS ITS MY LAST QUESTION IN GEOMETRY
aliina [53]

Answer:

Example = ( 12, 1 )

Step-by-step explanation:

There are many possible solutions to this equation, and all you would have to do to determine the ordered pair, is satisfy the following criteria -

x - 6y = 6,\\- 6y = - x + 6,\\y = 1 / 6x - 1

Take a look at a graph of the line y = 1 / 6x - 1. Any ordered pair that lies on this line is a solution to the equation. The " criteria " is that the ordered pair must lie on this line. Let me give you an example, ( 12, 1 ).

<u><em>Hope that helps!</em></u>

7 0
3 years ago
HEEEEELP!!!!!!!!!!!!!11 I WILL MARK BRAINLIEST kx-3y=4 4x-5y=7 k is constant and x and y are variables, for what value of k will
Studentka2010 [4]

Answer:

<h2>               A.  ¹²/₅</h2>

Step-by-step explanation:

There is no solution for system of equations:  a_1x+b_1y=c_1\\a_2y+b_2y=c_2

if:     a_1=a_2\quad and\quad b_1=b_2\quad and\quad \bold{c_1\ne c_2}

so first, we we need to transform the equations to the form where the coefficients at y will be the same:

kx-3y=4\\ 4x-5y=7\\\\ (kx-3y)\cdot5=4\cdot5\\ (4x-5y)\cdot3=7\cdot3\\\\ 5kx-15y=20\\ 12x-15y=21

Now we have  b₁=b₂ and c₁≠c₂ so the system has no solution if a₁=a₂

5k = 12

÷5     ÷5

k = ¹²/₅

7 0
3 years ago
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