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Nikolay [14]
3 years ago
11

3x+2y=11 y=5x-1 SUBSTITUTION WORTH 10 POINTS

Mathematics
1 answer:
cupoosta [38]3 years ago
5 0
X=1 and y=4. first I did 
3x+2y=11
y=5x-1
and 3x+2(5x-1)=11
=3x+10x-2=11
=13x-2=11
cancel 2 out
13x=13
x=1. then simply add x in the original y equation,
y=5x-1
y=5(1)-1
y=5-1
y=4.
check your work
3(1)+2(4)=11
4=5(1)-1 
again X=1 AND Y=4 
Hope this helped
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Write a quadratic function for each graph described.
user100 [1]

Answer:

y=2x^2-\frac{4}{3}x-\frac{10}{3}

Step-by-step explanation:

we know that

The roots of the quadratic function (x-intercepts) are

x=-1 and x=5/3

so

we can write the equation of the parabola as

y=a(x+1)(x-\frac{5}{3})

where

a is a coefficient

Remember that

The parabola pass through the point (5,40)

substitute the value of x and the value of y of the ordered pair in the quadratic equation and solve for a

x=5, y=40

40=a(5+1)(5-\frac{5}{3})

40=a(6)(\frac{10}{3})

40=20a\\a=2

substitute

y=2(x+1)(x-\frac{5}{3})

apply distributive property

y=2(x^2-\frac{5}{3}x+x-\frac{5}{3})\\\\y=2(x^2-\frac{2}{3}x-\frac{5}{3})\\\\y=2x^2-\frac{4}{3}x-\frac{10}{3}

see the attached figure to better understand the problem

6 0
3 years ago
The length of a rectangle is 2 cm longer than its width if the perimetier of the rectangle is 44cm, find its area
Natasha2012 [34]
L=w+2 and 2w+2l=44

substitute equation for length in perimeter equation

2w+2(w+2)=44
2w+2w+4=44
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substitute width into length equation to get length

l=10+2
l=12

a=lw
a=12*10
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7 0
3 years ago
A jet can carry up to 200,000 liters of fuel.it used 130,000 liters of fuel during a flight. What percentage of the fuel capacit
Komok [63]
Percentage=(fuel consumed/ full fuel capacity) *100%
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3 0
2 years ago
1
Alja [10]

Hi!

<h2>Your answer will be: <em>180.</em></h2>

Reason being: <em>183-3=180. Also, 180+3=183.</em>

Hope this helps!

<em></em>

<em>Yours Truly,</em>

<h2><em></em></h2><h2><em>~~~PicklePoppers~~~</em></h2>

4 0
3 years ago
Which of the following represents a 1.00 M (M = mol/dm3) aqueous solution of glucose (C6H12O6)?
tino4ka555 [31]

Answer:

Option C. 90.0 g glucose per 500 cm3 water

Step-by-step explanation:

1.00 M (mol/dm3) aqueous solution of glucose (C6H12O6) means that 1 mole of glucose is dissolved in 1dm³ (1000cm³) of water.

The concentration of a substance is defined as the mole of solute per unit volume of solvent (water) in dm³.

Option A is wrong since it state: 180g of C6H12O6 per 1000cm³ of solution. Concentration is always expressed as per 1000cm³ or dm³ of water.

For option B:

We shall determine the number of mole of C6H12O6, followed by the concentration.

This is illustrated below:

Molar mass of C6H12O6 = (12x6) + (12x1) + (16x6) = 180g/mol

Mass of C6H12O6 = 10g

Mole = Mass /Molar Mass

Mole of C6H12O6 = 10/180 = 0.056 mole.

Volume of water = 10cm³ = 10/1000 = 0.01dm³

Concentration = mole /Volume

Concentration = 0.056/0.01 = 5.6mol/dm³

Option B is wrong

For option C:

We shall determine the number of mole of C6H12O6, followed by the concentration.

This is illustrated below:

Molar mass of C6H12O6 = 180g/mol

Mass of C6H12O6 = 90g

Mole = Mass /Molar Mass

Mole of C6H12O6 = 90/180 = 0.5mole

Volume of water = 10cm³ = 500/1000 = 0.5dm³

Concentration = mole /Volume

Concentration = 0.5/0.5 = 1mol/dm³

Option C is correct.

For option D is wrong as well as it states 0.1g of C6H12O6 per cm³ of solution. Concentration is always per 1000cm³ or dm³ of water.

From the above illustrations, option C is the correct answer.

7 0
3 years ago
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