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expeople1 [14]
3 years ago
7

Bill and susan have ages that are consecutive odd integers.The product of their ages 438.Which equation could be used to find bi

ll age b if he is the younger one
Mathematics
1 answer:
zhuklara [117]3 years ago
5 0

Answer:

4 x^{2} + 8 x - 435=0

Step-by-step explanation:

Let x be any natural number

Let the age of Bill= 2 x+ 1  years (Since age is odd integer)

Than, age of a Susan= 2 x + 3 years

Since ages are consecutive odd integers

According to question

(2 x+1)(2 x+ 3)= 438

4 x^{2} + 8 x + 3 = 438

4 x^{2} + 8 x + 3 - 438 =0

4 x^{2} + 8 x - 435=0

This equation can be used to find the ages of both

Hence, the correct answer is 4 x^{2} + 8 x - 435=0

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3 years ago
Solve the system using elimination: Help me please<br><br><br>12x - 8y = -12 <br><br>6x + 4y = -30
LUCKY_DIMON [66]

12x - 8y = -12  

6x + 4y = -30

Multiply the 2nd equation by 2, to make the Y coefficients opposite:

6x + 4y = -30 x 2 = 12x + 8y = -60

Now add the two equations:

12x -8y = -12 + 12x +8y = -60

=  24x = -72

Divide bothe sides by 24 to solve for x:

x = -72/24

x = -3

Now replace x with -3 in the first equation to solve for y:

12(-3) - 8y = -12

-36 - 8y = -12

Add 36 to each side:

-8y = 24

Divide both sides by -8 to solve for y:

y = 24 / -8

y = -3

X = -3 and y = -3

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4 0
3 years ago
Quadratic 6X2 + 5X = -7
poizon [28]
Ok, so here your being asked to solve 6x2<span> + 5x = -7 

The procedure that I did was using this formula it led me to get the following:
</span>Using the formula:

x = -(-5) ± √(-5)² - 4(6)(-6)/ 2(6)
x = 5 ± √ 25 + 144 / 12
x = 5 ± √ 169 / 12
x = 5 ± 13/12

x1 = 5 + 13/12
x1 = 18/12
x1 = 3/2

x2 = 5 - 13/12
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Hope this helped :)</span>
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3 years ago
There are 1,657 souvenir paperweights that need to be packed in boxes. Each box will hold 17
blondinia [14]

Answer:

Therefore we can say that 98 boxes will be needed to hold all the paperweights.

Step-by-step explanation:

i) there are 1657 souvenir paperweights that need to be packed in boxes.

ii) each box will hold 17 paperweights

iii) therefore the number of boxes that will be needed are

   = \dfrac{total\hspace{0.15cm} number\hspace{0.15cm} souvenir\hspace{0.15cm} paperweights}{number\hspace{0.15cm} of\hspace{0.15cm} paperweights\hspace{0.15cm} per\hspace{0.15cm} box}  = \dfrac{1657}{17} = 97.471

iv) Therefore we can say that 98 boxes will be needed to hold all the paperweights.

4 0
3 years ago
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