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3 years ago

Fifteen is greater than 3

Mathematics

2 answers:

lbvjy [14]3 years ago

7 0

15>3
If you mean as an inequality

gayaneshka [121]3 years ago

3 0

Yes 15 is greater than 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 see it is greater

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The perimeter of a rectangle is 162 ft. The ratio of the length to the width is 5:4. Find the length of the rectangle.

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3 years ago

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Joy has ₱500.00 to spend for food. She spent ₱98.75 on Monday, ₱73.75 on Tuesday and ₱50.25 on Wednesday. How much money was lef

Alex787 [66]

Answer:

₱277.25

Step-by-step explanation:

₱500.00 - ₱98.75 - ₱73.75 - ₱50.25 = ₱277.25

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2 years ago

Please help me solve this problem... I will mark you brainliest

AVprozaik [17]

Answer:

(8,1)

Step-by-step explanation:

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3 years ago

Two problems here I need solved! I need every step, so please have that with your answers!!

Free_Kalibri [48]

QUESTION 1

We want to solve,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-6x+8}$

We factor the denominator of the fraction on the right hand side to get,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-4x - 2x+8}.$

This implies
$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x(x-4) - 2(x - 4)}.$

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{(x-4)(x - 2)}$

We multiply through by LCM of
$(x-4)(x - 2)$

$(x - 2) + x(x-4) = 2$

We expand to get,

$x - 2 + {x}^{2} - 4x= 2$

We group like terms and equate everything to zero,

${x}^{2} + x - 4x - 2 - 2 = 0$

We split the middle term,

${x}^{2} + - 3x - 4 = 0$

We factor to get,

${x}^{2} + x - 4x- 4 = 0$

$x(x + 1) - 4(x + 1) = 0$

$(x + 1)(x - 4) = 0$

$x + 1 = 0 \: or \: x - 4 = 0$

$x = - 1 \: or \: x = 4$

But
$x = 4$
is not in the domain of the given equation.

It is an extraneous solution.

$\therefore \: x = - 1$
is the only solution.

QUESTION 2

$\sqrt{x+11} -x=-1$

We add x to both sides,

$\sqrt{x+11} =x-1$

We square both sides,

$x + 11 = (x - 1)^{2}$

We expand to get,

$x + 11 = {x}^{2} - 2x + 1$

This implies,

${x}^{2} - 3x - 10 = 0$

We solve this quadratic equation by factorization,

${x}^{2} - 5x + 2x - 10 = 0$

$x(x - 5) + 2(x - 5) = 0$

$(x + 2)(x - 5) = 0$

$x + 2 = 0 \: or \: x - 5 = 0$

$x = - 2 \: or \: x = 5$

But
$x = - 2$
is an extraneous solution

$\therefore \: x = 5$

7 0

3 years ago

Solve:<br> (a+b+c) (a+b-c)

Katen [24]

Answer:

Step-by-step explanation:

Solve:

(a+b+c) (a+b-c)=

(a²+ab-ac+ab+b²-bc+ac+bc-c²)=

a²+ab-ac+ab+b²-bc+ac+bc-c²=

a²+2ab+0ac+b²+0bc-c²=

a²+2ab+b²-c²

7 0

2 years ago

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