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3 years ago

Jamal spent $6.75 on wire. Wire costs $0.45 per foot. How many feet of wire did Jamal buy?

Mathematics

2 answers:

kiruha [24]3 years ago

8 0

Answer:

Step-by-step explanation:

blondinia [14]3 years ago

5 0

Answer:15

Step-by-step explanation:

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Jully bought 3 slices of cheese pizza and 4 slices of pepperoni pizza for $18.00. Christian bought 3 slices of cheese pizza and

andrew11 [14]

Answer:

$2.00

Step-by-step explanation:

You can form an equation with these statements, x being the price of cheese pizza and y being the price of pepperoni. Making the equation,

3x+4y=18

3x+2y=12

Solving the equation,

3x+4y=18

6x+4y=24

-3x=-6

x=2

y=3

Solving another way,

2y=6

y=3

x=2

Therefore, the answer is $2.00

8 0

3 years ago

Alex is 12 years older than George. Carl is Three times older than Alex The sum of their ages is 68 Find the ratio of George’s a

earnstyle [38]

Ratio of George’s age to Carl’s age to Alex’s age is 1 : 12 : 4

Solution:

Let "a" be the age of Alex

Let "g" be the age of George

Let "c" be the age of Carl

Alex is 12 years older than George

Therefore,

Age of Alex = 12 + Age of George

a = 12 + g

g = a - 12 ---- eqn 1

Carl is Three times older than Alex

Age of Carl = 3 x Age of Alex

c = 3a -------- eqn 2

The sum of their ages is 68

age of Alex + age of George + age of Carl = 68

a + g + c = 68 ----------- eqn 3

Substitute eqn 1 and eqn 2 in eqn 3

a + a - 12 + 3a = 68

5a - 12 = 68

5a = 12 + 68

5a = 80

a = 16

Substitute a = 16 in eqn 2

c = 3(16) = 48

c = 48

Substitute a = 16 in eqn 1

g = 16 - 12 = 4

g = 4

Find the ratio of George’s age to Carl’s age to Alex’s age

Ratio of George’s age to Carl’s age to Alex’s age = g : c : a

g : c : a = 4 : 48 : 16

Divide the above ratio by 4 to reduce to lowest terms

g : c : a - 1 : 12 : 4

Thus ratio of George’s age to Carl’s age to Alex’s age is 1 : 12 : 4

6 0

4 years ago

Evaluate the following

IRINA_888 [86]

(a) $[\frac{9}{2.6} - \frac{2.5^{2} }{2.5} ]^{2}$

Answer:

$[\frac{9}{2.6} - \frac{2.5^{2} }{2.5} ]^{2}$

= $[\frac{9}{2.6} - \frac{2.5*2.5 }{2.5} ]^{2}$

= $[\frac{9}{2.6} - \frac{2.5}{1} ]^{2}$

*canceling 2.5 in numerator and denominator*

$= [\frac{9-(2.5)(2.6)}{2.6} ]^2\\*Using L.C.M of 2.6 and 1 which comes out to be '2.6'= [\frac{9-(6.5)}{2.6} ]^2\\= [\frac{2.5}{2.6} ]^2\\*multiplying and dividing by '10'= [\frac{2.5*10}{2.6*10} ]^2\\= [\frac{25}{26} ]^2\\= \frac{25^2}{26^2}\\= \frac{625}{676}\\= 0.925$

Properties used:

Cancellation property of fractions

Least Common Multiplier(LCM)

The least or smallest common multiple of any two or more given natural numbers are termed as LCM. For example, LCM of 10, 15, and 20 is 60.

(b) $[[\frac{3x^{a}y^{b}} {-3x^{a} y^{b} } ]^{3} ] ^{2}$

Answer:

$[[\frac{3x^{a}y^{b}} {-3x^{a} y^{b} } ]^{3}] ^{2}\\$

*using $[x^{a}]^b = x^{ab}$ *

$= [\frac{3x^{3a}y^{3b}} {-3x^{3a} y^{3b} }] ^{2}$

*Again, using $[x^{a}]^b = x^{ab}$ *

$= \frac{3x^{2*3a}y^{2*3b}} {-3x^{2*3a} y^{2*3b} } \\= (-1)\frac{3x^{6a}y^{6b}} {3x^{6a} y^{6b} }\\[\tex]*taking -1 common, denominator and numerator are equal*[tex]= -(1)\frac{1}{1}\\= -1$