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sdas [7]
2 years ago
5

Macy bought some records. She spent a total of $45 after a discount of $11 off of her total purchase. Each record cost $8. How m

any records did she buy?
Select an equation that could be used to answer the question above. Let r represent the number of records.


8r - 11 = 45

45r - 11 = 8

11r + 8 = 45

8r + 45 = 11
Mathematics
1 answer:
Rzqust [24]2 years ago
7 0

Answer:

7 records. and the answer would be 8r-11=45

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Drako found an emerald in a cave at a depth between Negative one-half and Negative 1 and two-thirds meters. Which number could r
8090 [49]

Options :

Negative 2 and one-third meters

Negative three-fourths meters

Negative one-fourth meters

Negative 1 and

StartFraction 5 Over 6 EndFraction meters

Answer: Negative three-fourths meters

Step-by-step explanation:

Given the following :

Depth Location of Emerald :

Between - 1/2 and - 1 2/3

The answer will be the option which lies between the two depth points.

-1/2 = - 0.5 ( upper bound)

-1 2/3 = - 1.667 ( lower bound)

The option which lies inbetween :

- 1.667 and - 0.5

Taking each option :

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Negative three-fourths meters = - 3/4 = - 0.75

Negative one-fourth meters = - 1/4 = - 0.25

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Hence, the only depth position which lies inbetween (-1.667 and - 0.5) is - 0.75

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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:

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

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

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

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Property used: 'Power of a power'

We can raise a power to a power

(x^2)4=(x⋅x)⋅(x⋅x)⋅(x⋅x)⋅(x⋅x)=x^8

This is called the power of a power property and says that to find a power of a power you just have to multiply the exponents.

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