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3 years ago

Maggie's brother is 6 years younger than four times her age. The sum of their ages is 69.

Mathematics

2 answers:

Tomtit [17]3 years ago

6 0

Answer:

Step-by-step explanation:

If Maggies age is, for example, x, we can say that her brothers age is 4x-6. We know from the problem that Maggies age (x) plus her bothers age, (4x-6) is 69. So,

4x-6+x=69

5x-6=69

5x=75

x=15

So, Maggie is 15.

Airida [17]3 years ago

5 0

Answer: Maggie is 15 years old

Step-by-step explanation:

69 = (4m - 6) + m --- get an equation

69 - m = 4m - 6 --- lets subtract m from both sides first

75 - m = 4m --- add 6 on both sides

75 = 5m --- add m on both sides

15 = m --- divide by 5 on both sides

m = 15, so Maggie is 15 years old

Let's check:

69 = (4(15)-6) +15

69 = (60 - 6) + 15

69 = 54 + 15

69 = 69

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Rewrite the expression with a rational exponent as a radical expression.

Nuetrik [128]

Answer:

Rewriting the expression $(3^\frac{2}{3})^\frac{1}{6}$ with a rational exponent as a radical expression we get $\mathbf{\sqrt[9]{3} }$

Step-by-step explanation:

We need to rewrite the expression $(3^\frac{2}{3})^\frac{1}{6}$ with a rational exponent as a radical expression.

The expression given is:

$(3^\frac{2}{3})^\frac{1}{6}$

First we will simply the expression using exponent rule $(a^m)^n=a^{mn}$

$(3^\frac{2}{3})^\frac{1}{6}\\=(3^\frac{2}{18})$

As we know 2 and 18 are both divisible by 2, we can write

$=(3^\frac{1}{9})$

Now we know that $a^\frac{1}{9}=\sqrt[9]{a}$

Using this we get

$=\sqrt[9]{3}$

So, rewriting the expression $(3^\frac{2}{3})^\frac{1}{6}$ with a rational exponent as a radical expression we get $\mathbf{\sqrt[9]{3} }$

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3 years ago

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DaniilM [7]

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3 years ago

Evaluate 12- (16) + (-8) + (-6) - 2 - (10) + (-5)

Murrr4er [49]

Answer:

$-35$

Step-by-step explanation:

Follow PEMDAS order of operations:

Parentheses: ( )

Exponents: a²

Multiplication: ×

Division: ÷

Addition: +

Subtraction: -

$12- (16) + (-8) + (-6) - 2 - (10) + (-5)$

Add and subtract:

$12-\left(16\right)=-4$

$-4+\left(-8\right)+\left(-6\right)-2-\left(10\right)+\left(-5\right)$

Apply rule: $+\left(-a\right)\:=\:-a$

$-4-8+\left(-6\right)-2-\left(10\right)+\left(-5\right)$

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$-12+\left(-6\right)-2-\left(10\right)+\left(-5\right)$

Rule: $+\left(-a\right)\:=\:-a$

$+\left(-6\right)=-6$

$-12-6-2-\left(10\right)+\left(-5\right)$

$-12-6=-18$

$-18-2-\left(10\right)+\left(-5\right)$

$-18-2=-20$

$-20-\left(10\right)+\left(-5\right)$

$-20-\left(10\right)=-30$

$-30+\left(-5\right)$

Apply rule: $+\left(-a\right)\:=\:-a$

$+\left(-5\right)=-5$

$-30-5$

$-35$

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Mashutka [201]

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