Which of the following numbers doesn't belong? 64 16 36 32 8 4

I need help on 5 - 11

nalin [4]

5-11.

Positive 5 minus negative 11=

11-5=6

Now is 11 greater or 5, 11 is. And what is the integer of 11? A negative.

Therefore, the answer is -6.

4 0

2 years ago

An Item has a listed price of $30. If the sales tax rate is 5%, how much is the sales tax (in doll

DochEvi [55]

Answer:

sales tax is 1.5

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Match the numerical expressions to their simplest forms.

Aloiza [94]

Answer:

$(a^6b^1^2)^\frac{1}{3}$ = $a^2b^4$

$\frac{(a^5b^3)^\frac{1}{2}}{(ab)^-^\frac{1}{2}}$ = $a^3b^2$

$(\frac{a^5}{a^-^3b^-^4})^\frac{1}{4}$ = $a^2b$

$(\frac{a^3}{ab^-^6})^\frac{1}{2}$ = $ab^3$

Step-by-step explanation:

Simplify each of the expressions:

1

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

Distribute the exponent. Multiply the exponent of the term outside of the parenthesis by the exponents of the variable.

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

$a^6^*^\frac{1}{3}b^1^2^*^\frac{1}{3}$

Simplify,

$a^2b^4$

2

Use a similar technique to solve this problem. Remember, a fractional exponent is the same as a radical, if the denominator is (2), then the operation is taking the square root of the number.

$\frac{(a^5b^3)^\frac{1}{2}}{(ab)^-^\frac{1}{2}}$

Rewrite as square roots:

$\frac{\sqrt{a^5b^3}}{\sqrt{(ab)}^-^1}$

A negative exponent indicates one needs to take the reciprocal of the number. Apply this here:

$\frac{\sqrt{a^5b^3}}{\frac{1}{\sqrt{ab}}}$

Simplify,

$\sqrt{a^5b^3}*\sqrt{ab}$

Since both numbers are under a radical, one can rewrite them such that they are under the same radical,

$\sqrt{a^5b^3*ab}$

Simplify,

$\sqrt{a^6b^4}$

Since this operation is taking the square root, divide the exponents in half to do this operation:

$a^3b^2$

3

$(\frac{a^5}{a^-^3b^-^4})^\frac{1}{4}$

Simplify, to simplify the expression in the numerator and the denominator, the base must be the same. Remember, the base is the number that is being raised to the exponent. One subtracts the exponent of the number in the denominator from the exponent of the like base in the numerator. This only works if all terms in both the numerator and the denominator have the operation of multiplication between them:

$(\frac{a^8}{b^-^4})^\frac{1}{4}$

Bring the negative exponent to the numerator. Change the sign of the exponent and rewrite it in the numerator,

$(a^8b^4)^\frac{1}{4}$

This expression to the power of the one forth. This is the same as taking the quartic root of the expression. Rewrite the expression with such,

$\sqrt[4]{a^8b^4}$

SImplify, divide the exponents by (4) to simulate taking the quartic root,

$a^2b$

4

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

Using all of the rules mentioned above, simplify the fraction. The only operation happening between the numbers in both the numerator and the denominator is multiplication. Therefore, one can subtract the exponents of the terms with the like base. The term in the denomaintor can be rewritten in the numerator with its exponent times negative (1).

$(a^3^-^1b^(^-^6^*^(^-^1^)^))^\frac{1}{2}$

$(a^2b^6)^\frac{1}{2}$

Rewrite to the half-power as a square root,

$\sqrt{a^2b^6}$

Simplify, divide all of the exponents by (2),

$ab^3$

7 0

3 years ago

Fine the area of the shaped trapezium in the rectangle below

Ksenya-84 [330]

At first find the area of rectangle:
(x + 4)*(x + 1) = x^2 + 5x + 4
Then find the area of triangle:
$\frac{(x+1)(x-3)}{2} = \frac{x^2-2x-3}{2}$
Then you must do like this
Area of rectangle - Area of triangle = Area of the shaded
x^2+5x+4 - ( $\frac{x^2-2x-3}{2}$ )