All categories

1 year ago

Parville surveys every 15th person in the

Mathematics

1 answer:

nlexa [21]1 year ago

3 0

Answer:

2677.4

Step-by-step explanation:

40161/15

You might be interested in

An amount of $700 was invested at 7% for 7 months. What is the interest? Round your

Levart [38]

Answer:

49 dollars

Step-by-step explanation:

700 times .07

7 0

3 years ago

Need help ASAP

joja [24]

Part (1) : The solution is $729$

Part (2): The solution is $\frac{1}{16 x^{8}}$

Part (3): The solution is $\frac{2 x^{2}}{3 y z^{7}}$

Explanation:

Part (1): The expression is $3^{2} \cdot3^{4}$

Applying the exponent rule, $a^{b} \cdot a^{c}=a^{b+c}$ , we get,

$3^{2} \cdot 3^{4}=3^{2+4}$

Adding the exponent, we get,

$3^{2} \cdot3^{4}=3^6=729$

Thus, the simplified value of the expression is $729$

Part (2): The expression is $\left(2 x^{2}\right)^{-4}$

Applying the exponent rule, $a^{-b}=\frac{1}{a^{b}}$ , we have,

$\left(2 x^{2}\right)^{-4}=\frac{1}{\left(2 x^{2}\right)^{4}}$

Simplifying the expression, we have,

$\frac{1}{2^4x^8}$

Thus, we have,

$\frac{1}{16 x^{8}}$

Thus, the value of the expression is $\frac{1}{16 x^{8}}$

Part (3): The expression is $\frac{2 x^{4} y^{-4} z^{-3}}{3 x^{2} y^{-3} z^{4}}$

Applying the exponent rule, $\frac{x^{a}}{x^{b}}=x^{a-b}$ , we have,

$\frac{2x^{4-2}y^{-4+3}z^{-3-4}}{3}$

Adding the powers, we get,

$\frac{2x^{2}y^{-1}z^{-7}}{3}$

Applying the exponent rule, $a^{-b}=\frac{1}{a^{b}}$ , we have,

$\frac{2 x^{2}}{3 y z^{7}}$

Thus, the value of the expression is $\frac{2 x^{2}}{3 y z^{7}}$

8 0

3 years ago

9>d, if d=3 <br> True or false

Strike441 [17]

Answer:

This is true. 9 is greater than 3.

4 0

2 years ago

Read 2 more answers

-12x - 12y = 4 3x + 3y = 0

Svetradugi [14.3K]

Answer:

no Solution

Step-by-step explanation:

-12x-12y=4\\ 3x+3y=0

12x-12y=4

add 12y to both sides

12x-12y+12y=4+12y

divid both sides by -12

\frac{-12x}{-12}=\frac{4}{-12}+\frac{12y}{-12}

simplfy

x=-\frac{1+3y}{3}

\mathrm{Substitute\:}x=-\frac{1+3y}{3}

\begin{bmatrix}3\left(-\frac{1+3y}{3}\right)+3y=0\end{bmatrix}

\begin{bmatrix}-1=0\end{bmatrix}

7 0

3 years ago

Quadrilateral EFGH has the vertices E (-1,5), F (-2,3), G (-9,7), H (-3,6). What quadrant would all the vertices of quadrilatera

LenaWriter [7]

Answer:

Quadrant III

Step-by-step explanation:

5 0

3 years ago