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3 years ago

Which two inequalities to compare |-5| and 4?

Mathematics

1 answer:

lilavasa [31]3 years ago

7 0

Answer:

|-5|>4

Step-by-step explanation:

Negative 5 in absolute value is 5 since any number in absolute value turns out positive or at the very least 0. 5 is greater than 4, so |-5|>4.

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Prove the divisibility:<br><br>45^45·15^15 by 75^30

garri49 [273]

Answer:

$3^{75}$ .

Step-by-step explanation:

We have been an division problem: $\frac{45^{45}*15^{15}}{75^{30}}$ .

We will simplify our division problem using rules of exponents.

Using product rule of exponents $(a*b)^n=a^n*b^n$ we can write:

$45^{45}=(9*5)^{45}=9^{45}*5^{45}$

$15^{15}=(3*5)^{15}=3^{15}*5^{15}$

$75^{30}=(15*5)^{30}=15^{30}*5^{30}$

Substituting these values in our division problem we will get,

$\frac{9^{45}*5^{45}*3^{15}*5^{15}}{15^{30}*5^{30}}$

Using power rule of exponents $a^n*a^m=a^{n+m}$ we will get,

$\frac{9^{45}*5^{(45+15)}*3^{15}}{15^{30}*5^{30}}$

$\frac{9^{45}*5^{60}*3^{15}}{15^{30}*5^{30}}$

Using product rule of exponents $(a*b)^n=a^n*b^n$ we will get,

$\frac{(3*3)^{45}*5^{60}*3^{15}}{(3*5)^{30}*5^{30}}$

$\frac{3^{45}*3^{45}*5^{60}*3^{15}}{3^{30}*5^{30}*5^{30}}$

Using power rule of exponents $a^n*a^m=a^{n+m}$ we will get,

$\frac{3^{(45+45+15)}*5^{60}}{3^{30}*5^{(30+30)}}$

$\frac{3^{105}*5^{60}}{3^{30}*5^{60}}$

$\frac{3^{105}}{3^{30}}$

Using quotient rule of exponent $\frac{a^m}{a^n}=a^{m-n}$ we will get,

$\frac{3^{105}}{3^{30}}=3^{105-30}$

$3^{105-30}=3^{75}$

Therefore, our resulting quotient will be $3^{75}$ .

7 0

3 years ago

Pls help is due for today pls

pentagon [3]

Answer:

Part A: The rate of change was greater in case of the graph.

Part B: The figure shows the graph of function A in Redline and Function B in blue line

Part C : Initial value of function B is greater than Function A.

Step-by-step explanation:

Here, Function A show graph of a line and Function show table show a set of value of x and y.

Part A). Which function show greater rate of change.

The slope of line is given by m = $\frac{Y2-Y1}{X2-X1}$

For function A:

Given graph passes through points (04) and (2,12)

The slope of line is m= $\frac{12-4}{2-0}$

= $\frac{8}{2}$

= 4

For function B:

Given the table has points (012) and (2,14)

The slope of line is m= $\frac{14-12}{2-0}$

= $\frac{2}{2}$

= 1

Therefore, The rate of change was greater in case of graph.

Part B). Plot the graph of a function B onto the graph of a function A

The figure shows the graph of function A in Redline and Function B in blueline

Line of function A is more tilted toward the y-axis than Line of function B.

Therefore, Rate of change of function A is greater than Function B.

Part C). Which function has greater initial value?

For function A:

From the graph of part B,

When x=0 y=4

For function B:

From the graph of part B,

When x=0 y=12

Therefore, Initial value of function B is greater than Function A.

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3 years ago

calvin and 4 of his friends want to share 4 pounds of nuts equally.write an expression to show the fraction of the nuts each fri

PilotLPTM [1.2K]

4/5

The 4 is how many pounds of nuts there are and dividing by 5 shows that you are sharing the nuts equally amongst the 5 friends.

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