All categories

3 years ago

All numbers that end in 5 or 0 are divisible by 5 and 10 True or False?! PLZ HELP

Mathematics

2 answers:

xxTIMURxx [149]3 years ago

5 0

All numbers that end in 5 are not divisible by 10. So this is false.

lawyer [7]3 years ago

4 0

False because 5 isnt divisible by 10 and 0 isnt divisible by anything but if you mean positive numbers with multiple digits than its true

You might be interested in

Estimate £19.65 x 395

vladimir2022 [97]

Answer:

£7761.75

Step-by-step explanation:

Calculator..?

5 0

3 years ago

Read 2 more answers

Djxjicjcjcicifiddiciicifididididjdjhhh

Sedbober [7]

Answer:

Step-by-step explanation:

Let the unknown number be = x

Then,

x + 16 = 34

=> x = 34 - 16

=> x = 18 (Ans)

7 0

3 years ago

Read 2 more answers

What is the value of

LiRa [457]

Answer:

Step-by-step explanation:

-8(17-12)

------------

-2(8-(-2)

= -8(5)

--------

-2(10)

= -40

----------

-20

4 0

3 years ago

What is the correct answer to the last box with a red x on it?

ahrayia [7]

Answer:

good question

did i ask shut man stop cheating

4 0

2 years ago

Help ASAP show work please thanksss!!!!

Llana [10]

Answer:

$\displaystyle log_\frac{1}{2}(64)=-6$

Step-by-step explanation:

<u>Properties of Logarithms</u>

We'll recall below the basic properties of logarithms:

$log_b(1) = 0$

Logarithm of the base:

$log_b(b) = 1$

Product rule:

$log_b(xy) = log_b(x) + log_b(y)$

Division rule:

$\displaystyle log_b(\frac{x}{y}) = log_b(x) - log_b(y)$

Power rule:

$log_b(x^n) = n\cdot log_b(x)$

Change of base:

$\displaystyle log_b(x) = \frac{ log_a(x)}{log_a(b)}$

Simplifying logarithms often requires the application of one or more of the above properties.

Simplify

$\displaystyle log_\frac{1}{2}(64)$

Factoring $64=2^6$ .

$\displaystyle log_\frac{1}{2}(64)=\displaystyle log_\frac{1}{2}(2^6)$

Applying the power rule:

$\displaystyle log_\frac{1}{2}(64)=6\cdot log_\frac{1}{2}(2)$

Since

$\displaystyle 2=(1/2)^{-1}$

$\displaystyle log_\frac{1}{2}(64)=6\cdot log_\frac{1}{2}((1/2)^{-1})$

Applying the power rule:

$\displaystyle log_\frac{1}{2}(64)=-6\cdot log_\frac{1}{2}(\frac{1}{2})$

Applying the logarithm of the base:

$\mathbf{\displaystyle log_\frac{1}{2}(64)=-6}$

5 0

2 years ago