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3 years ago

Which place value is 1/10 the value of the ones place?

Mathematics

1 answer:

marishachu [46]3 years ago

4 0

Answer:

the tenth place (0.1)

Step-by-step explanation:

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Choose the expressions that are equivalent to m∠AOB. Select all that apply.

Tanzania [10]

Answer:

A, B, D, G, and H should be correct

Step-by-step explanation:

I've done the same problem and those were the answers shown

6 0

3 years ago

Read 2 more answers

John borrowed $ from bank for 3 years at a rate of 10%

garik1379 [7]

<h3>Given:</h3>

P= $12500
R= 10%
T= 3 years

P= Principal amount
R= Rate of interest
T= Time period

The simple interest
The total amount paid

<h3>Solution:</h3>

$\large\boxed{Formula: I= \frac{PRT}{100}}$

First, we'll have to multiply, principal amount (12500), rate (10) and time period (3).

$12500×10×3$

$=375000$

Now, we'll have to divide the amount (375000) by 100.

$\frac{375000}{100}$

I=$3750

Now, we can find the total amount paid.

$\large\boxed{A=SI+P}$

Let's substitute according to the formula.

$A=3750+12500$

A=$16250

Therefore, simple interest is $3750 and $16250 was paid in total.

5 0

1 year ago

Calcule a segunda derivada )a y=√x

serious [3.7K]

The answer to this question would be x=a2y2

6 0

3 years ago

There are 180 cars in the parking lot 30% are red cars how many red cards are there

Yuri [45]

There are 54 red cars

7 0

2 years ago

Read 2 more answers

If log2 5 = k, determine an expression for log32 5 in terms of k.

lukranit [14]

Answer:

$log_3_2(5)=\frac{1}{5} k$

Step-by-step explanation:

Let's start by using change of base property:

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

So, for $log_2(5)$

$log_2(5)=k=\frac{log(5)}{log(2)}\hspace{10}(1)$

Now, using change of base for $log_3_2(5)$

$log_3_2(5)=\frac{log(5)}{log(32)}$

You can express $32$ as:

$2^5$

Using reduction of power property:

$log_z(x^y)=ylog_z(x)$

$log(32)=log(2^5)=5log(2)$

Therefore:

$log_3_2(5)=\frac{log(5)}{5*log(2)}=\frac{1}{5} \frac{log(5)}{log(2)}\hspace{10}(2)$

As you can see the only difference between (1) and (2) is the coefficient $\frac{1}{5}$ :

So:

$\frac{log(5)}{log(2)} =k\\$

$log_3_2(5)=\frac{1}{5} \frac{log(5)}{log(2)} =\frac{1}{5} k$

6 0

3 years ago