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3 years ago

Rename 120,000 = ten thousands

1 answer:

4 0

<span>The answer is twelve ten thousands. Let's find how many ten thousands are in 120,000. Ten thousand in a number form is 10,000. Now, divide 120,000 by 10,000 (ten thousand). 120,000 : 10,000 = 12. In other words, there are twelve ten thousands in 120,000.Hope this helps. Let me know if you need additional help!</span>

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100/32 can someone help me how much would this be in fraction

kap26 [50]

Your answer would be 3 and 1/8

7 0

2 years ago

A, B & C form the vertices of a triangle.

Savatey [412]

Answer:

BC = 17.2.

Step-by-step explanation:

Since CAB = 90º, this is a right triangle.

The triangle format is given below:

A B

We have that AB = 8.6, and angle B measures 60º.

Length of BC:

BC is the hypotenyse.

Side AB is adjacent to angle B = 60º.

In a right triangle, angle $\alpha$ , it's adjacent side with length s and the hypotenuse h are related by the cosine of the angle, that is:

$\cos{\alpha} = \frac{s}{h}$

In this question, we have an angle of 60º, with has cosine 0.5. We also have that side AB = s = 8.6, and the hypotenuse h is side BC. So

$\cos{\alpha} = \frac{s}{h}$

$0.5 = \frac{8.6}{h}$

$0.5h = 8.6$

$h = \frac{8.6}{0.5}$

$h = 17.2$

BC = 17.2.

4 0

2 years ago

Multiple choice, please help

Marina86 [1]

I think it A but I’m not 100% sure

5 0

3 years ago

Which is the simplified form of the expression ((2 Superscript negative 2 Baseline) (3 Superscript 4 Baseline)) Superscript nega

AlekseyPX

Answer:

The option "StartFraction 1 Over 3 Superscript 8" is correct

That is $\frac{1}{3^8}$ is correct answer

Therefore $[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2=\frac{1}{3^8}$

Step-by-step explanation:

Given expression is ((2 Superscript negative 2 Baseline) (3 Superscript 4 Baseline)) Superscript negative 3 Baseline times ((2 Superscript negative 3 Baseline) (3 squared)) squared

The given expression can be written as

$[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2$

To find the simplified form of the given expression :

$[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2$

$=(2^{-2})^{-3}(3^4)^{-3}\times (2^{-3})^2(3^2)^2$ ( using the property $(ab)^m=a^m.b^m$ )

$=(2^6)(3^{-12})\times (2^{-6})(3^4)$ ( using the property $(a^m)^n=a^{mn}$

$=(2^6)(2^{-6})(3^{-12})(3^4)$ ( combining the like powers )

$=2^{6-6}3^{-12+4}$ ( using the property $a^m.a^n=a^{m+n}$ )

$=2^03^{-8}$

$=\frac{1}{3^8}$ ( using the property $a^{-m}=\frac{1}{a^m}$ )

Therefore $[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2=\frac{1}{3^8}$

Therefore option "StartFraction 1 Over 3 Superscript 8" is correct

That is $\frac{1}{3^8}$ is correct answer

6 0

3 years ago

Read 2 more answers

The figures are similar. Find the missing length.

saw5 [17]

Step-by-step explanation:

<h2>Answer:-</h2><h3>Given ,</h3>

The figure is Similar.

Observation:-

Similar figures have similar sides. If we see carefully in smaller triangle, 3 has been added to each side and they are similar. We need to find y.

We have :-

5+3=8 as similar sides.

So, applying same algorithm,

$y + 3 = 5$

$y = 5 - 3$

$\boxed{ \tt{y = 2 \ cm}}$

is the answer.

Hope it helps :)

4 0

2 years ago