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3 years ago

What is 0.003is 1/10 of

1 answer:

6 0

0.003 is 1/10 of 0.03.

Just multiply the 0.003 by 10 and you'll get the result. The easiest way to multiply such numbers is to move the dot one position to the right (it would be 2 positions if multiplying by 100, 3 by 1000 etc.). You can do the same with dividing, just remember to move to the left then.

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the second longest bridge in the world, the tianjin grand bridge, is 373,000 feet long. Its length is 167,700 feet less than the

Mademuasel [1]

If something is 167,700 feet less than something then the value of the longer object would be the that much more than the length of that shorter object. In this case it would be 373,000 + 167,700 = 540,700 ft.

1 1 <------- Carried Values
37300
167700
+
_______
540700

4 0

2 years ago

Notebooks cost $1.39 each. What are the possible numbers of notebooks that can be purchased with $10

murzikaleks [220]

Answer:

Step-by-step explanation:

HOPE THIS HELPS

PLZ MARK BRAINLIEST

4 0

2 years ago

Read 2 more answers

FI have no points left please answer, FLIP THE IMAGE OVER PLS

Nezavi [6.7K]

Answer:

Step-by-step explanation:

b. 1000 messages: 30 dollars

1,725 messages: 51.75

c. 2,000

2,500

600

b. 37.5

I cant do the rest because they have graph stuff :)

8 0

2 years ago

The triangular-shaped bases of a prism are 24 cm in length and have a height of 35 cm. Each side is 37 cm. The height of the pri

Savatey [412]

Answer:

The surface area of Triangular base Prism = 3682 cm²

Step-by-step explanation:

Given in question as :

For a Triangular base prism ,

The base of prism (b) = 24 cm

The height of prism (l) = 29 cm

Each side length (s) = 37 cm

The height of base triangle ( h ) = 35 cm

Hence , we know that The surface area of Triangular base prism is

= (b×h) + (2×l×s) + (l×b)

= (24×35) + (2×29×37) + (29×24)

= (840) + (2146) + (696)

= 3682 cm²

Hence The surface area of Triangular base Prism is 3682 cm² Answer

3 0

2 years ago

Can someone solve this?

ladessa [460]

First question:

Recall that $\cos^2x+\sin^2x=1$ and $\sqrt{x^2}=|x|$ for all $x$ . So

$\sqrt{1-\cos^2x}=\sqrt{\sin^2x}=|\sin x|$

$\sqrt{1-\sin^2x}=\sqrt{\cos^2x}=|\cos x|$

For $0$ , we expect both $\cos x>0$ and $\sin x>0$ (i.e. the sine and cosine of any angle that lies in the first quadrant must be positive). By definition of absolute value, $|x|=x$ if $x>0$ .

So we have

$\dfrac{\sqrt{1-\cos^2x}}{\sin x}+\dfrac{\sqrt{1-\sin^2x}}{\cos x}=\dfrac{\sin x}{\sin x}+\dfrac{\cos x}{\cos x}=1+1=\boxed{2}$

making H the answer.

Second question:

C is always true, because the inequality reduces to x > y.

6 0

3 years ago