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3 years ago

What is 120.70 rounded to the nearest tenth

Mathematics

2 answers:

vodomira [7]3 years ago

7 0

120.7 is 120.70 rounded to the nearest tenth.

Elan Coil [88]3 years ago

5 0

Since the tenth would be the first number from the right of the decimal, the correct answer would be 120.7.

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How to put percent into decimal

fiasKO [112]

Answer:

There are multiple ways to do so:

You can divide the number by 100 to convert it to a decimal

You could take your percentage and add a decimal two places to the left.

Examples of both:

For division its as simple as 75 divided by 100 = 0.75

If we are going to use the decimal method, lets say we have 45%, so we can do this:

45% =45.00

Move the decimal two places to the left.

0.45

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2 years ago

Which set of ordered pairs represents a function? {(0, 1), (1, 3), (1, 5), (2, 8)} {(0, 0), (1, 2), (2, 6), (2, 8)} {(0, 0), (0,

11111nata11111 [884]

Answer: {(0, 2), (1, 4), (2, 6), (3, 6)}

Step-by-step explanation:

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Since all the other sets of ordered pairs feature points with two x-values with different y-values, the set above is the only function of the provided options.

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2 years ago

What is the product? (-2x-9y^2)(-4x-3)

telo118 [61]

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Step-by-step explanation:

using the FOIL method

6 0

3 years ago

Read 2 more answers

PLEASE PLEASE ANSWER!

muminat

The area of the shaded region is $$8(\pi \ -\sqrt{3})\ \text{cm}^2$ .

Solution:

Given radius = 4 cm

Diameter = 2 × 4 = 8 cm

Let us first find the area of the semi-circle.

Area of the semi-circle = $\frac{1}{2}\times \pi r^2$

$$=\frac{1}{2}\times \pi\times 4^2$

$$=\frac{1}{2}\times \pi\times 16$

Area of the semi-circle = $$8\pi$ cm²

Angle in a semi-circle is always 90º.

∠C = 90°

So, ABC is a right angled triangle.

Using Pythagoras theorem, we can find base of the triangle.

$AC^2+BC^2=AB^2$

$AC^2+4^2=8^2$

$AC^2=64-16$

$AC^2=48$

$AC=4\sqrt{3}$ cm

Base of the triangle ABC = $4\sqrt{3}$ cm

Height of the triangle = 4 cm

Area of the triangle ABC = $\frac{1}{2}\times b \times h$

$$=\frac{1}{2}\times 4\sqrt{3} \times 4$

Area of the triangle ABC = $8\sqrt{3}$ cm²

Area of the shaded region

= Area of the semi-circle – Area of the triangle ABC

= $$8\pi \ \text{cm}^2-8\sqrt{3}\ \text{cm}^2$

= $$8(\pi \ -\sqrt{3})\ \text{cm}^2$

Hence the area of the shaded region is $$8(\pi \ -\sqrt{3})\ \text{cm}^2$ .

3 0

2 years ago

What is the square root of 10,000

Setler [38]

100 and -100 are the two solutions for the square root of 10,000

7 0

2 years ago

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