All categories

2 years ago

This model is composed of two shapes. What is the area of this figure?

Mathematics

2 answers:

mote1985 [20]2 years ago

7 0

Measurements of 30 I believe yw

pav-90 [236]2 years ago

7 0

Answer:

69,27 cm^2

Step-by-step explanation:

area of rectangle: 10 x 3 = 30 cm^2

area of the semicircle: (5^2π)/2 ≃ 39,27 cm^2

total area = 30 + 39,27 ≃ 69,27 cm^2

You might be interested in

10 points

Ket [755]

Answer:

9/50 or 0.18

Step-by-step explanation:

7 0

3 years ago

What are the domain and range of the function

atroni [7]

Answer:

Domain (-∞,∞)

Range (0, ∞)

Step-by-step explanation:

The domain of a function is always the set of all real numbers therefore the domain is going to be (-∞,∞). The range is (0, ∞) because 0 is one of the outputs for the function

3 0

3 years ago

Can you answer this in picture form, please?

babunello [35]

Answer:

It may be Arrow diagram yes or not

8 0

2 years ago

Read 2 more answers

Given that log 2 = 0.3010 and log 3 = 0.4771 , how can we find log 6 ?

bearhunter [10]

Answer:

$\sf \log_{10}6=0.7781$

Step-by-step explanation:

<u>Given</u>:

$\sf \log_{10} 2 = 0.3010$

$\sf \log_{10} 3 = 0.4771$

To find log₁₀ 6, first rewrite 6 as 3 · 2:

$\sf \implies \log_{10}6=\log_{10}(3 \cdot 2)$

$\textsf{Apply the log product law}: \quad \log_axy=\log_ax + \log_ay$

$\implies \sf \log_{10}(3 \cdot 2)=\log_{10}3+\log_{10}2$

Substituting the given values for log₁₀ 3 and log₁₀ 2:

$\begin{aligned} \sf \implies \log_{10}3+\log_{10}2 & = \sf 0.4771+0.3010\\ & = \sf 0.7781 \end{aligned}$

Therefore:

$\sf \log_{10}6=0.7781$

Learn more about logarithm laws here:

brainly.com/question/27953288

brainly.com/question/27963321

5 0

1 year ago

Read 2 more answers

How would I find a? What formula would I use?

xenn [34]

Answer:

You can use either of the following to find "a":

Pythagorean theorem
Law of Cosines

Step-by-step explanation:

It looks like you have an isosceles trapezoid with one base 12.6 ft and a height of 15 ft.

I find it reasonably convenient to find the length of x using the sine of the 70° angle:

x = (15 ft)/sin(70°)

x ≈ 15.96 ft

That is not what you asked, but this value is sufficiently different from what is marked on your diagram, that I thought it might be helpful.

Consider the diagram below. The relation between DE and AE can be written as ...

DE/AE = tan(70°)

AE = DE/tan(70°) = DE·tan(20°)

AE = 15·tan(20°) ≈ 5.459554

Then the length EC is ...

EC = AC - AE

EC = 6.3 - DE·tan(20°) ≈ 0.840446

Now, we can find DC using the Pythagorean theorem:

DC² = DE² + EC²

DC = √(15² +0.840446²) ≈ 15.023527

a ≈ 15.02 ft

_____

You can also make use of the Law of Cosines and the lengths x=AD and AC to find "a". (Do not round intermediate values from calculations.)

DC² = AD² + AC² - 2·AD·AC·cos(A)

a² = x² +6.3² -2·6.3x·cos(70°) ≈ 225.70635

a = √225.70635 ≈ 15.0235 . . . feet

3 0

3 years ago