All categories

2 years ago

Can anyone help me with this please!!? I’ll mark you brainliest!!

Mathematics

1 answer:

djverab [1.8K]2 years ago

3 0

Answer:

each square is 5". the base is 4x4 squares- so 20 inches square. 20x20= 400sq inches.

the semicircle at the top is 20 inches in diameter, which makes the radius 10 inches.

pi× 10^2= 314.16.

duvide that by 2 to get a semicircle- 157.8

so, the total area is the 400+ 157.8, for a total area of 557.8 square inches

You might be interested in

Evaluate the expression when C=1/2 and D=1/12<br> 3(1/4+C²)+2D<br> Please hurry!

Mkey [24]

Answer is 1 1/12
3(1/4 + 1/2^2)+ (2x1/12)
3/4+1/4+2/12 (you get to this part by distributing 3 throughout the parentheses)
9/12+3/12+2/12= 1 1/12 hope this helps

4 0

3 years ago

Read 2 more answers

Ive been on this question for a while as homework<br> and im stuck<br> help?

iogann1982 [59]

Answer:

please can you indicate the question

7 0

3 years ago

CAN SOMEONE PLEASE HELP ME

Degger [83]

Answer:

38 degrees

Step-by-step explanation:

this is an isosceles triangle, meaning that the base angles are congruent.

the total triangle is equal to 180

so the equation would be: x=180 - (71+71)

evaluate:

x=180-142

x=38 degrees

Hope this helps :)

5 0

2 years ago

Write 2x+y=17 in slope-intercept form

Harlamova29_29 [7]

Answer:

y=-2x+17

Step-by-step explanation:

6 0

2 years ago

For what value of constant c is the function k(x) continuous at x = 0 if k =

nlexa [21]

The value of constant c for which the function k(x) is continuous is zero.

<h3>What is the limit of a function?</h3>

The limit of a function at a point k in its field is the value that the function approaches as its parameter approaches k.

To determine the value of constant c for which the function of k(x) is continuous, we take the limit of the parameter as follows:

$\mathbf{ \lim_{x \to 0^-} k(x) = \lim_{x \to 0^+} k(x) = 0 }$

$\mathbf{\implies \lim_{x \to 0 } \ \ \dfrac{sec \ x - 1}{x}= c }$

Provided that:

$\mathbf{\implies \lim_{x \to 0 } \ \ \dfrac{sec \ x - 1}{x}= \dfrac{0}{0} \ (form) }$

Using l'Hospital's rule:

$\mathbf{\implies \lim_{x \to 0} \ \ \dfrac{\dfrac{d}{dx}(sec \ x - 1)}{\dfrac{d}{dx}(x)}= \lim_{x \to 0} sec \ x \ tan \ x = 0}$

Therefore:

$\mathbf{\implies \lim_{x \to 0 } \ \ \dfrac{sec \ x - 1}{x}=0 }$

Hence; c = 0

Learn more about the limit of a function x here:

brainly.com/question/8131777

#SPJ1

5 0

2 years ago