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EleoNora [17]
3 years ago
15

The figure consists of a quarter circle and a parallelogram. What is the area of the composite figure? Use 3.14 for pi. Round to

the nearest whole number. please answer as soon as poossible
70 in
84 in
154 in
224 in

Mathematics
2 answers:
Andru [333]3 years ago
7 0
The area of a full circle is given by
  A = π*r^2
so the area of a 1/4 circle will be 1/4 of that:
  A = (π/4)*r^2

The area of a parallelogram is the product of base length and height:
  A = bh

Your composite figure will have an area that is the sum of the areas of the parts.
  A = (quarter circle area) + (parallelogram area)
  A ≈ (3.14/4)*(14 in)^2 +(14 in)*(5 in)
  A ≈ 223.86 in^2 ≈ 224 in^2

The most appropriate selection is the last one,
  224 in^2
Andreyy893 years ago
6 0

Answer:

the answer is d

Step-by-step explanation:

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Select the correct answer. What is the domain of the function f(x) = 4x − 16? all real numbers all positive real numbers all pos
bagirrra123 [75]
Hello!

Since the function has no undefined points or domain constants the answer is all real numbers

The answer is all real numbers

Hope this helps!
8 0
3 years ago
Read 2 more answers
Let u= <4,3>. Find the unit vector in the direction of u, and write your answer in component form.
aksik [14]
Take the vector u = <ux, uy> = <4, 3>.

Find the magnitude of u:

||u|| = sqrt[ (ux)^2 + (uy)^2]

||u|| = sqrt[ 4^2 + 3^2 ]

||u|| = sqrt[ 16 + 9 ]

||u|| = sqrt[ 25 ]

||u|| = 5

To find the unit vector in the direction of u, and also with the same sign, just divide each coordinate of u by ||u||. So the vector you are looking for is

u/||u||

u * (1/||u||)

= <4, 3> * (1/5)

= <4/5, 3/5>

and there it is.

Writing it in component form:

= (4/5) * i + (3/5) * j

I hope this helps. =)
3 0
3 years ago
Will give brainlest if right! Two plus two?
Temka [501]
4??? fish??? if it's a trick question, then the answer is probably fish :o)
4 0
3 years ago
Read 2 more answers
I have nooooo clue, please help
Aleksandr-060686 [28]

Answer:

case a) x^{2}=3y ----> open up

case b) x^{2}=-10y ----> open down

case c) y^{2}=-2x ----> open left

case d) y^{2}=6x ----> open right

Step-by-step explanation:

we know that

1) The general equation of a vertical parabola is equal to

y=a(x-h)^{2}+k

where

a is a coefficient

(h,k) is the vertex

If a>0 ----> the parabola open upward and the vertex is a minimum

If a<0 ----> the parabola open downward and the vertex is a maximum

2) The general equation of a horizontal parabola is equal to

x=a(y-k)^{2}+h

where

a is a coefficient

(h,k) is the vertex

If a>0 ----> the parabola open to the right

If a<0 ----> the parabola open to the left

Verify each case

case a) we have

x^{2}=3y

so

y=(1/3)x^{2}

a=(1/3)

so

a>0

therefore

The parabola open up

case b) we have

x^{2}=-10y

so

y=-(1/10)x^{2}

a=-(1/10)

a

therefore

The parabola open down

case c) we have

y^{2}=-2x

so

x=-(1/2)y^{2}

a=-(1/2)

a

therefore

The parabola open to the left

case d) we have

y^{2}=6x

so

x=(1/6)y^{2}

a=(1/6)

a>0

therefore

The parabola open to the right

3 0
3 years ago
This is math please help !!!!!!!
aev [14]

Answer:

Hello I'lll solve this problem.

Step-by-step explanation:

So the answer is actually C. I did the work.

3 0
3 years ago
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