Answer asap

Mathematics

1 answer:

docker41 [41]3 years ago

5 0

<h2>˜”*°•.˜”*°• Question •°*”˜.•°*”˜</h2>

<em>The library had 102 shelves of books. Each shelf had 14 books. Librarians checked out 807 books to students. how many books does the library have remaining?</em>

<em />

<h2>˜”*°•.˜”*°• Answer •°*”˜.•°*”˜</h2>

627 Books

<h2> </h2><h2>˜”*°•.˜”*°• Explanation •°*”˜.•°*”˜</h2>

Let's go through this together.

First, let's figure out how many books are in the library in all

We'll multiply s times b

- S being the shelves

- B being the books per shelf.

102(14) = 1,428

So, there are 1,428 books in the library.

We also know that 801 of the 1,428 books were checked out.

Let's subtract the total number of books by the amount checked out.

1,428 - 801 = 627

Therefore, there are 627 books still in the library after checkout.

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The number of 8th graders in choir is three times the number of the 7th graders. if there are 48 8th graders choir how.many 7th

stira [4]

Divide 48 by 3, for every three 8th graders, there is one 7th grader in choir
48÷3=16
There's 16 7th graders in choir !!

7 0

3 years ago

Phillip claimed that the expression − p + 5 + p is positive for any value of p. Determine whether Phillip’s statement is always

DIA [1.3K]

Answer:

always true.

the expression will always be equal to +5 because -p+p is zero

at any given value to p, the value is constantly +5

8 0

3 years ago

Read 2 more answers

I will mark brainliest IF CORRECT

umka2103 [35]

Answer:

1 - (π/6) - (√3/4)

Step-by-step explanation:

First, calculate the area of ABCD. That's just $1*1 = 1 un^{2}$

Then, we are told that arcs BD and AC are circular, meaning they are portions of a circle. Therefore, let's redraw this image a little. Look below for the image.

Essentially, what I've done is created an equilateral triangle and two sectors. You may be wondering why the triangle is equilateral. Well, we firstly know it's isosceles because the two original arcs are circular and congruent. Their overlap creates two congruent lines that can be extend from point M to the vertices A and D. These two lines also happen to be radii of the quarter circles because the two sectors have AM AND AD as radii. AM and AD = 1, so AMD is an equilateral triangle.

Draw altitude OD. This is a right angle. Since equilateral triangles are also isosceles triangles, and isosceles triangles have the property of sharing the median to base and altitude to base, we can say that OD = $\frac{1}{2}$ AD = $\frac{1}{2}$ . MD = 1, so this triangle is a 30-60-90 triangle. Why? This is because of the 30º angle converse theorem. The shorter leg is half of the hypotenuse MD, so <OMD is a 30º angle, making <ODM a 60º angle and inevitably <MDC a 30º angle (because <D is a right angle) and <ABM as well because the two sectors are congruent.

Because m<ABM = m<MDC = 30º, these two sectors are 30/360 of the individual circles or 1/12 of the original circles, which have areas of π (because 1^2*π = π). Each is 1/12π, combined they are 1/6π.

The area of the equilateral triangle is √3/8. This is because MO can be calculated, using the 30-60-90 ratios, to be √3/2. (1 * √3/2)/2 = (√3/2)/2 = √3/4; this uses the area of a triangle. Now, time to calculate the shaded area.

The area of the square is 1, the area of the sectors π/6, and the equilateral triangle √3/4. Subtract the equilateral triangle and sectors' areas from the area of the square to be left with the remaining shaded part. 1-(π/6)-(√3/4) is your final answer. To put that a little more cleanly below in the image.

Hope this helps, have a great day.