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3 years ago

8

Each morning librarians place 560 books on a book shelf in the school library. Each student that visits library that day takes e

xactly 3 books from this shelf. Let s be the number of students which visit library on a particular day. Let b be the number of books left on the shelf at the end of that day.
c
How many books were left on the shelf the day 87 students visited the library?

1 answer:

Anarel [89]3 years ago

7 0

Answer:

the answer is 299 books

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Find the value of the combination. From a committee consisting of 8 men and 5 women, a subcommittee is formed consisting of 4 me

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3 years ago

Find the f^-1(x) and it’s domain

borishaifa [10]

Answer:

$f^{-1}(x) = (x + 8)^2$

$x \ge -8$

Step-by-step explanation:

Given

$f(x) = \sqrt x - 8$

Solving (a): $f^{-1}(x)$

We have:

$f(x) = \sqrt x - 8$

Express f(x) as y

$y = \sqrt x - 8$

Swap x and y

$x = \sqrt y - 8$

Add 8 to $both\ sides$

$x + 8 = \sqrt y - 8 + 8$

$x + 8 = \sqrt y$

Square both sides

$(x + 8)^2 = y$

Rewrite as:

$y = (x + 8)^2$

Express y as: $f^{-1}(x)$

$f^{-1}(x) = (x + 8)^2$

To determine the domain, we have:

The original function is $f(x) = \sqrt x - 8$

The range of this is: $f(x) \ge -8$

The $domain$ of the $inverse$ function is the $range$ of the $original$ function.

<em>Hence, the domain is:</em>

$x \ge -8$

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3 years ago

Please help me solve this problem ASAP

DiKsa [7]

$\bold{\huge{\blue{\underline{ Solution }}}}$

<h3><u>Given </u><u>:</u><u>-</u></h3>

<u>The </u><u>right </u><u>angled </u><u>below </u><u>is </u><u>formed </u><u>by </u><u>3</u><u> </u><u>squares </u><u>A</u><u>, </u><u> </u><u>B </u><u>and </u><u>C</u>
<u>The </u><u>area </u><u>of </u><u>square </u><u>B</u><u> </u><u>has </u><u>an </u><u>area </u><u>of </u><u>1</u><u>4</u><u>4</u><u> </u><u>inches </u><u>²</u>
<u>The </u><u>area </u><u>of </u><u>square </u><u>C </u><u>has </u><u>an </u><u>of </u><u>1</u><u>6</u><u>9</u><u> </u><u>inches </u><u>²</u>

<h3><u>To </u><u>Find </u><u>:</u><u>-</u></h3>

<u>We </u><u>have </u><u>to </u><u>find </u><u>the </u><u>area </u><u>of </u><u>square </u><u>A</u><u>? </u>

<h3><u>Let's </u><u>Begin </u><u>:</u><u>-</u><u> </u></h3>

The right angled triangle is formed by 3 squares

<u>We </u><u>have</u><u>, </u>

Area of square B is 144 inches²
Area of square C is 169 inches²

<u>We </u><u>know </u><u>that</u><u>, </u>

$\bold{ Area \: of \: square = Side × Side }$

Let the side of square B be x

<u>Subsitute </u><u>the </u><u>required </u><u>values</u><u>, </u>

$\sf{ 144 = x × x }$

$\sf{ 144 = x² }$

$\sf{ x = √144}$

$\bold{\red{ x = 12\: inches }}$

Thus, The dimension of square B is 12 inches

<h3><u>Now, </u></h3>

Area of square C = 169 inches

Let the side of square C be y

<u>Subsitute </u><u>the </u><u>required </u><u>values</u><u>, </u>

$\sf{ 169 = y × y }$

$\sf{ 169 = y² }$

$\sf{ y = √169}$

$\bold{\green{ y = 13\: inches }}$

Thus, The dimension of square C is 13 inches.

<h3><u>Now, </u></h3>

It is mentioned in the question that, the right angled triangle is formed by 3 squares

The dimensions of square be is x and y

Let the dimensions of square A be z

<h3><u>Therefore</u><u>, </u><u>By </u><u>using </u><u>Pythagoras </u><u>theorem</u><u>, </u></h3>

<u>The </u><u>sum </u><u>of </u><u>squares </u><u>of </u><u>base </u><u>and </u><u>perpendicular </u><u>height </u><u>equal </u><u>to </u><u>the </u><u>square </u><u>of </u><u>hypotenuse </u>

<u>That </u><u>is</u><u>, </u>

$\bold{\pink{ (Perpendicular)² + (Base)² = (Hypotenuse)² }}$

<u>Here</u><u>, </u>

Base = x = 12 inches
Perpendicular = z
Hypotenuse = y = 13 inches

<u>Subsitute </u><u>the </u><u>required </u><u>values</u><u>, </u>

$\sf{ (z)² + (x)² = (y)² }$

$\sf{ (z)² + (12)² = (169)² }$

$\sf{ (z)² + 144 = 169}$

$\sf{ (z)² = 169 - 144 }$

$\sf{ (z)² = 25}$

$\bold{\blue{ z = 5 }}$

Thus, The dimensions of square A is 5 inches

<h3><u>Therefore</u><u>,</u></h3>

Area of square

$\sf{ = Side × Side }$

$\sf{ = 5 × 5 }$

$\bold{\orange{ = 25\: inches }}$

Hence, The area of square A is 25 inches.

6 0

2 years ago