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4 years ago

10

The math class is 45 minutes long. If announcements are 5 minutes long, the warm up is 7 minutes long and the notes take 11 minu

tes to complete, how long do you have in class for individual work?

1 answer:

atroni [7]4 years ago

7 0

Answer:

22?

Step-by-step explanation:

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What is the area of this figure?

Elanso [62]

First of all we can draw a parallel line to divide the figure into a triangle and a rectangle as shown in the figure. To find the area of our rectangle, remember that the area of a rectangle is length times width, so $A_{r} =lw$ . Since we know for our figure that the length and width of our rectangle are 13cm and 6cm respectively, lets replace those values in our formula to get its area:
$A _{r} =(13cm)(6cm)$
$A=78cm^{2}$
Similarly, the area of a triangle is one half times base times height, so $At=( \frac{1}{2})bh$ . Since we know that our base is 8cm and our height 6cm, lets replace those values in our equation to find the area of our triangle:
$A_{t} =( \frac{1}{2})(8cm)(6cm)$
$A_{t} =24cm^{2}$

Now the only thing left is add our areas:
$A_{total} =78cm^{2}+24cm^{2} =102cm^{2}$

We can conclude that the correct answer is <span>A. 102</span>

4 0

3 years ago

How many different linear arrangements are there of the letters a, b,c, d, e for which: (a a is last in line? (b a is before d?

inna [77]

A) Since a is last in line, we can disregard a, and concentrate on the remaining letters.
Let's start by drawing out a representation:

_ _ _ _ a

Since the other letters don't matter, then the number of ways simply becomes 4! = 24 ways

b) Since a is before d, we need to account for all of the possible cases.

Case 1: a d _ _ _
Case 2: a _ d _ _
Case 3: a _ _ d _
Case 4: a _ _ _ d

Let's start with case 1.
Since there are four different arrangements they can make, we also need to account for the remaining 4 letters.
$\text{Case 1: } 4 \cdot 4!$

Now, for case 2:
Let's group the three terms together. They can appear in: 3 spaces.
$\text{Case 2: } 3 \cdot 4!$

Case 3:
Exactly, the same process. Account for how many times this can happen, and multiply by 4!, since there are no specifics for the remaining letters.
$\text{Case 3: } 2 \cdot 4!$

$\text{Case 4: } 1 \cdot 4!$

$\text{Total arrangements}: 4 \cdot 4! + 3 \cdot 4! + 2 \cdot 4! + 1 \cdot 4! = 240$

c) Let's start by dealing with the restrictions.
By visually representing it, then we can see some obvious patterns.

a b c _ _

We know that this isn't the only arrangement that they can make.
From the previous question, we know that they can also sit in these positions:

_ a b c _
_ _ a b c

So, we have three possible arrangements. Now, we can say:
a c b _ _ or c a b _ _
and they are together.

In fact, they can swap in 3! ways. Thus, we need to account for these extra 3! and 2! (since the d and e can swap as well).

$\text{Total arrangements: } 3 \cdot 3! \cdot 2! = 36$

7 0

3 years ago

Dwayne is trying to find the fraction represented by the repeating decimal 0.5. He completes the three steps shown:

Studentka2010 [4]

Answer:

<h2>C. 5/9</h2><h2 />

Step-by-step explanation:

10x = 5.5

<u> -x = 0.5</u>

9x = 5

x = 5/9

therefore, the answer is C. 5/9

8 0

3 years ago

What the vertex of the graph of y=-x^2

Neporo4naja [7]

First let's define vertex: A point on the curve with a local minimum or maximum of curvature. If we look for the minimum and maximum value of the equation y=x^2+5: minimum value X=0, substitute in the equation to get the maximum value of Y y = 0^2 + 5y = 0 + 5y= 5 so the ordered pair is (0,5) Hope That Helped =D

3 0

3 years ago

Read 2 more answers

1,200 and 11 and you will get 13,200

VLD [36.1K]

Answer:

That is true!

$1,200 * 11 = 13,200$

Let me know if you need more help with multiplication. I may be able to help.

Have a great day.

7 0

2 years ago