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emmasim [6.3K]
3 years ago
5

(solve the following problem using integers. show ALL your work.)

Mathematics
1 answer:
yulyashka [42]3 years ago
8 0

Answer:

$4

Step-by-step explanation:

30+20=50

50-48=2

2+20+=22

22-18=4

if she has 30 then gets 20 she has 50. Then she spends 48 dollars leaving her with only 2 dollars. She then makes another 20 dollars so she now has 22 dollars. She spends 18 dollars leaving her with 4 dollars.

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find the value of the polygon. the polygon has 6 interior angles 2 of them are y and 4 of them are (2Y-20)
Anna35 [415]
First we need to determine what the 6 angles must add to. Turns out we use this formula

S = 180(n-2)

where S is the sum of the angles (result of adding them all up) and n is the number of sides. In this case, n = 6. So let's plug that in to get

S = 180(n-2)
S = 180(6-2)
S = 180(4)
S = 720

The six angles, whatever they are individually, add to 720 degrees. The six angles are y, y, 2y-20, 2y-20,  2y-20,  2y-20, <span> 

They add up and must be equal to 720, so let's set up the equation to get...

(y)+(y)+(</span>2y-20)+(2y-20)+(2y-20)+(<span>2y-20) = 720

Let's solve for y

</span>y+y+2y-20+2y-20+2y-20+2y-20 = 720

10y-80 = 720

10y-80+80 = 720+80

<span>10y = 800
</span>
10y/10 = 800/10

y = 80

Now that we know the value of y, we can figure out the six angles

angle1 = y = 80 degrees
<span>angle2 = y = 80 degrees
</span><span>angle3 = 2y-20 = 2*80-20 = 140 degrees
</span>angle4 = 2y-20 = 2*80-20 =<span> 140 degrees
</span><span>angle5 = 2y-20 = 2*80-20 = 140 degrees
</span>angle6 = 2y-20 = 2*80-20 =<span> 140 degrees
</span>
and that's all there is to it
3 0
3 years ago
An 8.0 kg block is moving at 3.2 m/s. A net force of 10 N is constantly applied on the block in the direction of its movement, u
Softa [21]

Answer:

5.02 m/s

Step-by-step explanation:

We are given that

Mass of block,m=8 kg

Initial velocity,u=3.2 m/s

Net force ,F=10 N

Distance,s=6 m

We have to find the approximate final velocity of the block.

We know that a=\frac{F}{m}

Using the formula

a=\frac{10}{8}=\frac{5}{4} m/s^2

We know that

v=\sqrt{u^2+2as}

Using the formula

v=\sqrt{(3.2)^2+2\times \frac{5}{4}\times 6}

v=5.02 m/s

8 0
3 years ago
can some one also please help my one that i should do after i finds the area of the square aka how to find the area of the trian
Andre45 [30]

Answer:

6 square units

Step-by-step explanation:

Given shape is of a trapezoid:

area \: of \: trapezoid  \\ =  \frac{1}{2}  \times (4 + 2) \times 2 \\  \\  = 6 \:  {units}^{2}

Area of shape = area of square + area of triangle

= 2^2 + \frac{1}{2} \times 2\times 2\\= 4 + 2\\= 6\: units^2 \\

4 0
3 years ago
Suppose you have developed a scale that indicates the brightness of sunlight. Each category in the table is 4 times brighter tha
raketka [301]

According to the developed scale, a radiant day is <u>16 times</u> brighter than a dim day.

We assume the brightness of a dim day to be x.

According to the developed scale, the brightness of an illuminated day will be 4 times that of a dim day.

Thus, the brightness of an illuminated day = 4*the brightness of a dim day = 4x.

According to the developed scale, the brightness of a radiant day will be 4 times that of an illuminated day.

Thus, the brightness of a radiant day = 4*the brightness of an illuminated day = 4*4x = 16x.

Now, the ratio of the brightness of a radiant day to the brightness of a dim day = 16x:x = 16x/x = 16:1.

Thus, according to the developed scale, a radiant day is <u>16 times</u> brighter than a dim day.

Learn more about the developed scale at

brainly.com/question/4970963

#SPJ1

The question provided is incomplete. The complete question is:

"Suppose you have developed a scale that indicates the brightness of sunlight. Each category in the table is 4 times brighter than the next lower category. For example, a day that is dazzling is 4 times brighter than a day that is radiant. How many times brighter is a radiant day than a dim day?

Dim=2

Illuminated=3

Radiant=4

Dazzling=5"

8 0
1 year ago
Which fraction does point P represent on the number line below?
yarga [219]
The answer is C because it is at the first of four intervals between 0 and 1           
8 0
3 years ago
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