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Readme [11.4K]
3 years ago
13

Simplify -2.8f+0.9f-12-4

Mathematics
2 answers:
bogdanovich [222]3 years ago
3 0

Answer:

  • -1.9f - 16

Step-by-step explanation:

  • -2.8f + 0.9f - 12 - 4 =
  • (-2.8 + 0.9)f - (12 + 4) =
  • -1.9f - 16
Cloud [144]3 years ago
3 0

Answer:

- 1.9f - 16

Step-by-step explanation:

1. Collect like terms.

( - 2.8f + 0.9f) + ( - 12 - 4)

2. simplify.

- 1.9f - 16

Therefor, the answer is, -1.9f - 16

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Simple way to see it is to multiply the value opposite of the axis you're flipping it over by -1. For example the point (-5,7) would become (5,7) when reflected over the y-axis, because the point is moving from the left to the right. This means the triangle would be (1,3) (5,3) and (5,7) when reflected over the y-axis. For reflection over the x-axis it would be (-1,-3) (-5,-3) and (-5,-7) because each point is moving down.
8 0
3 years ago
Can I get help solving this graph please?
Daniel [21]
see the attached figure with the letters

1) find m(x) in the interval A,B
A (0,100)  B(50,40) -------------- > p=(y2-y1(/(x2-x1)=(40-100)/(50-0)=-6/5
m=px+b---------- > 100=(-6/5)*0 +b------------- > b=100
mAB=(-6/5)x+100

2) find m(x) in the interval B,C
B(50,40)  C(100,100) -------------- > p=(y2-y1(/(x2-x1)=(100-40)/(100-50)=6/5
m=px+b---------- > 40=(6/5)*50 +b------------- > b=-20
mBC=(6/5)x-20

3) find n(x) in the interval A,B
A (0,0)  B(50,60) -------------- > p=(y2-y1(/(x2-x1)=(60)/(50)=6/5
n=px+b---------- > 0=(6/5)*0 +b------------- > b=0
nAB=(6/5)x

4) find n(x) in the interval B,C
B(50,60)  C(100,90) -------------- > p=(y2-y1(/(x2-x1)=(90-60)/(100-50)=3/5
n=px+b---------- > 60=(3/5)*50 +b------------- > b=30
nBC=(3/5)x+30

5) find h(x) = n(m(x)) in the interval A,B
mAB=(-6/5)x+100
nAB=(6/5)x
then 
n(m(x))=(6/5)*[(-6/5)x+100]=(-36/25)x+120
h(x)=(-36/25)x+120
find <span>h'(x) 
</span>h'(x)=-36/25=-1.44

6) find h(x) = n(m(x)) in the interval B,C
mBC=(6/5)x-20
nBC=(3/5)x+30
then 
n(m(x))=(3/5)*[(6/5)x-20]+30 =(18/25)x-12+30=(18/25)x+18
h(x)=(18/25)x+18
find h'(x) 
h'(x)=18/25=0.72 

for the interval (A,B) h'(x)=-1.44
for the interval (B,C) h'(x)= 0.72

<span> h'(x) = 1.44 ------------ > not exist</span>

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Sarah needs 10 feet of fabric for a project she is working on, but the store only sells the fabric in meters. One meter of fabri
Aleks [24]
<h2>Hello!</h2>

The answer is:

The fabric will cost $4.58

D.$4.58

<h2>Why?</h2>

Since it's a conversion exercise, we need to pay special attention to the conversion procedure.

From the statement we know that:

1ft=0.305m

So, if she needs 10 feet of fabric, we need to convert it to meters, and then, calculate its cost.

We have that:

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Now, we have that each meter of fabric cost $1.50, so, 3.05m of fabric will cost:

\frac{1m}{1.50(dollars)}=\frac{3.05m}{cost}\\\\cost=\frac{3.05m*1.50(dollars)}{1m}=4.575(dollars)=4.58(dollars)

Hence, we have that the cost of the fabric will be $4.58.

The correct option is: D.$4.58​

Have a nice day!

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Answer:

78º

Step-by-step explanation:

<u>Note</u> : -

In mathematics, the reference angle is defined as the acute angle and it is measuring less than 90 degrees. It is always the smallest angle, and it makes the terminal side of an angle with the

x - axis.

<u>S</u><u>t</u><u>e</u><u>p</u><u> </u><u>1</u> : -

Find where 192º lies in which quadrant.

192º lies in 3rd quadrant ( between

180º and 270º )

<u>S</u><u>t</u><u>e</u><u>p</u><u> </u><u>2</u> : -

Subtract : 270 - 192

270 - 192 = 78º

( Since, 192º is greater than 180º, is subtracted from 270º )

<u>A</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u> : -

78º is the reference angle for 192º.

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49.

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