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Katyanochek1 [597]
3 years ago
7

Identify 1 and 2. Select all that apply of the following terms: acute, right, obtuse, adjacent, vertical, complementary, supplem

entary. SELECT ALL THAT APPLY!
acute
right
obtuse
adjacent
vertical
complementary
supplementary

Mathematics
2 answers:
Grace [21]3 years ago
4 0
Right angle and complementary. this cold be wrong.
Pani-rosa [81]3 years ago
4 0

Answer:

Given: An Angle in the figure is right angle. Its measure = 90°

So,

∠2 = 90° ( vertically opposite angles are equal )

∠1 + ∠2 = 180° ( Linear Pair )

∠1 = 180 - 90

∠1 = 90°

So, ∠1 and ∠2 are

Right angles because measure of both angles are 90°.

Adjacent angles because both a common arm and a common vertex.

Supplementary angles because sum of both angles is 180°.

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Han is cycling at a speed of -8 miles per hour; if he starts at the same zero point, what will his position be after 45 minutes?
baherus [9]

Answer:

-6mph

Step-by-step explanation:

45 is 3/4hour so -8 times 3/4= -24/4=-6mph

7 0
3 years ago
A cable installer charges $30.00 per hour plus a $50.00 service charge. Your father's firm hires him to hookup his company's int
vladimir1956 [14]
30*8.5=255 255+50=$305
4 0
3 years ago
Read 2 more answers
Pls Help!!! Worth 99 Pts!!
Alenkasestr [34]
First, for end behavior, the highest power of x is x^3 and it is positive. So towards infinity, the graph will be positive, and towards negative infinity the graph will be negative (because this is a cubic graph)

To find the zeros, you set the equation equal to 0 and solve for x
x^3+2x^2-8x=0
x(x^2+2x-8)=0
x(x+4)(x-2)=0
x=0   x=-4   x=2

So the zeros are at 0, -4, and 2. Therefore, you can plot the points (0,0), (-4,0) and (2,0)

And we can plug values into the original that are between each of the zeros to see which intervals are positive or negative.
Plugging in a -5 gets us -35
-1 gets us 9
1 gets us -5 
3 gets us 21

So now you know end behavior, zeroes, and signs of intervals

Hope this helps<span />
6 0
3 years ago
Read 2 more answers
In 2000 the total amount of gamma ray bursts was recorded at 6.4 million for a city. In 2005, the same survey was made and the t
ratelena [41]

Let's assume

It started in 2000

so, t=0 in 2000

P_0=6.4million

we can use formula

P(t)=P_0 e^{rt}

we can plug value

P(t)=6.4 e^{rt}

In 2005, the same survey was made and the total amount of gamma ray bursts was 7.3 million

so, at t=2005-2000=5

P(t)=7.3 million

we can plug value and then we can solve for r

7.3=6.4 e^{r*5}

r=0.02632

now, we can plug back

P(t)=6.4 e^{0.02632t}

now, we have

P(t)=1 billion =1000 million

so, we can set it and then we can solve for t

1000=6.4 e^{0.02632t}

t=191.924

approximately

t=192

Year is 2000+192

year is 2192................Answer

4 0
3 years ago
Given f(x)=4x^2−5 and g(x)=x+3 . What is (fg)(x) ?
zmey [24]
(fg)(x)=f(g(x))=4(g(x))²-5=4(x+3)²-5=4x²+24x+31
4 0
3 years ago
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