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victus00 [196]
3 years ago
5

PLEASE HELP GEOMETRY EXPLAIN PLEASE I WILL GIVE BRAINLIEST!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics
1 answer:
expeople1 [14]3 years ago
6 0
Angle DCE and angle BCA
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The two lines graphed below are parallel. How many solutions are there to the
NISA [10]
The is going to be infinite answers
7 0
2 years ago
A country's population in 1995 was 164 million. in 2001 it was 169 million. estimate the population in 2015 using the exponentia
Vaselesa [24]
1995.          164 million
2001.          169 million
t = 2001 - 1995 = 6
169 =164 *  p^{6}
 p^{6} = \frac{169}{164}= 1.0304878 \\  p =  \sqrt[6]{1.0304878}
p = 1.00502
t 1 = 2015 - 2001 = 14
f ( t1 ) = 169 * 1.00502^{14}
f (t1) = 169 * 1.0762 = 181.27
Answer: a country´s population in 2015 will be 181 million.
4 0
3 years ago
Help me please!!!!!!!!!!!!!!!!!!
Stella [2.4K]

Answer:

16/81

Step-by-step explanation:

(2/3)^4  = (2/3)*(2/3)*(2/3)*(2/3)

             = 16/81

4 0
2 years ago
Consider the following data set:
Komok [63]

Answer:

+5 Range; The range is now 25

Step-by-step explanation:

The original range would be calculated by subtracting 20 from 40, giving you 20 as the range. However, with the point 15 added, there would be a new lowest number, making the new range be 40-15, which is 25.

3 0
3 years ago
A projectile is launched into the air. The function h(t) = –16t2 + 32t + 128 gives the height, h, in feet, of the projectile t s
Zinaida [17]

Answer:

t = 4 seconds

Step-by-step explanation:

The height of the projectile after it is launched is given by the function :

h(t)=-16t^2+32t+128

t is time in seconds

We need to find after how many seconds will the projectile land back on the ground. When it land, h(t)=0

So,

-16t^2+32t+128=0

The above is a quadratic equation. It can be solved by the formula as follows :

t=\dfrac{-b\pm \sqrt{b^2-4ac} }{2a}

Here, a = -16, b = 32 and c = 128

t=\dfrac{-32\pm \sqrt{(32)^2-4\times (-16)(128)} }{2\times (-16)}\\\\t=\dfrac{-32+ \sqrt{(32)^2-4\times (-16)(128)} }{2\times (-16)}, \dfrac{-32\- \sqrt{(32)^2-4\times (-16)(128)} }{2\times (-16)}\\\\t=-2\ s\ \text{and}\ 4\ s

Neglecting negative value, the projectile will land after 4 seconds.

4 0
3 years ago
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