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Andrei [34K]
2 years ago
8

No anyone know these answers ?

Mathematics
2 answers:
JulsSmile [24]2 years ago
6 0

Answer:

Step-by-step explanation:

y -intercept

for g(x)=2

for f(x)=0

for h(x)=-2

so g(x) has the largest y-intercept.

2.

decreasing linear ,it goes to -∞

exponential decay →0

positive quadratic has minimum  a finite value.

negative quadratic has no minimum value.

3.

y=-3(3)^x+4

it is an exponential function.

Pepsi [2]2 years ago
4 0

Answer: Positive Quadratic

Step-by-step explanation: A positive quadratic will have a minimum value, whereas a negative quadratic will have a maximum value

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The difference of 4 minus 17 less the sum of 6 and 1
steposvetlana [31]

Answer:

4-(-10)

Step-by-step explanation:

17 less than 6+1 is negative 10 so 4 minus negative 10

5 0
3 years ago
√225x^7y^9
igor_vitrenko [27]

Answer:

√(225x^7y^9) = 15 sqrt(x^7 y^9)

Step-by-step explanation:

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3 years ago
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oksian1 [2.3K]

Given :-

  • a² - 2a - b² = 0
  • 2b + 2ab = 0

To find :-

  • Value of a and b .

Solution :-

<u>Taking</u><u> </u><u>second</u><u> </u><u>equation</u><u>:</u><u>-</u>

  • 2b + 2ab = 0
  • 2b ( 1 + a ) = 0
  • 2b = 0 or (1+a) = 0
  • b = 0 , a = -1

<u>Substitute</u><u> </u><u>in </u><u>first </u><u>equation</u><u> </u><u>:</u><u>-</u><u> </u>

  • a² - 2a - b² = 0

<u>When </u><u>b </u><u>=</u><u> </u><u>0</u><u> </u><u>,</u>

  • a² - 2a - 0² = 0
  • a² - a = 0
  • a( a -1) =0
  • a = 0 , 1

<u>When </u><u>a </u><u>=</u><u> </u><u>-</u><u>1</u><u> </u><u>,</u>

  • (-1)² - 2*(-1) - b² = 0
  • 1 + 2 - b² = 0
  • b² = 3
  • b = ±√3

<u>Answer </u><u>:</u><u>-</u><u> </u>

  • a = 0,1 ; b = 0
  • a = -1 , b = ±√3
8 0
2 years ago
Simplifying algebraic expressions 7d+d use d=9
Lostsunrise [7]
Hey there Titachely1p09ewg,

Answer:

7d + d = 7 (9) + 9
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           = 72

Hope this helps :D

<em>~Natasha♥</em>
5 0
3 years ago
L.c.m of 84 ,27,35,45 <br>please help​
den301095 [7]

Answer:

3780

Step-by-step explanation:

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27 = 3^3

35 = 5 * 7

45 = 3^2 * 5

lcm = 2^2 * 3^3 * 5 * 7 = 3780

8 0
3 years ago
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