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Sergio039 [100]
3 years ago
14

Without using a calculator, determine the number of real zeros of the function

Mathematics
1 answer:
Lena [83]3 years ago
4 0
X= 1, -3, -2 or -7 and 7
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Rewrite the sum of 30 and 42 by factoring out of the greatest common factor
kvasek [131]

Answer:

12

Step-by-step explanation:

Factors of 30: 1,2,3,5,(6),10,15,30

Factors of 42: 1,2,3,(6),7,14,21,42

30 divided by 6, 42 divided by 6

(5) + (7)= 12

8 0
2 years ago
A 9 kilogram bag of animal feed costs $45.45 . ​ What is the unit price per kilogram?
blsea [12.9K]

Answer:

5.05

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Can someone help me with number 3 please??!!
ipn [44]
There are 2 way to solve this.

one using Pythagoras theorem and 2nd using trigonometry

so lets solve it by both

using Pythagoras theorem we know

base^2 + perpendicular^2 = hypotanes^2

6^2 + x^2 = 12^2

36 + x^2 = 144

x^2 = 144- 36 = 108

x = √(108) = √( 2×2×3×3×3)

= (2×3) √ (3) = 6 √3

so answer is option 2

bow lets use trigonometry

we know
sin theta = perpendicular / hypotanes
sin 60 = x /12
x = 12 × sin 60
we kNow sin 60 = √3/ 2
so
x = 12×√3 /2 = 6√3
7 0
3 years ago
Match each equation with its solution set.
taurus [48]

Answer:

1- The solution of I2x + 5I = 9 is {-7 , 2}

2- The solution of I2x + 7I + 2 = 11 is {-8 , 1}

3- The solution of I5 - xI = 6 is {-1 , 11}

4- The solution of I6x - 8I + 7 = 5 is ∅

5- The solution of Ix + 3I = 12 is {-15 , 9}

6- The solution of Ix - 3I = -12 is ∅

Step-by-step explanation:

* At first lets explain the meaning of IxI = a

- If IxI = a ⇒ then x = a or x = -a

- IxI never give a negative answer, because IxI means the

 magnitude of x is always positive

Ex: I-2I is 2

* Now lets find the solution of each equation

1- ∵ I2x + 5I = 9

∴ 2x + 5 = 9 ⇒ subtract 5 from both sides

∴ 2x = 4 ⇒ divide both sides by 2

∴ x = 2

OR

∴ 2x + 5 = -9 ⇒ subtract 5 from both sides

∴ 2x = -14 ⇒ divide both sides by 2

∴ x = -7

* The solution of I2x + 5I = 9 is {-7 , 2}

2- ∵ I2x + 7I + 2 = 11 ⇒ Subtract 2 from both sides

∴ I2x + 7I = 9

∴ 2x + 7 = 9 ⇒ subtract 7 from both sides

∴ 2x = 2 ⇒ divide both sides by 2

∴ x = 1

OR

∴ 2x + 7 = -9 ⇒ subtract 7 from both sides

∴ 2x = -16 ⇒ divide both sides by 2

∴ x = -8

* The solution of I2x + 7I + 2 = 11 is {-8 , 1}

3- ∵ I5 - xI = 6

∴ 5 -x = 6 ⇒ subtract 5 from both sides

∴ -x = 1 ⇒ divide both sides by -1

∴ x = -1

OR

∴ 5 -x = -6 ⇒ subtract 5 from both sides

∴ -x = -11 ⇒ divide both sides by -1

∴ x = 11

* The solution of I5 - xI = 6 is {-1 , 11}

4- ∵ I6x - 8I + 7 = 5 ⇒ Subtract 7 from both sides

∴ I6x - 8I = -2

- I  I never give negative answer

* The solution of I6x - 8I + 7 = 5 is ∅

5- ∵ Ix + 3I = 12

∴ x + 3 = 12 ⇒ subtract 3 from both sides

∴ x = 9

OR

∴ x + 3 = -12 ⇒ subtract 3 from both sides

∴ x = -15

* The solution of Ix + 3I = 12 is {-15 , 9}

6- ∵ Ix - 3I = -12

- I  I never give negative answer

* The solution of Ix - 3I = -12 is ∅

4 0
3 years ago
How many zeros are in one million?
Julli [10]

Answer:

six

Step-by-step explanation:

1,000,000

5 0
3 years ago
Read 2 more answers
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