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astraxan [27]
3 years ago
9

The product of odd numbers is always even true or false ?

Mathematics
2 answers:
mars1129 [50]3 years ago
8 0

Answer:

false

Step-by-step explanation:

If we can show a counter example, then it would be false

3*5  ( both are odd numbers)

15  ( that is odd)

So it is not a true statement

likoan [24]3 years ago
6 0

Answer:

false

Step-by-step explanation:

let's say we have two odd numbers

2k+1 and 2t+1

(2k+1)(2t+1)\\\\=4kt+2k+2t+1\\\\=2(2kt+k+t)+1

so we see it's odd

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The point (0, 0) is a solution to which of these inequalities? A. y – 4 < 3x – 1 B. y – 1 < 3x – 4 C. y + 4 < 3x – 1 D.
miskamm [114]

Answer:

A

Step-by-step explanation:

Replace the point (0,0) in each inequality

A. y - 4 < 3x - 1

B. y - 1 < 3x - 4

C. y + 4 < 3x - 1

D. y + 4 < 3x + 1

A. 0 - 4 < 3(0) - 1

- 4 < - 1

True

B. 0 - 1 < 3(0) - 4

- 1 < - 4

False

C. y + 4 < 3x - 1

0 + 4 < 3(0) - 1

4 < - 1

False

D. y + 4 < 3x + 1

0 + 4 < 3(0) + 1

4 < 1

False

5 0
3 years ago
Find the sum and product of the roots.
xz_007 [3.2K]

Answer:

sum= -3 product=-4

Step-by-step explanation:

2x^2-2x+8x-8

2x(x-1)+8(x-1)

(2x+8)(x-1)

x=-4 x=1

8 0
3 years ago
Read 2 more answers
Will mixes 1/4 gallon of milk and 1/2 gallon of chocolate syrup to make chocolate milk. Each serving is 1/16 gallon. How many se
Dennis_Churaev [7]

Answer:

12 servings!

Step-by-step explanation:

1/4 equals 0.25 gallons, 1/2 gallons is 0.5, so Will will have a mix of 0.75 gallons. If each serving is 1/16 and 1/16 equals 0.0625, divide 0.75 by 0.0625 = 12 servings.

6 0
2 years ago
Please answer asap!!!
ASHA 777 [7]

Answer:

15/3

Step-by-step explanation:

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6 0
3 years ago
Solve y'' + 10y' + 25y = 0, y(0) = -2, y'(0) = 11 y(t) = Preview
svetlana [45]

Answer:  The required solution is

y=(-2+t)e^{-5t}.

Step-by-step explanation:   We are given to solve the following differential equation :

y^{\prime\prime}+10y^\prime+25y=0,~~~~~~~y(0)=-2,~~y^\prime(0)=11~~~~~~~~~~~~~~~~~~~~~~~~(i)

Let us consider that

y=e^{mt} be an auxiliary solution of equation (i).

Then, we have

y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.

Substituting these values in equation (i), we get

m^2e^{mt}+10me^{mt}+25e^{mt}=0\\\\\Rightarrow (m^2+10y+25)e^{mt}=0\\\\\Rightarrow m^2+10m+25=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow m^2+2\times m\times5+5^2=0\\\\\Rightarrow (m+5)^2=0\\\\\Rightarrow m=-5,-5.

So, the general solution of the given equation is

y(t)=(A+Bt)e^{-5t}.

Differentiating with respect to t, we get

y^\prime(t)=-5e^{-5t}(A+Bt)+Be^{-5t}.

According to the given conditions, we have

y(0)=-2\\\\\Rightarrow A=-2

and

y^\prime(0)=11\\\\\Rightarrow -5(A+B\times0)+B=11\\\\\Rightarrow -5A+B=11\\\\\Rightarrow (-5)\times(-2)+B=11\\\\\Rightarrow 10+B=11\\\\\Rightarrow B=11-10\\\\\Rightarrow B=1.

Thus, the required solution is

y(t)=(-2+1\times t)e^{-5t}\\\\\Rightarrow y(t)=(-2+t)e^{-5t}.

6 0
3 years ago
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