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choli [55]
3 years ago
10

p: 3 is an odd number q: 9 is an even number Which of the following statements is false? p ∨ q p ∧ q q → p

Mathematics
2 answers:
8_murik_8 [283]3 years ago
6 0

Answer:

p ^ q

Step-by-step explanation:

its stating that 9 it an equality with 3.

Margarita [4]3 years ago
5 0

⇒p=3 is an odd number------(True Statement)→T

⇒q=9 is an even number------(False Statement)→F

 Truth Table

p               q                p∧q               p∨q

T                 T                 T                    T

<u>T                  F                 F                    T</u>

F                  T                  F                    T

F                   F                 F                     F

⇒<u>p∧q is False</u>, when p is true and q is False.

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Pleas help me find the composition of goh and it’s domain
Ksivusya [100]

Solution:

The function is given below as

\begin{gathered} g(x)=\frac{x+4}{x-1} \\ h(x)=2x-1 \end{gathered}

To figure out

(goh)(x)

To do this , we will substitute x= 2x-1 in g(x)

\begin{gathered} g(x)=\frac{x+4}{x-1} \\ g(h)(x)=\frac{2x-1+4}{2x-1-1} \\ g(h)(x)=\frac{2x+3}{2x-2} \end{gathered}

Hence,

The composte function will be

(goh)(x)=\frac{2x+3}{2x-2}

Step 2:

To figure out the domain,

In mathematics, the domain of a function is the set of inputs accepted by the function.

Hence,

The domain of the function is

\begin{bmatrix}\mathrm{Solution:}\:&\:x1\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-\infty \:,\:1\right)\cup \left(1,\:\infty \:\right)\end{bmatrix}

5 0
1 year ago
Jamal hiked on two trails. The first trail was 7 1/3 miles long, and the second trail was 1 9/11 times as long as the first trai
algol [13]

9514 1404 393

Answer:

  20 2/3 miles

Step-by-step explanation:

Altogether Jamal hiked 1 + 1 9/11 times the length of the first trail:

  (2 9/11)(7 1/3) = (31/11)(22/3) = 62/3 = 20 2/3

Jamal hiked 20 2/3 miles.

7 0
3 years ago
Simplify the expression. <br><br> [90 – (12 + 13)](15 ÷ 3)
slega [8]
[90-(12+13)](15:3)= \\\\=(90-25)*5=\\\\= 65*5=\\\\=\boxed{325}
6 0
3 years ago
Read 2 more answers
ank A contains 80 gallons of water in which 20 pounds of salt has been dissolved. Tank B contains30 gallons of water in which 5
adell [148]

Answer:

\frac{dx}{dt}=2-\frac{3}{40}x+\frac{y}{15}

\frac{dy}{dt}=\frac{3}{40}x-\frac{y}{5}

x(0)=20,y(0)=5

Step-by-step explanation:

We are given that

Tank A contains water=80 gallons

x(0)=20,y(0)=5

Tank B contains water=30 gallons

Rate=4 gallon/min

Concentration of salt  is pumped into tank=0.5 pound /gallon of water

Solution pumped from tank A to tank B at the rate=6 gallons/min

Solution pumped from tank B to tank A at the rate=2gallon/min

Solution from tank B is pumped out of the system at the rate=4 gallon/min

We have to find the DE at time t

For x

Rate in=0.5\times 4+y(t)/30\times 2

Rate in=2+y/15

Rate out=x/80\times 6

Rate out=3x/40[/tex]

\frac{dx}{dt}=Rate in-Rate out

\frac{dx}{dt}=2+y/15-3x/40

\frac{dx}{dt}=2-\frac{3}{40}x+\frac{y}{15}

For y

Rate in=x/80\times 6

Rate in=3/40x

Rate out=y/30\times (4+2)

Rate out=y/5

\frac{dy}{dt}=Rate in-Rate out

\frac{dy}{dt}=\frac{3}{40}x-\frac{y}{5}

8 0
3 years ago
In a parenting magazine, Mrs. Smith read an advertisement about all the different ways that bleach can be used to kill bacteria.
babymother [125]

Question:

(a) How many bacteria would be found in 24 hours

(b) How many bacteria would be found in 2 days

(c) How long for 1000 bacteria to be found

Answer:

(a) 282429536481 bacteria

(b) 7.97*10^{22 bacteria

(c) 6 days

Step-by-step explanation:

The question illustrates an exponential function

y = ab^x

Where

a = Initial\ Value = 1

b=Rate = 3 --- i.e. triples

x = number\ of\ hours

Solving (a): Bacteria in 24 hours

In this case:

x = 24

Substitute x = 24, b = 3 and a = 1

So:

y = ab^x

y = 1 * 3^{24

y = 1 * 282429536481

y = 282429536481 bacteria

Solving (b): Bacteria in 2 days

2\ days =48\ hours

So:

x = 48

Substitute x = 48, b = 3 and a = 1

So:

y = ab^x

y = 1 * 3^{48

y =7.97*10^{22 bacteria

Solving (c): How long for 1000 bacteria

In this case:

y = 1000

Substitute y = 1000, b = 3 and a = 1

So:

y = ab^x

1000 = 1 *3^x

1000 = 3^x

Take Log of both sides

Log1000 = Log3^x

This gives:

Log1000 = xLog3

Make x the subject

x = \frac{Log1000}{Log3}

x \approx 6

<em>Hence: It takes 6 days</em>

8 0
3 years ago
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