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2 years ago

If a and b are real numbers and a>0 and b>0 , then ab ___0

Mathematics

1 answer:

Pepsi [2]2 years ago

4 0

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Answer: $\textsf{ab} > \textsf{0}$

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Given: $\textsf{a} > \textsf{0 and b} > \textsf{0}$

Find: $\textsf{ab } \_\_\_ \textsf{ 0}$

Solution: Since we know both a and b are positive this means that if they are multiplied against each other they would produce another positive. Therefore, this would cause the statement to be ab > 0.

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What two pencils do i need to move to make three triangles?

asambeis [7]

Would this work? Hope it helps

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3 years ago

Simplify the expression below. 2^3 x 2^2

Zigmanuir [339]

Answer:

$32$ $or~2^5$

Step-by-step explanation:

→ $(2^3)(2^2)$

$=2^3*2^2$

$=(2*2*2)*(2*2)$

$=2*2*2*2*2$

$=32$

You can also put this in another way:

 $2^5=2^3*2^2=2^3^+^2=2^5$ 

7 0

4 years ago

Read 2 more answers

Solve for x in the following: 8x^2 + 6x = -5

artcher [175]

Answer:

No solution or

(-3± $2i\sqrt{31}$ )/8

Step-by-step explanation:

To get the solution, move everything to the left side of the equation:

8x²+6x+5 = 0

If you try to factor the equation or graph it, you will see that it will never equal 0.

If you need an imaginary solution you can plug it into the quadratic equation:

-6± $\sqrt{36-160}$ over 16

Simplify and you get

-6± $i\sqrt{124}$ over 16

-6± $4i\sqrt{31}$ over 16

Simplify the fraction and you can get

(-3± $2i\sqrt{31}$ )/8

8 0

3 years ago

A library subscribes to two different weekly news magazines, each of which is supposed to arrive in Wednesday�s mail. In actuali

Softa [21]

Answer:

P(Y = 0) = 0.09

P(Y = 1) = 0.4

P(Y = 2) = 0.32

P(Y = 3) = 0.19

Step-by-step explanation:

Let the events be:

$W =$ Wednesday

$T =$ Thursday

$F =$ Friday

$S =$ Saturday

Their corresponding probabilities are

$P(W) = 0.3\\P(T) = 0.4\\P(F) = 0.2\\P(S) = 0.1$

Since $Y$ = number of days beyond Wednesday that it takes for both magazines to arrive(so possible $Y$ values are 0, 1, 2 or 3)

The possible number of outcomes are therefore $4^2 = 16\\(W, W), (W, T), (W, F), (W, S)\\(T, W), (T, T), (T, F), (T, S)\\(F, W), (F, T), (F, F), (F, S)\\(S, W), (S, T), (S, F), (S, S)$

The values associated for each of the outcomes are as follows:

$Y(W, W) = 0, Y(W, T) = 1, Y(W, F) = 2, Y(W, S) = 3\\Y(T, W) = 1, Y(T, T) = 1, Y(T, F) = 2, Y(T, S) = 3\\Y(F, W) = 2, Y(F, T) = 2, Y(F, F) = 2, Y(F, S) = 3\\Y(S, W) = 3, Y(S, T) = 3, Y(S, F) = 3, Y(S, S) = 3$

The probability mass function of $Y$ is,

$P(Y = 0) = 0.3(0.3) = 0.09\\P(Y = 1) = P[(W, T) or (T, W) or (T, T)]\\= [0.3(0.4) + 0.3(0.4) + 0.4(0.4)]\\= 0.4\\\\P(Y = 2) = P[(W, F) or (T, F) or (F, W) or (F, T) or (F, F)]\\= [0.3(0.2) + 0.4(0.2) + 0.2(0.3) + 0.2(0.4) + 0.2(0.2)]\\= 0.32\\\\P(Y = 3) = P[(W, S) or (T, S) or (F, S) or (S, W) or (S, T) or (S, F) or (S, S)]\\= [0.3(0.1) + 0.4(0.1) + 0.2(0.1) + 0.1(0.3) 0.1(0.4) + 0.1(0.2) + 0.1(0.1)]\\= 0.19$

7 0

3 years ago

What is the midpoint of line AC

pishuonlain [190]

Answer:

N/A

Step-by-step explanation:

What is line AC, was there supposed to be a picture?

4 0

3 years ago