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2 years ago

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Using the distributive property, Marta multiplied the binomial (2x + 3) by the trinomial (x2 + x – 2) and got the expression bel

ow.
(2x)(x2) + (2x)(x) + (2x)(–2) + (3)(x2) + (3)(x) + (3)(–2)

Which is the simplified product?

2x3 + 6x2 – x – 6
2x3 + x2 – x – 6
2x3 + 5x2 – x – 6
2x3 – x2 – 7x – 6

2 answers:

Crazy boy [7]2 years ago

5 0

Answer:

option C is correct

$2x^3+5x^2-x-6$

Step-by-step explanation:

The distributive property states that:

$a \cdot (x+y+z) = a \cdot x +a \cdot y+a \cdot z$

Given that:

Marta multiplied the binomial (2x + 3) by the trinomial $(x^2 + x -2)$

i.e,

$(2x+3)(x^2+x-2)$

Distribute first term of the binomial to the trinomial expression and second term of the binomial to the trinomial expression.

$2x(x^2+x-2)+3(x^2+x-2)$

Apply the distributive property:

$(2x)(x^2)+(2x)(x)-(2x)(2)+(3)(x^2)+(3)(x)-(3)(2)$

Simplify:

$2x^3+2x^2-4x+3x^2+3x-6$

Combine like terms:

$2x^3+5x^2-x-6$

Therefore, the simplified product of the given expression is: $2x^3+5x^2-x-6$

densk [106]2 years ago

3 0

Marta got the expression

$2x\cdot x^2 + 2x\cdot x + 2x\cdot (-2) + 3\cdot x^2 + 3\cdot x + 3\cdot (-2).$

First, note that

$2x\cdot x^2=2x^3,\\2x\cdot x=2x^2,\\2x\cdot (-2)=-4x,\\3\cdot x^2=3x^2,\\ 3\cdot x=3x,\\3\cdot (-2)=-6.$

Then add terms with the same degree:

$2x^2+3x^2=5x^2,\\-4x+3x=-x.$

At last,

$(2x+3)(x^2 + x - 2)=2x^3+5x^2-x-6.$

Answer: correct choice is C.

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2 years ago

PLS HELP ASAP I WILL IVE BRAINLIST ANSWER!!!!!Leon verified that the side lengths 21, 28, 35 form a Pythagorean triple using thi

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Answer:

Leon is correct. (Option 1)

Step-by-step explanation:

Given that Leon verified that the side lengths 21, 28, 35 form a Pythagorean triple using this procedure.

Step 1: Find the greatest common factor of the given lengths: 7

Step 2: Divide the given lengths by the greatest common factor: 3, 4, 5

Step 3: Verify that the lengths found in step 2 form a Pythagorean triple.

we have to explain whether or not Leon is correct.

As, 3,4,5 forms a Pythagorean triplet i.e satisfies the Pythagoras theorem

$Hypotenuse^2=Base^2+Perpendicular^2$

⇒ $5^2=3^2+4^2$

Let a, b, c forms a Pythagorean triplet

$a^2+b^2=c^2$

Multiplied by 4 on both sides

⇒ $4a^2+4b^2=4c^2$

⇒ ${2a}^2+{2b}^2={2c}^2$

Hence, we say 4a, 4b and 4c also forms a Pythagorean triplet.

∴ multiplying every length of a Pythagorean triple by the same whole number results in a Pythagorean triple.

Hence, Leon is correct.

4 0

1 year ago

Read 2 more answers

The manager of a symphony in a large city wants to investigate music preferences for adults and students in the city. Let pA rep

monitta

Answer:

Answer E

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6 0

1 year ago

A bakery sells rolls in units of a dozen. The demand X (in 1000 units) for rolls has a gamma distribution with parameters α = 3,

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Answer:

The value is $E(X) = \$ 1.7067$

Step-by-step explanation:

From the question we are told that

The parameters are α = 3, θ = 0.5

The cost of making a unit on the first day is c = $2

The selling price of a unit on the first day is s = $5

The selling price of a leftover unit on the second day is v = $ 1

Generally the profit of a unit on the first day is

$p_1 = 5 - 2$

$p_1 = \$3$

The profit of a unit on the second day is

$p_2 = 1 - 2$

=> $p_2 = - \$1$

Generally the probability of making profit greater than $ 1 is mathematically represented as

$P(X > 1 ) = Gamma (X ,\alpha , \theta)$

=> $P(X > 1 ) = Gamma (1 ,3 , 0.5)$

Now from the gamma distribution table we have that

$P(X > 1 ) = 0.67668$

Generally the probability of making profit less than or equal to $ 1 is mathematically represented as

$P(X \le 1 ) = 1 - P(X > 1 )$

=> $P(X \le 1 ) = 1 - 0.67668$

=> $P(X \le 1 ) = 0.32332$

So the probability of making $3 is $P(X > 1 ) = 0.67668$

and the probability of making -$1 is $P(X \le 1 ) = 0.32332$

Generally the value of profit per day is mathematically represented as

$E(X) = 3 * P(X > 1 ) + (-1 * P(X \le 1 ) )$

=> $E(X) = 3 * 0.67668 + (-1 * 0.32332 )$

=> $E(X) = \$ 1.7067$

4 0

1 year ago

A pond forms as water collects in a conical depression of radius a and depth h. Suppose that water flows in at a constant rate k

Scrat [10]

Answer:

a. dV/dt = K - ∝π(3a/πh)^⅔V^⅔

b. V = (hk^3/2)/[(∝^3/2.π^½.(3a))]

The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.

c. πa² ≥ k/∝

Step-by-step explanation:

a.

The rate of volume of water in the pond is calculated by

The rate of water entering - The rate of water leaving the pond.

Given

k = Rate of Water flows in

The surface of the pond and that's where evaporation occurs.

The area of a circle is πr² with ∝ as the coefficient of evaporation.

Rate of volume of water in pond with time = k - ∝πr²

dV/dt = k - ∝πr² ----- equation 1

The volume of the conical pond is calculated by πr²L/3

Where L = height of the cone

L = hr/a where h is the height of water in the pond

So, V = πr²(hr/a)/3

V = πr³h/3a ------ Make r the subject of formula

3aV = πr³h

r³ = 3aV/πh

r = ∛(3aV/πh)

Substitute ∛(3aV/πh) for r in equation 1

dV/dt = k - ∝π(∛(3aV/πh))²

dV/dt = k - ∝π((3aV/πh)^⅓)²

dV/dt = K - ∝π(3aV/πh)^⅔

dV/dt = K - ∝π(3a/πh)^⅔V^⅔

b. Equilibrium depth of water

The equilibrium depth of water is when the differential equation is 0

i.e. dV/dt = K - ∝π(3a/πh)^⅔V^⅔ = 0

k - ∝π(3a/πh)^⅔V^⅔ = 0

∝π(3a/πh)^⅔V^⅔ = k ------ make V the subject of formula

V^⅔ = k/∝π(3a/πh)^⅔ -------- find the 3/2th root of both sides

V^(⅔ * 3/2) = k^3/2 / [∝π(3a/πh)^⅔]^3/2

V = (k^3/2)/[(∝π.π^-⅔(3a/h)^⅔)]^3/2

V = (k^3/2)/[(∝π^⅓(3a/h)^⅔)]^3/2

V = (k^3/2)/[(∝^3/2.π^½.(3a/h))]

V = (hk^3/2)/[(∝^3/2.π^½.(3a))]

The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.

c. Condition that must be satisfied

If we continue adding water to the pond after the rate of water flow becomes 0, the pond will overflow.

i.e. dV/dt = k - ∝πr² but r = a and the rate is now ≤ 0.

So, we have

k - ∝πa² ≤ 0 ---- subtract k from both w

- ∝πa² ≤ -k divide both sides by - ∝

πa² ≥ k/∝

5 0

2 years ago