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Marizza181 [45]

3 years ago

14

Prepare the journal entry to record the issuance of 5,000 shares of $20 par common stock for $25 per share

1 answer:

vagabundo [1.1K]3 years ago

7 0

Answer:

value preferred stock for $65,250 cash

<h2> HOPE THIS HELP PLEASE TOUCH THANKS.</h2>

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F(x)=4/x<br> g(x)=4/x <br><br> Find f(g(x))

qaws [65]

The answer of the question is x.

4 0

3 years ago

neonofarm [45]

Answer:

multiply the left side of the constant vector by the inverse matrix

Step-by-step explanation:

The matrix equation ...

AX = B

is solved by left-multiplying by the inverse of A:

A⁻¹AX = A⁻¹B

IX = A⁻¹B . . . . . the result of multiplying A⁻¹A is the identity matrix

X = A⁻¹B . . . . . B needs to be multiplied by the inverse matrix

$\left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{cc}-4&1\\3&2\end{array}\right]^{-1}\left[\begin{array}{c}9&7\end{array}\right]=\dfrac{1}{11}\left[\begin{array}{cc}-2&1\\3&4\end{array}\right]\left[\begin{array}{c}9&7\end{array}\right]=\left[\begin{array}{c}-1&5\end{array}\right]$

4 0

3 years ago

Width of a rectangle with the length of x+6 and area x^3+2x^3-5x-66

In-s [12.5K]

Hello :
A= W×L
W =A/L
W = (<span>x^3+2x^3-5x-66) / (x+6).......
</span>Review your statement........

4 0

3 years ago

Hello:) anyone able to explain how to solve the equestrian for part (d) ? Thank youu

Anna007 [38]

Answer:

see explanation

Step-by-step explanation:

Given

4 $a^{4}$ - 5a² + 1 = 0

Use the substitution u = a², then equation is

4u² - 5u + 1 = 0

Consider the product of the coefficient of the u² term and the constant term

product = 4 × 1 = 4 and sum = - 5

The factors are - 4 and - 1

Use these factors to split the u- term

4u² - 4u - u + 1 = 0 ( factor the first/second and third/fourth terms )

4u(u - 1) - 1(u - 1) = 0 ← factor out (u - 1) from each term

(u - 1)(4u - 1) = 0

Equate each factor to zero and solve for u

u - 1 = 0 ⇒ u = 1

4u - 1 = 0 ⇒ 4u = 1 ⇒ u = $\frac{1}{4}$

Convert u back into terms of a, that is

a² = 1 ⇒ a = ± 1

a² = $\frac{1}{4}$ ⇒ a = ± $\frac{1}{2}$

Solutions are a = ± 1 , a = ± $\frac{1}{2}$

7 0

4 years ago

Learning Task 3

ASHA 777 [7]

Answer:

1. The sides are: 38.5, 39.5, 40.5 and 41.5

2. Presently, Inzo is 15 years old and Miz is 10 years old

3. $Distance = 3682km$

4. Possible pairs are: (12, 14), (16, 18) and (20, 24)

5. $L < 38m$

Step-by-step explanation:

The question has lots of missing details. See the comment section for complete question

Solving (a):

Given

Shape: Quadrilateral

Sides: Consecutive

$Perimeter = 160$

Let one of the sides be x. The other sides will be x + 1, x + 2 and x + 3.

So, the perimeter is:

$Perimeter = x + x + 1 + x + 2 + x + 3$

This gives:

$160 = x + x + 1 + x + 2 + x + 3$

Collect Like Terms

$160 -1-2-3= x+x+x+x$

$154= 4x$

Divide both sides by 4

$38.5 = x$

$x =38.5$

<em>So, the sides are: 38.5, 39.5, 40.5 and 41.5</em>

<em></em>

Solving (b):

Given

Presently: $Inzo = 10 + Miz$

Five years ago: $Miz - 5= \frac{1}{3} * (Inzo-5)$ --- (Missing from the question)

Required: Determine their present ages

Substitute $10 + Miz$ for Inzo in the second equation

$Miz - 5 = \frac{1}{3} * (10 + Miz - 5)$

$Miz - 5 = \frac{1}{3} * (Miz + 5)$

Multiply both sides by 3

$3Miz - 15 = Miz + 5$

Collect Like Terms

$3Miz -Miz = 15 + 5$

$2Miz= 20$

Divide both sides by 2

$Miz = 10$

Substitute 10 for Miz in $Inzo = 10 + Miz$

$Inzo = 10 + 5$

$Inzo = 15$

<em>So presently, Inzo is 15 years old and Miz is 10 years old</em>

Solving (3):

Given

Express Train:

$Speed = 100kph$

$Time = t$

Local Train:

$Speed = 45kph$

$Time = t + 45$

Required: Determine the distance between A and B

Distance is calculated as:

$Distance = Speed * Time$

For the express train:

$Distance = 100 * t$

For the local train

$Distance = 45 * (45 + t)$

Distance between both stations are equal; So,

$100 * t = 45 * (45 + t)$

$100 t = 2025 + 45t$

Collect Like Terms

$100 t -45t= 2025$

$55t= 2025$

Divide by 55

$t = 36.82$

Substitute 36.82 for t in $Distance = 100 * t$

$Distance = 100 * 36.82$

$Distance = 3682km$

Solving (4):

Let the even numbers be: n and n + 2.

Sum greater than 24: $n + n + 2 > 24$

Sum less than 60: $n + n + 2 < 60$

So, we have:

$n + n + 2 > 24$

$2n + 2 > 24$

Collect Like Terms

$2n > 24-2$

$2n > 22$

$n > 11$

$n + n + 2 < 60$

$2n + 2< 60$

Collect Like Terms

$2n < 60-2$

$2n < 58$

Divide by 2

$n$

So, we have:

$n > 11$ and $n < 29$

So, possible pairs are: (12, 14), (16, 18) and (20, 24)

Solving (5):

Given

$Perimeter < 152m$

Required: Find the possible lengths

Let the length be L.

Perimeter is calculated as:

$Perimeter =4 * L$

$Perimeter =4 L$

So, we have:

$4L < 152m$

Divide through by 4

$L < 38m$

Hence, length is less than 38m (e.g. 37m)

5 0

3 years ago