All categories

3 years ago

Choose the correct simplification of the expression: (5x^4y^2z^2)(3x^4y^3z^5)

1 answer:

5 0

15x^8y^5z^7
Multiply the like terms(x goes with x,y goes with y)and add the exponents of the like terms together to simplfy.

You might be interested in

What is the quotient? for 692 divided by 16

Rufina [12.5K]

Answer:

43.25

Step-by-step explanation:

7 0

3 years ago

If a laptop computer regularly costs $979 and is on sale for $679, what percentage discount is being given? Round your answer to

Lady_Fox [76]

Answer:

Look at the attachment

4 0

3 years ago

Read 2 more answers

a_sh-v [17]

Answer:

I could be wrong, but I believe that the answer is D, none. This is because there is only proof of one congruent side and angle.

Step-by-step explanation:

4 0

2 years ago

On a given day, 36 of the 445 students in a school were absent. What was the appproximate absentee rate that day?

Nata [24]

Answer: The approximate absentee rate that day would be 8.09%.

Step-by-step explanation:

Since we have given that

Number of students who were absent = 36

Total number of students = 445

We need to find the approximate absentee rate that day :

Rate of absentee of that day would be

$\dfrac{\text{Number of absentee}}{\text{Total number of students}}\times 100\\\\=\dfrac{36}{445}\times 100\\\\=8.09\%$

Hence, the approximate absentee rate that day would be 8.09%.

5 0

3 years ago

Use triangle ABC drawn below & only the sides labeled. Find the side of length AB in terms of side a, side b & angle C o

Brrunno [24]

Answer:

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

Step-by-step explanation:

Given:

The above triangle

Required

Solve for AB in terms of a, b and angle C

Considering right angled triangle BOC where O is the point between b-x and x

From BOC, we have that:

$Sin\ C = \frac{h}{a}$

Make h the subject:

$h = aSin\ C$

Also, in BOC (Using Pythagoras)

$a^2 = h^2 + x^2$

Make $x^2$ the subject

$x^2 = a^2 - h^2$

Substitute $aSin\ C$ for h

$x^2 = a^2 - h^2$ becomes

$x^2 = a^2 - (aSin\ C)^2$

$x^2 = a^2 - a^2Sin^2\ C$

Factorize

$x^2 = a^2 (1 - Sin^2\ C)$

In trigonometry:

$Cos^2C = 1-Sin^2C$

So, we have that:

$x^2 = a^2 Cos^2\ C$

Take square roots of both sides

$x= aCos\ C$

In triangle BOA, applying Pythagoras theorem, we have that:

$AB^2 = h^2 + (b-x)^2$

Open bracket

$AB^2 = h^2 + b^2-2bx+x^2$

Substitute $x= aCos\ C$ and $h = aSin\ C$ in $AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = (aSin\ C)^2 + b^2-2b(aCos\ C)+(aCos\ C)^2$

Open Bracket

$AB^2 = a^2Sin^2\ C + b^2-2abCos\ C+a^2Cos^2\ C$

Reorder

$AB^2 = a^2Sin^2\ C +a^2Cos^2\ C + b^2-2abCos\ C$

Factorize:

$AB^2 = a^2(Sin^2\ C +Cos^2\ C) + b^2-2abCos\ C$

In trigonometry:

$Sin^2C + Cos^2 = 1$

So, we have that:

$AB^2 = a^2 * 1 + b^2-2abCos\ C$

$AB^2 = a^2 + b^2-2abCos\ C$

Take square roots of both sides

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

6 0

3 years ago