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3 years ago

Use the FOIL method to find the product below.

1 answer:

6 0

(8x^5 + 9)(8x^3 + 9)

FOIL...first outside inside last

F....multiply first terms in each set.....8x^5 * 8x^3 = 64x^8
O...multiply outside terms in each set....8x^5 * 9 = 72x^5
I....multiply inside terms in each set....9 * 8x^3 = 72x^3
L...multiply last terms in each set.....9 * 9 = 81

now we have : 64x^8 + 72x^5 + 72x^3 + 81...and this cant be simplified...so ur answer is C

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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

3 2/3+ 8 7/12<br><br> I need the work

castortr0y [4]

Answer:

$\frac{49}{4}$

Step-by-step explanation:

Convert each to a improper fraction

$3 \frac{2}{3} = \frac{11}{3}$

$8\frac{7}{12} = \frac{103}{12}$

$\frac{11}{3} +\frac{103}{12} = \frac{49}{4}$

4 0

3 years ago

Read 2 more answers

cube A has volume V. the edges of cube b are three times as long as the edges of cube A. What is the volume of cube b in terms o

Mariulka [41]

Answer:

V=a cube

If i am wrong let me know!

4 0

3 years ago

Find each percent decrease of 80 to 64

Anastasy [175]

80-64=24
24 is 30% of 80 so a decrease of 30%

4 0

3 years ago

Can I have halp with all and an explanation because I am terrible at properties.

Salsk061 [2.6K]

1. 4 • (–3) • 5
so 4 x (-3) = -12 and then -12 x 5 = -60
2. (2.25 x 23) x 4
so (2.25 x 23) 51.75 and then 51.75 x4 = 207
4. 5 x 12 x (-2)
so 5 x 12 = 60 and then 60 x (-2) = -120
5. 35(26)(0) =
so 35 x 26 = 910 and then 910 x 0 = 0

4 0

4 years ago