What percent of 45.4 is 6.81

Show twelve in four different ways use words pictures and numbers

Hunter-Best [27]

24/4: 24-12: 12+0: 12/1: 48/4: 72/6:

3 0

3 years ago

What is the expansion of (3+x)^4

Vlad1618 [11]

Answer:

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

Step-by-step explanation:

Considering the expression

$\left(3+x\right)^4$

Lets determine the expansion of the expression

$\left(3+x\right)^4$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=3,\:\:b=x$

$=\sum _{i=0}^4\binom{4}{i}\cdot \:3^{\left(4-i\right)}x^i$

Expanding summation

$\binom{n}{i}=\frac{n!}{i!\left(n-i\right)!}$

$i=0\quad :\quad \frac{4!}{0!\left(4-0\right)!}3^4x^0$

$i=1\quad :\quad \frac{4!}{1!\left(4-1\right)!}3^3x^1$

$i=2\quad :\quad \frac{4!}{2!\left(4-2\right)!}3^2x^2$

$i=3\quad :\quad \frac{4!}{3!\left(4-3\right)!}3^1x^3$

$i=4\quad :\quad \frac{4!}{4!\left(4-4\right)!}3^0x^4$

$=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4$

as

$\frac{4!}{0!\left(4-0\right)!}\cdot \:\:3^4x^0:\:\:\:\:\:\:81$

$\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1:\quad 108x$

$\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2:\quad 54x^2$

$\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3:\quad 12x^3$

$\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4:\quad x^4$

so equation becomes

$=81+108x+54x^2+12x^3+x^4$

$=x^4+12x^3+54x^2+108x+81$

Therefore,

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

6 0

3 years ago

PYTHAGORAS THEROM AND TRIGONOMETRY RATIO

marysya [2.9K]

Answer:

1) ΔACD is a right triangle at C

=> sin 32° = AC/15

⇔ AC = sin 32°.15 ≈ 7.9 (cm)

2) ΔABC is a right triangle at C, using Pythagoras theorem, we have:

AB² = AC² + BC²

⇔ AB² = 7.9² + 9.7² = 156.5

⇒ AB = 12.5 (cm)

3) ΔABC is a right triangle at C

=> sin ∠BAC = BC/AB

⇔ sin ∠BAC = 9.7/12.5 = 0.776

⇒ ∠BAC ≈ 50.9°

4) ΔACD is a right triangle at C

=> cos 32° = CD/15

⇔ CD = cos32°.15

⇒ CD ≈ 12.72 (cm)

Step-by-step explanation:

7 0

3 years ago

Each pair of figures below is similar. Find the lengths of all unknown sides

lions [1.4K]

Answer:

a) w = 8; y = 5.25

b) x = 10; z = 7.2

Step-by-step explanation:

a) Dimensions on the smaller figure are FA/F'A' = 3/4 times those on the larger figure.

6 = (3/4)w

w = 24/3 = 8

y = (3/4)·7 = 21/4 = 5.25

__

b) Dimensions on the smaller figure are ER/E'R' = 9/15 = 3/5 times those on the larger figure.

6 = (3/5)x

x = 30/3 = 10

z = (3/5)12 = 36/5 = 7.2

7 0

3 years ago

Really need help with this .

lions [1.4K]

Answer:

Step-by-step explanation:

The attached photo shows the diagram of quadrilateral QRST with more illustrations.

Line RT divides the quadrilateral into 2 congruent triangles QRT and SRT. The sum of the angles in each triangle is 180 degrees(98 + 50 + 32)

The area of the quadrilateral = 2 × area of triangle QRT = 2 × area of triangle SRT

Using sine rule,

q/SinQ = t/SinT = r/SinR

24/sin98 = QT/sin50

QT = r = sin50 × 24.24 = 18.57

Also

24/sin98 = QR/sin32

QR = t = sin32 × 24.24 = 12.84

Let us find area of triangle QRT

Area of a triangle

= 1/2 abSinC = 1/2 rtSinQ

Area of triangle QRT

= 1/2 × 18.57 × 12.84Sin98

= 118.06

Therefore, area of quadrilateral QRST = 2 × 118.06 = 236.12

6 0

3 years ago

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