Quadrilateral A'B'C'D' is the image of quadrilateral ABCD under a translation.

Helpppp please!!!!!!

uysha [10]

Answer:

x-8

Step-by-step explanation:

since we dont know how much gas is in the tank, lets make that X.

and since we subtract 8 gallons from how much is in the tank, we make that X-8

5 0

3 years ago

Read 2 more answers

What is the equation of a line that contains the point (4, 0) and is parallel to x + y = 2?

gogolik [260]

Answer:

x + y = 4 or y = -x + 4

Step-by-step explanation:

0 = -1[4] + b

4 = b

y = -x + 4

If you want it in Standard Form:

y = -x + 4

+x +x

_________

x + y = 4 >> Line in Standard Form

* Since the rate of change [slope] is -1 and that parallel lines have SIMILAR SLOPES, -1 remains the same.

I am joyous to assist you anytime.

6 0

3 years ago

The largest prime number that is a factor of 35 is multiplied by the smallest prime number

Lubov Fominskaja [6]

Answer:

the largest factor of 35 is 35 and the smallest prime number is 2 so 35 times two is 70

Step-by-step explanation:

6 0

2 years ago

Can someone answer number 5 plz

Georgia [21]

Answer:

C) 113 m^2

Step-by-step explanation:

diameter/2 =radius = 12/2 = 6

area = 6^2(3.14) = 6x6x3.14 = 113 m^2

5 0

2 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