The price of a television set on scale is $360. This two thirds of the regular price. Find the regular price.

What are the next three terms in the sequence?

Ne4ueva [31]

-1 + 10 = 9 +10 = 19 + 10 = 29 so the next would be 39 because you add ten to it and there is only one answer with 39 so it's C

7 0

3 years ago

Read 2 more answers

How much will $850 amount to be in three years if it is invested at 8% interest compounded quarterly for 3 years? A. $886.47 B.

Dimas [21]

$\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{compounded amount}\\
P=\textit{original amount deposited}\to &\$850\\
r=rate\to 8\%\to \frac{8}{100}\to &0.08\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{quarterly, so four times}
\end{array}\to &4\\

t=years\to &3
\end{cases}$

7 0

3 years ago

Which two values of x are roots of the polynomial below?<br> 3x2-3x+1

QveST [7]

Answer:

The roots are

$x=\frac{1}{6} [3+ i\sqrt{3}]$

$x=\frac{1}{6} [3- i\sqrt{3}]$

Step-by-step explanation:

we have

$3x^2-3x+1$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$3x^2-3x+1=0$

so

$a=3\\b=-3\\c=1$

substitute in the formula

$x=\frac{-(-3)\pm\sqrt{-3^{2}-4(3)(1)}} {2(3)}$

$x=\frac{3\pm\sqrt{-3}} {6}$

Remember that

$i=\sqrt{-1}$

so

$x=\frac{1}{6} [3\pm i\sqrt{3}]$

The roots are

$x=\frac{1}{6} [3+ i\sqrt{3}]$

$x=\frac{1}{6} [3- i\sqrt{3}]$

6 0

2 years ago

If you have a room with four walls and a ceiling, how would you label the walls (plane, line, or point)? How would you label

Sergio039 [100]

Walls = Plane

Intersection = line

Corner = Point

3 0

3 years ago

Does anyone knows how to do this?

dedylja [7]

Answer:

(a) i) Vector BC = 3/2 a + 5b

ii) Vector AM = 15/4 a + 5/2 b

(b) Vector QP = -15/4 b where k = -15/4

Step-by-step explanation:

* Lets explain how to solve this problem

∵ ABCD is a trapezium

∵ AB // DC

∵ The vector AB = 3a

∵ Vector DC = 3/2 vector AB

∴ Vector DC = 3/2 × 3a = 9/2 a

∵ Vector AD = 5b

(a)

i) ∵ Vector BC = vector BA + vector AD + vector DC

∵ Vector AB = 3a , then vector BA = -3a

∵ Vector AD = 5b , vector DC = 9/2 a

∴ Vector BC = -3a + 5b + 9/2 a = (-3a + 9/2 a) + 5b

∴ Vector BC = 3/2 a + 5b

ii) ∵ Vector AM = vector AB + vector BM

∵ M is the mid-point of BC

∴ Vector BM = 1/2 vector BC

∵ Vector BC = 3/2 a + 5b

∴ Vector BM = 1/2(3/2 a + 5b) = (1/2 × 3/2) a + (1/2 × 5) b

∴ Vector BM = 3/4 a + 5/2 b

∴ Vector AB = 3a

∴ Vector AM = 3a + 3/4 a + 5/2 b = (3a + 3/4 a ) + 5/2 b

∴ Vector AM = 15/4 a + 5/2 b

(2)

∵ 7 DQ = 5 QC ⇒ divide both sides by 7

∴ DQ = 5/7 DC

∴ The line DC = 7 + 5 = 12 parts ⇒ DQ 5 parts and QC 7 parts

∵ DQ = 5/12 DC

∵ Vector DC = 9/2 a

∴ Vector DQ = 5/12 (9/2 a) = 45/24 a ⇒ divide up and down by 3

∴ Vector DQ = 15/8 a

∵ P is the mid point of AM

∴ Vector AP = 1/2(15/4 a + 5/2 b) = (1/2 × 15/4) a + (1/2 × 5/2) b

∴ Vector AP = 15/8 a + 5/4 b

∵ Vector QP = QD + DA + AP

∵ Vector DQ = 15/8 , then vector QD = -15/8 a

∵ Vector AD = 5b , then vector DA = -5b

∴ Vector QP = -15/8 a + -5b + 15/8 a + 5/4 b

∴ Vector QP = (-15/8 a + 15/8 a) + (-5b + 5/4 b)

∴ Vector QP = -15/4 b

∵ -15/4 is constant

∴ Vector QP = k b ⇒ proved

8 0

2 years ago