Suppose that a sequence is defined as follows.

A landscaping company ordered 13 plants and 4 trees for a total of $487. If plants are $8 each, write and solve a linear equatio

oksian1 [2.3K]

Answer:

trees are $95.75

Step-by-step explanation:

13(8) + 4t =487

104 + 4t= 487

104 -104 + 4t = 487 -104

4t = 383

4/4t = 383/4

t = 95.75

5 0

3 years ago

A sequence has an initial value (I) of -17 and its fourth term (a4) is 35. What is its general equation? Its 14th term?

aleksley [76]

Answer:

The sequence is an Arthemtic Progression

An=A1+(n-1)d

A1 is first term, An is nth term, n is number of term, and d is common difference

therefore

A4=35, A1= -17

A4=A1+(4-1)d

35= -17+3d

35+17=3d

52=3d

52/3=3d/3

14=d

common diffrence(d)=14

The general solution is given by

An= -17+(n-1)14

An= -17+14n-14

An= -31+14n

An= 14n-31

A14 term, means n=14

From An=A1+(n-1)d

A14= -17+(14-1)14

= -17+(13×14)

= -17+182

= 165.

Therfore, the 14th term is 165.

2. A sequence has a CR of 4/5 and its eighth term (a8) is (393216/3125). What is its general equation? Its 3rd term?

solution

common ratio(r)=4/5

eighth term(G8)=393216/3125

From Gn= G1r^(n-1)

G8 means n=8

G8=G1r^(n-1)

393216/3125=G1(4/5)^(8-1)

393216/3125=G1(4/5)^7

G1=(393216/3125)/(4/5)^7

G1=600

The first term is given by G1=600

Therefore

The General equation is given by

The General equation is given by Gn= 600(4/5)^(n-1)

3rd term (G3)

G3= G1(4/5)^(3-1) where n=3,

=600(4/5)^2

=600(16/25)

=384

Therefore, the 3rd term is given by G3= 384.

I have made some Corrections i messed up somewhere.

7 0

3 years ago

Could someone help me with this?? thanks!

ANEK [815]

Answer:

Step-by-step explanation:

1/4 ( x+10) = x -4

one fourth the sum of x and 10 is identical to x minus 4

5 0

3 years ago

Number two please ASAP

miv72 [106K]

The solutions for $-5 +2x^{2} = -6x$ is option 3. $x= \frac{-6 \pm \sqrt{36-4 (2)(-5)}}{4} .$

Step-by-step explanation:

Step 1:

First, we must bring the equation to the form of $ax^{2} +bx +c =0.$

So $-5 +2x^{2} = -6x$ becomes $2x^{2} +6x-5=0.$

The value of a is the coefficient of the $x^{2}$ term, the value of b is the coefficient of x term and c is the coefficient of the constant term.

Comparing the above equation to $ax^{2} +bx +c =0,$ we get $a = 2, b = 6,$ and $c=-5.$

We have the formula $x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}.$

Step 2:

By substituting the known values, we get

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} =\frac{-6 \pm \sqrt{6^{2}-4 (2)(-5)}}{2 (2)}.$

$\frac{-6 \pm \sqrt{6^{2}-4 (2)(-5)}}{2 (2)} = \frac{-6 \pm \sqrt{36-4 (2)(-5)}}{4} .$

This is the third option.

8 0

4 years ago

What is the standard form of five and three hundred fifty two thousands

suter [353]

The number "seven hundred and four thousand, six hundred and twenty" is written as 704,620 in numeric or standard form. In converting a number expressed in words to numerals, it is important to write each digit in its proper place value.

7 0

3 years ago