Really really need this answer. Please help!

Solve the inequality 3/5x+30 is less than or equal to 150

FrozenT [24]

Answer: less than

Step-by-step explanation:

8 0

2 years ago

1. You are given the 3rd and 5th term of an arithmetic sequence. Describe in words how to determine the general term.

kozerog [31]

Step-by-step explanation:

1. In an arithmetic sequence, the general term can be written as

xₙ = y + d(a-1), where xₐ represents the ath term, y is the first value, and d is the common difference.

Given the third term and the fifth term, and knowing that the difference between each term is d, we can say that the 4th term is x₃+d and the fifth term is the fourth term plus d, or (x₃+d)+d =

x₃+2d. =x₅ Given x₃ and x₅, we can subtract x₃ from both sides to get

x₅-x₃ = 2d

divide by 2 to isolate d

(x₅-x₃)/2 = d

This lets us solve for d. Given d, we can say that

x₃ = y+d(2)

subtract 2*d from both sides to isolate the y

x₃ -2*d = y

Therefore, because we know x₃ and d at this point, we can solve for y, letting us plug y and d into our original equation of

xₙ = y + d(a-1)

2.

Given the third and fifth term, with a common ratio of r, we can say that the fourth term is x₃ * r. Then, the fifth term is

x₃* r * r

= x₃*r² = x₅

divide both sides by x₃ to isolate the r²

x₅/x₃ = r²

square root both sides

√(x₅/x₃) = ±r

One thing that is important to note is that we don't know whether r is positive or negative. For example, if x₃ = 4 and x₅ = 16, regardless of whether r is equal to 2 or -2, 4*r² = 16. I will be assuming that r is positive for this question.

Given the common ratio, we can find x₆ as x₅ * r, x₇ as x₅*r², and all the way up to x₁₀ = x₅*r⁵. We don't know the general term, but can still find the tenth term of the sequence

8 0

3 years ago

. For each of these intervals, list all its elements or explain why it is empty. a) [a, a] b) [a, a) c) (a, a] d) (a, a) e) (a,

Eva8 [605]

Answer:

Elements are of the form

(i) $[a,a]=\{[x,y] : a\leq x\leq a, a\leq y\leq a; a\in \mathbb R\}$

(ii) $[a,b)=\{[x,y) :a\leq x$

(iii) $(a,a]=\{(x,y] :a$

(iv) $(a,a)=\{(x,y): a$

(v) (a,b) where a>b= $\{(x,y) : a>x>b,a>y>b;a>b,a,b \in \mathbb R\}$

(vi) [a,b] where a>b= $\{[x,y] : a\geq x\geq b,a\geq y\geq b;a>b,a,b \in \mathbb R\}$

Step-by-step explanation:

Given intervals are,

(i) [a,a] (ii) [a,a) (iii) (a,a] (iv) (a,a) (v) (a,b) where a>b (vi) [a,b] where a>b.

To show all its elements,

(i) [a,a]

Imply the set including aa from left as well as right side.

Its elements are of the form.

$\{[a,a] : a\in \mathbb R\}=\{[0,0],[1, 1],[-1,-1],[2,2],[-2,-2],[3,3],[-3,-3],........\}$

Since there is a singleton element a of real numbers, this set is empty.

Because there is no increment so if $a\in \mathbb R$ then the set [a,a] represents singleton sets, and singleton sets are empty so is [a,a].

(ii) [a,a)

This means given interval containing a by left and exclude a by right.

Its elements are of the form.

$[ 1, 1),[-1,-1),[2,2),[-2,-2),[3,3),[-3,-3),........$

Since there is a singleton element a of real numbers withis the set, this set is empty.

Because there is no increment so if $a\in \mathbb R$ then the set [a,a) represents singleton sets, and singleton sets are empty so is [a.a).

(iii) (a,a]

It means the interval not taking a by left and include a by right.

Its elements are of the form.

$( 1, 1],(-1,-1],(2,2],(-2,-2],(3,3],(-3,-3],........$

Since there is a singleton element a of real numbers, this set is empty.

Because there is no increment so if $a\in \mathbb R$ then the set (a,a] represents singleton sets, and singleton sets are empty so is (a,a].

(iv) (a,a)

Means given set excluding a by left as well as right.

Since there is a singleton element a of real numbers, this set is empty.

Its elements are of the form.

$( 1, 1),(-1,-1],(2,2],(-2,-2],(3,3],(-3,-3],........$

Because there is no increment so if $a\in \mathbb R$ then the set (a,a) represents singleton sets, and singleton sets are empty, so is (a,a).

(v) (a,b) where a>b.

Which indicate the interval containing a, b such that increment of x is always greater than increment of y which not take x and y by any side of the interval.

That is the graph is bounded by value of a and it contains elements like it we fixed a=5 then,

$(a,b)=\{(5,0),(5,1),(5,2).....\}$ e.t.c

So this set is connected and we know singletons are connected in $\mathbb R$ . Hence given set is empty.

(vi) [a,b] where $a\leq b$ .

Which indicate the interval containing a, b such that increment of x is always greater than increment of y which include both x and y.

That is the graph is bounded by value of a and it contains elements like it we fixed a=5 then,

$[a,b]=\{[5,0],[5,1],[5,2].....\}$ e.t.c

So this set is connected and we know singletons are connected in $\mathbb R$ . Hence given set is empty.

8 0

3 years ago

Look at the Hexagon below

Mama L [17]

Answer:

Angle 9: 60°

Angle 10: 30°

Side = radius = 40sqrt(x/3)

Area = [800sqrt(3)]x

Step-by-step explanation:

Total angle in a hexagon:

(6 - 2) × 180

720

Each interior angle:

720/6 = 120

angle 9 = 120/2 = 60

Angle 10 = 60/2 = 30

sin(60) = 20sqrt(x)/r

r = 20sqrt(x) ÷ sqrt(3)/2

r = 40sqrt(x/3)

Side:

sin(30) = ½s/(40sqrt(x/3))

½s = 20sqrt(x/3)

s = 40sqrt(x/3)

Area = (3sqrt(3))/2 × s²

Area = 3sqrt(3)/2 × 1600x/3

Area = [800sqrt(3)]x

4 0

3 years ago

Which is greater 5/8 or -1/4

Tpy6a [65]

Answer:

5/8

Step-by-step explanation:

-1/4=-0.25

5/8=0.625

8 0

3 years ago

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