Find the interquartile range! PLEASE HELP!!!!!!

Please solve for x. 43x+1 = 45

Jet001 [13]

Answer:

x=44/43

Step-by-step explanation:

Subtract 1 from both sides

Simplify

Divide both sides by the same term

Cancel terms that are in both the numerator and the denominator

7 0

3 years ago

-9=k-6

Shkiper50 [21]

Answer:

k = -3

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Step-by-step explanation:

Step 1: Define

-9 = k - 6

Step 2: Solve for k

Add 6 to both sides: -3 = k
Rewrite: k = -3

3 0

3 years ago

If the numbers below were ordered from least to greatest, which number could you use to replace the blank? 0.20,2/5 □ , 3/4,8/10

faust18 [17]

1/2 would work for the blank space

because as shown in the image you could relate it to a whole for each one and put it to 100 to figure out what the blank could be

the 0.20 can be converted to 1/5 so that way you can put everything into the 100 group

8 0

4 years ago

Find the volume of the square pyramid showing. round to the nearest 10th if necessary

solong [7]

The volume of a square pyramid is (1/3)(area of base)(height of pyramid).

Here the area of the base is (10 ft)^2 = 100 ft^2.

13 ft is the height of one of the triangular sides, but not the height of the pyramid. To find the latter, draw another triangle whose upper vertex is connected to the middle of one of the four equal sides of the base by a diagonal of length 13 ft. That "middle" is 5 units straight down from the upper vertex. Thus, you have a triangle with known hypotenuse (13 ft) and known opposite side 5 feet (half of 10 ft). What is the height of the pyramid?
To find this, use the Pyth. Thm.: (5 ft)^2 + y^2 = (13 ft)^2. y = 12 ft.

Then the vol. of the pyramid is (1/3)(area of base)(height of pyramid) =
(1/3)(100 ft^2)(12 ft) = 400 ft^3 (answer)

5 0

4 years ago

Which of the following is equivalent to tan2θcos(2θ) for all values of θ for which tan2θcos(2θ) is defined?

Aloiza [94]

Answer:

2sin²θ - tan²θ

Step-by-step explanation:

Given

tan²θcos(2θ)

Required

Simplify

We start by simplifying cos(2θ)

cos(2θ) = cos(θ+θ)

From Cosine formula

cos(A+A) = cosAcosA - sinAsinA

cos(A+A) = cos²A - sin²A

By comparison

cos(2θ) = cos(θ+θ)

cos(2θ) = cos²θ - sin²θ ----- equation 1

Recall that cos²θ + sin²θ = 1

Make sin²θ the subject of formula

sin²θ = 1 - cos²θ

Substitute sin²θ = 1 - cos²θ in equation 1

cos(2θ) = cos²θ - (1 - cos²θ)

cos(2θ) = cos²θ - 1 +cos²θ

cos(2θ) = cos²θ + cos²θ - 1

cos(2θ) = 2cos²θ - 1

Substitute 2cos²θ - 1 for cos(2θ) in the given question

tan²θcos(2θ) becomes

tan²θ(2cos²θ - 1)

Open brackets

2cos²θtan²θ - tan²θ

------------------------

Simplify tan²θ

tan²θ = (tanθ)²

Recall that tanθ = sinθ/cosθ

So, we have

tan²θ = (sinθ/cosθ)²

tan²θ = sin²θ/cos²θ

------------------------

Substitute sin²θ/cos²θ for tan²θ

2cos²θtan²θ - tan²θ becomes

2cos²θ(sin²θ/cos²θ) - tan²θ

Open bracket (cos²θ will cancel out cos²θ) to give

2(sin²θ) - tan²θ

2sin²θ - tan²θ

Hence, the simplification of tan²θcos(2θ) is 2sin²θ - tan²θ

Option E is correct

7 0

3 years ago