Need help getting answer

4+3(15-2^(3)) Simplify the equation....

miskamm [114]

Answer:

Step-by-step explanation:

$4+3\left(15-2^3\right)\\\\3\left(15-2^3\right)=21\\\\=4+21\\\\=25$

6 0

3 years ago

Read 2 more answers

How do i solve 12×+9=10×1

s2008m [1.1K]

Answer:

Step-by-step explanation:

Unfortunately, your " 12×+9=10×1 " can be interpreted in more than one way. The symbol " × " should be used ONLY to indicate multiplication. Thus, " 12× " reads as ' 12x.' Then we have: 12x + 9 = 10. Unsure of what you meant by " 10×1". So I will treat your "10×1" as simply '10.'

Then we have 12x + 9 = 10, which reduces to 12x = 1 if we subtract 9 from both sides.

12x = 1 yields x = 1/12.

If this is not the result you expected, please go back to the original problem and share it as accurately as you can.

6 0

3 years ago

Find the distance between (4,6) and (-2, -2)

Nataly [62]

Answer:

$\displaystyle d = 10$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra II

Distance Formula: $\displaystyle d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Step-by-step explanation:

Step 1: Define

Point (4, 6)

Point (-2, -2)

Step 2: Find distance d

Simply plug in the 2 coordinates into the distance formula to find distance d

Substitute [DF]: $\displaystyle d = \sqrt{(-2-4)^2+(-2-6)^2}$
Subtract: $\displaystyle d = \sqrt{(-6)^2+(-8)^2}$
Exponents: $\displaystyle d = \sqrt{36+64}$
Add: $\displaystyle d = \sqrt{100}$
Evaluate: $\displaystyle d = 10$

8 0

3 years ago

Read 2 more answers

The lateral edge PA of a triangular pyramid ABCP is perpendicular to the base and PA = 2sqrt5 in. The base edges AB = 10 in, AC

DIA [1.3K]

Answer:

PQ = √84 = 2√21 in ≈ 9.165 in

Step-by-step explanation:

The base edges AB = 10 in, AC = 17 in, and BC = 21 in

Q ∈ BC and AQ is the altitude of the base.

Let BQ = x in ⇒ CQ = (21 - x) in

ΔAQB a right triangle at Q

∴ AQ² = AB² - BQ² = 10² - x² ⇒(1)

ΔAQC a right triangle at Q

∴ AQ² = AC² - CQ² = 17² - (21-x)² ⇒(2)

Equating (1) and (2)

∴ 10² - x² = 17² - (21-x)²

∴ 10² - x² = 17² - (21² - 42x + x²)

∴ 10² - x² = 17² - 21² + 42x - x²

∴ 10² - 17² + 21² = 42x

∴ 42x = 252

∴ x = 252/42 = 6

Substitute at (1)

∴ AQ² = AB² - BQ² = 10² - x² = 100 - 36 = 64 = 8²

∴ AQ = 8 in

∵ The lateral edge PA of a triangular pyramid ABCP is perpendicular to the base

∴ PA⊥AQ

∴ ΔPAQ is a right triangle at A,

PA = 2sqrt5 in and AQ = 8 in

∴ PQ (hypotenuse) = √(PA² + AQ²)

∴ PQ = √[(2√5)² + 8²] = √(20+64) = √84 = 2√21 in ≈ 9.165 in

6 0

3 years ago

Computer software generated 500 random numbers that should look like they are from the uniform distribution on the interval 0 to

Tanya [424]

Answer:

0.2 ; 100 ; 4.84

Step-by-step explanation:

Given that the probability of each of the 5 groups is the same :

Sum of probability = 1

Hence, Probability of each group = 1 / number of groups = 1 / 5 = 0.2

Expected number for each interval for a sample of 500 : ; X = 500

E(X) = X * P(x) = 500 * 0.2 = 100

Goodness of fit (X²) :

X² = Σ(X - E)² ÷ E

Groups :

113, 95, 108, 99, and 85

X : 113 ____ 95 ____ 108 ____ 99 _____ 85

(113 - 100)^2 / 100 = 1.69

(95 - 100)^2 / 100 = 0.25

(108 - 100)^2 / 100 = 0.64

(99 - 100)^2 / 100 = 0.01

(85 - 100)^2 / 100 = 2.25

(1.69 + 0.25 + 0.64 + 0.01 + 2.25) = 4.84

7 0

3 years ago