How would you write 0.00000.79512 in scientific notation

R, S, Q and P are the midpoints of OC, CB, BA and OA respectively.

Stolb23 [73]

Answer:

See below.

Step-by-step explanation:

Draw segment OB.

In triangle OBC, points R and S are the midpoints of sides OC and BC, respectively. That makes RS parallel to OB.

In triangle OBA, points P and Q are the midpoints of sides OA and BA, respectively. That makes PQ parallel to OB.

Since segments RS and PQ are parallel to segment OB, then RS and PQ are parallel to each other.

3 0

2 years ago

Teesha is in French club. There are 10 freshman, 12 sophomores, 15 juniors, and 30 seniors in the club. The advisor is going to

FrozenT [24]

We know that

In French club there are
10 freshman
12 sophomores
15 juniors
30 seniors

total of the members------> (10+12+15+30)=67
total freshman-----> 10

so
the probability that a freshman will be chosen=10/67
and
the probability that a freshman will not be chosen=(67-10)/67
the probability that a freshman will not be chosen=57/67---> 0.8507
0.8507= 85.07%

the answer is
the probability that a freshman will not be chosen is 85.07%

6 0

3 years ago

Last year , there were 30 male members and 20 female members in a Chess Club . This year , the number of male members is decreas

LekaFEV [45]

The r% of a quantity x is computed by dividing x in 100 parts, and considering r of such parts. So, the r% of the male is

$30\times\cfrac{r}{100} = 0.3 r$

and similarly, the r% of female is

$20\times\cfrac{r}{100} = 0.2 r$

The number of males decreased by this quantity, so now it is

$30 - 0.3r$

and the number of female increased by this quantity, so now it is

$20+0.2r$

we know that these two new counts are the same number, so we can build and solve the equality

$30 - 0.3r = 20+0.2r$

Subtract 20 and add 0.3r from both sides:

$10 = 0.5r$

Divide both sides by 0.5 to solve for r:

$r = 20$

Let's check the answer

The 20% of 30 is $30 \times \frac{20}{100} = 6$ , while the 20% of 20 is 4. So, we are stating that $30-6 = 20+4$ which is true because both expressions evaluate to 24.

7 0

3 years ago

Tell whether the angles are complementary or supplementary. Then find the value of x.

Genrish500 [490]

Answer:

x = 33

Step-by-step explanation:

Two angles are complementary as the sum of two angles is 90

x + 24 + x = 90 {Combine like terms}

2x + 24 = 90 {Subtract 24 from both sides}

2x = 90 - 24

2x = 66

x = 66/2

x = 33

7 0

2 years ago

The curve

kherson [118]

Answer:

Point N(4, 1)

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality

Algebra I

Coordinates (x, y)
Functions
Function Notation
Terms/Coefficients
Anything to the 0th power is 1
Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$
Exponential Rule [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

Calculus

Derivatives

Derivative Notation

Derivative of a constant is 0

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

Step 1: Define

$\displaystyle y = \sqrt{x - 3}$

$\displaystyle y' = \frac{1}{2}$

Step 2: Differentiate

[Function] Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle y = (x - 3)^{\frac{1}{2}}$
Chain Rule: $\displaystyle y' = \frac{d}{dx}[(x - 3)^{\frac{1}{2}}] \cdot \frac{d}{dx}[x - 3]$
Basic Power Rule: $\displaystyle y' = \frac{1}{2}(x - 3)^{\frac{1}{2} - 1} \cdot (1 \cdot x^{1 - 1} - 0)$
Simplify: $\displaystyle y' = \frac{1}{2}(x - 3)^{-\frac{1}{2}} \cdot 1$
Multiply: $\displaystyle y' = \frac{1}{2}(x - 3)^{-\frac{1}{2}}$
[Derivative] Rewrite [Exponential Rule - Rewrite]: $\displaystyle y' = \frac{1}{2(x - 3)^{\frac{1}{2}}}$
[Derivative] Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle y' = \frac{1}{2\sqrt{x - 3}}$

Step 3: Solve

Find coordinates

x-coordinate

Substitute in y' [Derivative]: $\displaystyle \frac{1}{2} = \frac{1}{2\sqrt{x - 3}}$
[Multiplication Property of Equality] Multiply 2 on both sides: $\displaystyle 1 = \frac{1}{\sqrt{x - 3}}$
[Multiplication Property of Equality] Multiply √(x - 3) on both sides: $\displaystyle \sqrt{x - 3} = 1$
[Equality Property] Square both sides: $\displaystyle x - 3 = 1$
[Addition Property of Equality] Add 3 on both sides: $\displaystyle x = 4$