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3 years ago

How does deposition help explain the layers that are often found in sedimentary rock?

Mathematics

1 answer:

julia-pushkina [17]3 years ago

6 0

Because it is deposition

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If (x) = x+8 and g(x) = -4x-3, find (f+ g)(x).

Ede4ka [16]

Answer:

Answer=B

(f+g)(x)=-3x-5

Step-by-step explanation:

f(x)=x+8.

g(x)=-4x-3.

(f+g)(x) means f(x)+g(x),So

(x+8)+(-4x-3)

(add all same terms together)

=(x+(-4x))+(8+(-3))

=-3x+5

so,

(f+g)(x)=-3x+5

8 0

3 years ago

PQR is a right angled triangle calculate the size of angle marked x

romanna [79]

Answer:

$\displaystyle x \approx 37.4^\circ$

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

Trigonometry

[Right Triangles Only] SOHCAHTOA
[Right Triangles Only] cosθ = adjacent over hypotenuse

Step-by-step explanation:

Step 1: Define

Angle θ = x

Adjacent Leg = 5.8

Hypotenuse = 7.3

Step 2: Solve for x

Substitute in variables [Cosine]: $\displaystyle cosx^\circ = \frac{5.8}{7.3}$
[Fraction] Divide: $\displaystyle cosx^\circ = 0.794521$
[Equality Property] Trig inverse: $\displaystyle x^\circ = cos^{-1}(0.794521)$
Evaluate trig inverse: $\displaystyle x = 37.39^\circ$
Round: $\displaystyle x \approx 37.4^\circ$

8 0

2 years ago

Find the x and y intercepts. Write your answer as x = y = 2x + 5y = 10

Ann [662]

The x-intercept is present where y = 0

2x + 5y - 10 = 0
2x - 10 = -5y
2x - 10 = -5(0)
2x = 10
x = 5

The y-intercept is present where x = 0

2x + 5y = 10
2x + 5y - 10 = 0
5y - 10 = -2(0)
5y = 10
y = 2

6 0

3 years ago

Cual es la derivada de ()=√x sin

fgiga [73]

Answer:

$f(x) =\sqrt{x} sin (x)$

And on this case we can use the product rule for a derivate given by:

$\frac{d}{dx} (f(x)* g(x)) = f'(x) g(x) +f(x) g'(x)$

Where $f(x) =\sqrt{x}$ and $g(x) =sin (x)$

And replacing we have this:

$f'(x)= \frac{1}{2\sqrt{x}} sin (x) + \sqrt{x}cos(x)$

Step-by-step explanation:

We assume that the function of interest is:

$f(x) =\sqrt{x} sin (x)$

And on this case we can use the product rule for a derivate given by:

$\frac{d}{dx} (f(x)* g(x)) = f'(x) g(x) +f(x) g'(x)$

Where $f(x) =\sqrt{x}$ and $g(x) =sin (x)$

And replacing we have this:

$f'(x)= \frac{1}{2\sqrt{x}} sin (x) + \sqrt{x}cos(x)$

3 0

3 years ago

Find the interest generated if one invests $6000 for 9 years at 7% per annum with interest compounded monthly

marysya [2.9K]

Answer:

principle= 6000

time =9 years

rate = 7%

so,

compound interest = p((1+r/100)^t. -1)

= 6000((1+0.07)^9 -1)

=6000(1.84-1)

=6000*0.84

=5040

8 0

3 years ago