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3 years ago

What is the solution to - 3 = w - 7 ? I need to show work please help

Mathematics

1 answer:

lisov135 [29]3 years ago

6 0

Answer:

w = 4

Step-by-step explanation:

You have to switch the negative 7 into a positive, and then add it to the negative 3

-3 = w - 7

+7 +7

4 = w

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Solve for x in each parallelogram. Do NOT include units (degrees, inches, feet, etc).

NemiM [27]

Answer:

x = 2

Step-by-step explanation:

As we can see that the diagonals bisects so

LN = 20

UN = 20 /2 = 10

So Now

UN = 5x

10 = 5x

x = 10 / 5

x = 2

7 0

4 years ago

I need some assistance with this

Norma-Jean [14]

A is 45° so if a to b is 8 cm than C to b is 16 and c to a is 14 so the answer is (30 for ur perimeter

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3 years ago

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HELP ASAP photo attached

ankoles [38]

Answer:

D. undefined

General Formulas and Concepts:

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

Trig Derivative: $\displaystyle \frac{d}{dx}[sinu] = u'cosu$

Derivatives of Parametrics: $\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

Step-by-step explanation:

Step 1: Define

$\displaystyle \frac{dx}{dt} = 5$

$\displaystyle \frac{dy}{dt} = sin(t^2)$

Step 2: Differentiate

[x Derivative] Basic Power Rule: $\displaystyle \frac{d^2x}{dt^2} = 0$
[y Derivative] Trig Derivative [Chain Rule]: $\displaystyle \frac{d^2y}{dt^2} = cos(t^2) \cdot \frac{d}{dt}[t^2]$
[y Derivative] Basic Power Rule: $\displaystyle \frac{d^2y}{dt^2} = cos(t^2) \cdot 2t^{2 - 1}$
[y Derivative] Simplify: $\displaystyle \frac{d^2y}{dt^2} = 2tcos(t^2)$
[Derivative] Rewrite: $\displaystyle \frac{d^2y}{dx^2} = \frac{2tcos(t^2)}{0}$