All categories

2 years ago

Solve c: P = a+b+c+d / 4

Mathematics

1 answer:

m_a_m_a [10]2 years ago

8 0

Answer:

as written: c = P - a - b - d/4
with parentheses: c = 4P - a - b - d

Step-by-step explanation:

The meaning of the given expression is ...

P = a + b + c + (d/4)

To solve for c, subtract all the terms on the right side not containing c.

P -(a + b + (d/4)) = c

c = P - a - b - (d/4)

_____

In such equations, parentheses are commonly missing. If that is the case here, then first we undo the division by 4, then we subtract the "not c" terms.

P = (a + b + c + d)/4 . . . . maybe what you meant

4P = a + b + c + d . . . . . . multiply by 4

4P - (a +b +d) = c

c = 4P -a -b -d

You might be interested in

What is the product of two odd numbers and a multiple of 2

ASHA 777 [7]

Answer: a number

Explanation: choose the odd numbers and I’ll answer but right now how bout you ask my hood friend google

5 0

3 years ago

Read 2 more answers

Parallelogram ABCD has vertices at A(-2,3), B(4,3), C(-1,1). What is the length of side AB? Answer choices ;

telo118 [61]

Answer:

D)6 units

Step-by-step explanation:

the answer is D because since side AB have the same y coordinate then you would have to either subtract or add the x. if one point on x is negative and the other is positive then you would add. but if they were both positive or both negative then you would subtract the x.

5 0

2 years ago

8n+?=0 plz solve i will mark brainliest

Ksivusya [100]

Answer:

negative 8

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Please helpppppppppppp

Gekata [30.6K]

D. The measure of TRU is 72

4 0

2 years ago

Given that cot θ = 1/√5, what is the value of (sec²θ - cosec²θ)/(sec²θ + cosec²θ) ?

Bogdan [553]

Step-by-step explanation:

$\mathsf{Given :\;\dfrac{{sec}^2\theta - co{sec}^2\theta}{{sec}^2\theta + co{sec}^2\theta}}$

$\bigstar\;\;\textsf{We know that : \large\boxed{\mathsf{{sec}\theta = \dfrac{1}{cos\theta}}}}$

$\bigstar\;\;\textsf{We know that : \large\boxed{\mathsf{co{sec}\theta = \dfrac{1}{sin\theta}}}}$

$\mathsf{\implies \dfrac{\dfrac{1}{cos^2\theta} - \dfrac{1}{sin^2\theta}}{\dfrac{1}{cos^2\theta} + \dfrac{1}{sin^2\theta}}}$

$\mathsf{\implies \dfrac{\dfrac{sin^2\theta - cos^2\theta}{sin^2\theta.cos^2\theta}}{\dfrac{sin^2\theta + cos^2\theta}{sin^2\theta.cos^2\theta}}}$

$\mathsf{\implies \dfrac{sin^2\theta - cos^2\theta}{sin^2\theta + cos^2\theta}}$

Taking sin²θ common in both numerator & denominator, We get :

$\mathsf{\implies \dfrac{sin^2\theta\left(1 - \dfrac{cos^2\theta}{sin^2\theta}\right)}{sin^2\theta\left(1 + \dfrac{cos^2\theta}{sin^2\theta}\right)}}$