All categories

2 years ago

The value of w is what?

Mathematics

1 answer:

icang [17]2 years ago

7 0

Answer:

value of w is 25...... .....

You might be interested in

Please help i'm so confused

kherson [118]

$\bf \begin{cases}
x=0\implies &(x)=0\\
x=2\implies x-2=0\implies &(x-2)=0\\
x=-5\implies x+5=0\implies &(x+5)=0
\end{cases}
\\\\\\
(x)(x-2)(x+5)=\textit{3rd degree polynomial}$

8 0

3 years ago

A point is randomly selected on the grid. What is the probability that the point will be located in the shaded region? Give your

ICE Princess25 [194]

Answer:

how big is the shaded region tell me that and i will edit my answer to the right one

Step-by-step explanation:

7 0

3 years ago

Read 3 more answers

Lamar pays 15.60 for a pack of 6 towels. Find the unit price in dollars per towel. If necessary, round your answer to the neares

soldi70 [24.7K]

Divide total price by quantity:

15.60 / 6 = 2.60

Unit price = $2.60

7 0

1 year ago

What is the answer 5x² – 33x - 14 = 0

KatRina [158]

Answer:

X=-2/5,7

Step-by-step explanation:

5x² – 33x - 14 = 0

5x^2-(35-2)x-14=0

5x^2-35x+2x-14=0

5x(x-7)+2(x-7)=0

(5x+2)(x-7)=0

X=-2/5,7

so the value of X is -2/5,7

8 0

3 years ago

zubka84 [21]

Answer:

See Explanation Below

Step-by-step explanation:

Given

$(sin x - cos x)^2 = sec^2x - tan^2x - 2sinx.cos x.$

Required

Prove

To start with; we open the bracket on the left hand side

$(sin x - cos x)^2 = (sin x - cos x)(sin x - cos x)$

$(sin x - cos x)^2 = (sin x )(sin x - cos x)- (cos x)(sin x - cos x)$

$(sin x - cos x)^2 = sin^2 x -sinx cos x - sin xcos x + cos^2 x$

$(sin x - cos x)^2 = sin^2 x -2sinx cos x + cos^2 x$

Reorder

$(sin x - cos x)^2 = sin^2 x + cos^2 x - 2sinx cos x$

From trigonometry;

$sin^2x + cos^2x = 1$

So;

$(sin x - cos x)^2 = sin^2 x + cos^2 x - 2sinx cos x$

becomes

$(sin x - cos x)^2 = 1 - 2sinx cos x$

Also from trigonometry;

$sec^2x - tan^2x = 1$

So,

$(sin x - cos x)^2 = 1 - 2sinx cos x$

becomes

$(sin x - cos x)^2 = sec^2x - tan^2x - 2sinx cos x$

Proved

3 0

2 years ago