All categories

3 years ago

Is brainly getting hacked?

Mathematics

2 answers:

Tems11 [23]3 years ago

8 0

I don’t think so
Hope that help

Anni [7]3 years ago

7 0

Answer:

I think so

Step-by-step explanation:

You might be interested in

Write an equation for the linear function whose graph contains the points (-1 ,-2) and (3, 10)?

Fofino [41]

Answer:

y+2 =3(x+1)

Step-by-step explanation:

Firstly I will find slope of this linear function:

m=(y-yo) / (x-xo)

m=(-2-10)/(-1-3)

m=-12/-4

m=3

Replacing A(-1-2) :

(y-yo)=m(x-xo)

3 0

3 years ago

547321 rounded to nearest one thousand

Sphinxa [80]

The nearest thousand is 547,000

5 0

3 years ago

I WILL GIVE BRAINLIEST PLS HELP PLS PLS PLS

bija089 [108]

Answer:

1. 1.1 times the weight of the puppy

2. 11/10

3. %110

Step-by-step explanation:

11 x 9 = 99

99/90=1.1

1.1 = 11/10 = %110

5 0

3 years ago

Read 2 more answers

Can you help solve the equation 2x=3y+10

zimovet [89]

3/2y+5 is the answer :)

8 0

3 years ago

Simplify the first trigonometric expression by writing the simplified form in terms of the second expression (1/1-cosx)-(cos/1+c

garik1379 [7]

$(\frac{1}{1-cos(x)})-(\frac{cos(x)}{1+cos(x)})$

cscx is 1/sinx

maybe they want us to use pythagorean identity

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

I notice we have 1-cos(x) and 1+cos(x), if we multiply them, we get $1-cos^2(x)$

and if we look at the pythagorean identity and minus cos^2(x) from both sides, we get

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

since $csc(x)=\frac{1}{sin(x)}$ , $csc^2(x)=\frac{1}{sin^2(x)}=\frac{1}{1-cos^2(x)}$

(recall that (a-b)(a+b)=a²-b²)

match the denomenators of the original fraction

multiply first fraction by $\frac{1+cos(x)}{1+cos(x)}$ and the 2nd by $\frac{1-cos(x)}{1-cos(x)}$

$(\frac{1+cos(x)}{1-cos^2(x)})-(\frac{cos(x)(1-cos(x))}{1-cos^2(x)})=$

$((csc^2(x))(1+cos(x)))-((csc^2(x))(cos(x)-cos^2(x)))=$

$csc^2(x)+csc^2(x)cos(x)-(csc^2(x)cos(x)-csc^2(x)cos^2(x)=$

$csc^2(x)+csc^2(x)cos(x)-csc^2(x)cos(x)+csc^2(x)cos^2(x)=$

$csc^2(x)+csc^2(x)cos^2(x)=$

$csc^2(x)(1+cos^2(x))$ , hmm, to get ride of those cos(x)

look to the pythagorean identity again

$sin^2(x)+cos^2(x)=1$ , force one side into form 1+cos^2(x)

$1+cos^2(x)=2-sin^2(x)$ , recall that since csc(x)=1/sin(x), sin(x)=1/csc(x) and sin^2(x)=1/(csc^2(x))

$1+cos^2(x)=2-\frac{1}{csc^2(x)}$

subsituting

$csc^2(x)(1+cos^2(x))=$

$(csc^2(x))(2-\frac{1}{csc^2(x)})=$ distributing

$2csc^2(x)-\frac{csc^2(x)}{csc^2(x)}=$

$2csc^2(x)-1$ is the simplified expression

8 0

3 years ago