All categories

3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWERS ONLY PLEASE!!! I CANNOT RETAKE THIS!!

Mathematics

1 answer:

Mama L [17]3 years ago

5 0

let b =x^2

b^2+10b+25

b^2+5b+5b+25

b(b+5)+5(b+5)

(b+5)^2

therefore

(x^2+5)^2

You might be interested in

Catron evaluates the expression (negative 9) (2 and two-fifths) using the steps below.

Vlad [161]

Answer:

Catron's error is

"She did not follow order of operations"

Step-by-step explanation:

Catron evaluates the expression (negative 9) (2 and two-fifths)

That expression can be written as below

$(-9)(2\frac{2}{5})$

Catron's error is

"She did not follow order of operations"

The corrected steps are

Step1: Given expression is $(-9)(2\frac{2}{5})$

Step2: Convert mixed fraction into improper fraction

$(-9)(2\frac{2}{5})=(-9)(\frac{12}{5})$

Step3: Multiplying the terms

$(-9)(2\frac{2}{5})=-7\frac{-108}{5}$

Therefore solution $(-9)(2\frac{2}{5})=-7\frac{-108}{5}$

6 0

3 years ago

Read 2 more answers

The temperature was 61 degrees at sunrise.

Olin [163]

The answer would be 73 degrees because 62 +12 = 73

5 0

3 years ago

Prove that : sinA/1+cosA + 1+cosA/sinA=2cosecA

tekilochka [14]

Step-by-step explanation:

$\frac{sinA}{1+cosA} +\frac{1+cos}{sinA} =2cosec\\\\\\\frac{sin^{2}A+(1+cos)^{2} }{sinA(1+cosA)} \\\\$

$\frac{sin^{2}A+1+cos^{2}+2cos}{sinA(1+cosA)} \\\\$ and we have $sin^{2}+cos^{2} =1$

so $\frac{1+1+2cosA }{sinA(1+cosA)} \\\\$

$\frac{2+2cosA }{sinA(1+cosA)} \\\\$

$\frac{2(1+cosA)}{sinA(1+cosA)} \\\\$

$\frac{2}{sinA}$ we have $\frac{1}{sin}=cosec\\$

Finally $2cosecA$

I really hope this helps coz it took much time

3 0

3 years ago

I need a help please

asambeis [7]

Answer:

Step-by-step explanation:

1+2->12^-4=1243

8 0

3 years ago

Fine the probability. I NEED THE STEPS IF POSSIBLE

miss Akunina [59]

Answer:

1/6x1/2=1/12

Step-by-step explanation:

3 0

3 years ago