All categories

4 years ago

Solve Please 4+8/2(6-3)

Mathematics

2 answers:

Aleksandr-060686 [28]4 years ago

7 0

Answer:

Step-by-step explanation:

4 + 8/2 x 3

4+4 x 3

4+12

Hope this helped

Brainliest is appreciated.

xeze [42]4 years ago

5 0

Answer: 2

Step-by-step explanation:

Solve equations in the parenthesis first: (6-3) = 3

4+8/2(3) - solve for 2(3) = 6

4+8/6 - solve for 4+8 = 12

12/6 = 2

You might be interested in

A student claims that 1,2,3 and 4 are the zeros of a cubic polynomial function. Explain why the student is mistaken. Give an exa

postnew [5]

A cubic function only has 3 solutions at which x = 0, never 4. For example the function y = $y = x^{3} + 3x^{2} + 2x$ gives the solutions:
x = 0, x = -1 and x = -2

8 0

4 years ago

Susie gains more weight over Christmas than any other time of the year. Christmas makes Susie gain weight. This conclusion is...

Dahasolnce [82]

Hello there!

Your answer is D) not justifiable, because there is another variable to consider.
While Susie may gain weight during Christmas, the holiday is not magically making her gain weight. It is more likely all of those candy canes.

I hope I helped!

Let me know if you need anything else!

~ Zoe

3 0

3 years ago

Read 2 more answers

What is the least common multiple of 4t³ and 6t⁴?

Rus_ich [418]

Answer:

i believe its 24 if i got it wrong sorry :c

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Sin(A+B) sin(A-B) /sin^A Cos^B=1-cot^A Tan^B

telo118 [61]

In order to prove

$\dfrac{\sin(x+y)\sin(x-y)}{\sin^2(x)\cos^2(x)}=1-\cot^2(x)\tan^2(y)$

Let's write both sides in terms of $\sin(x),\ \sin^2(x),\ \cos(x),\ \cos^2(x)$ only.

Let's start with the left hand side: we can use the formula for sum and subtraction of the sine to write

$\sin(x+y)=\cos(y)\sin(x)+\cos(x)\sin(y)$

and

$\sin(x-y)=\cos(y)\sin(x)-\cos(x)\sin(y)$

So, their multiplication is

$\sin(x+y)\sin(x-y)=(\cos(y)\sin(x))^2-(\cos(x)\sin(y))^2\\=\cos^2(y)\sin^2(x)-\cos^2(x)\sin^2(y)$

So, the left hand side simplifies to

$\dfrac{\cos^2(y)\sin^2(x)-\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}$

Now, on with the right hand side. We have

$1-\cot^2(x)\tan^2(y)=1-\dfrac{\cos^2(x)}{\sin^2(x)}\cdot\dfrac{\sin^2(y)}{\cos^2(y)} = 1-\dfrac{\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}$

Now simply make this expression one fraction:

$1-\dfrac{\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}=\dfrac{\sin^2(x)\cos^2(y)-\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}$

And as you can see, the two sides are equal.

6 0

4 years ago

WORTH 30 POINTS. Properties of Logarithm: Please help! I do not understand how to solve this kind of problem.

horrorfan [7]

Logb1=0 log b 1 = 0 . This follows from the fact that b0=1 b 0 = 1 .
logbb=1 log b b = 1 . This follows from the fact that b1=b b 1 = b .
logbbx=x log b b x = x . This can be generalized out to logbbf(x)=f(x) log b b f ( x ) = f ( x ) .

7 0

3 years ago