All categories

3 years ago

Help plz plz plz :)

Mathematics

1 answer:

murzikaleks [220]3 years ago

8 0

Answer:

x= 3/5 or x= 0.6

Step-by-step explanation:

You might be interested in

Which side of the equation 12.6v = 12 - 15.5v would you isolate the variable terms? Why?

lozanna [386]

The left side, because you only need to move one thing, and you don't need to subtract after that. (you can add, much more easier)

hope this helps

4 0

3 years ago

If cos A = 3 over 11, then which of the following is correct?

ziro4ka [17]

Answer: The answer is (a) sec A = 11 over 3.

Step-by-step explanation: Given that Cosine of an angle 'A' is 3 over 11,

i.e.,

$\cos A=\dfrac{3}{11}.$

And we need to find which one of the given four options is correct.

We have the following relations between cosine, secant and cosecant of an angle from trigonometry.

$\sec A=\dfrac{1}{\cos A}~~\textup{and}~~\csc A=\dfrac{1}{\sqrt{1-\cos^2 A}}.$

Therefore,

$\sec A=\dfrac{11}{3}$

and

$\csc A=\dfrac{1}{\sqrt{1-\frac{9}{121}}}=\dfrac{1}{\sqrt{\frac{112}{121}}}=\dfrac{11}{4\sqrt 7}.$

Thus, the correct option is (a) sec A = 11 over 3.

3 0

3 years ago

Can someone please help me I'm stuck on this question im in desperate need of help

kotegsom [21]

Answer:

Step-by-step explanation:

Point L's coordinates' difference is 7.

Point C's is 1.

Point H's is 2

And Point A's is 1

Hope that helps!

5 0

3 years ago

Read 2 more answers

Question 1,2, and 3 how do i factor those? Can you show the work and explain how?

DiKsa [7]

1: $3n^{2}+9n+6$

notice that each part is divisible by 3

$3n^{2}$ ÷ 3 = $n^{2}$

9n ÷ 3 = 3n

6 ÷ 3 = 2

so it becomes $3(n^{2} +3n+2)$

3n can be rewritten as 2n+n

-you want to rewrite it into two numbers that multiply to the number that's alone (in this case 2)

which would get you

$3(n^{2} +2n+n+2)$

Now that it's rewritten, you can factor out n + 2 from the equation.

the answer is

3(n+2)(n+1)

And you can check that by multiplying (n+2)(n+1) which is $n^{2} +2n+n+2$ and then each of those by 3, which is $3n^{2} +6n+3n+6$ or $3n^{2}+9n+6$ , our origional equation

2: $28+x^{2} -11x$

So I rewrote this as $x^{2} -11x+28$ (it's the same thing, just reordered using the commutative property)

now -11x can be rewritten as -4x-7x

(remember, the two numbers should multiply to equal 28, which is our constant.)

$x^{2} -4x-7x+28$

now we can factor out x from the first expression and -7 from the second

$x(x-4)-7(x-4)$

and lastly you factor out x-4,

which would give you

(x-4)(x-7)

Make sure to check your work and make sure it multiplies to $x^{2} -11x+28$

3: $9x^{2} -12x+4$

The first thing I notice when looking at this problem is that both 9 and 4 are perfect squares. Not only that, but they are the squares of 2 and 3, which are products of -12

So if you rewrite 9 as $3^{2}$ and 4 as $2^{2}$ , the equation becomes