All categories

2 years ago

43. ALGEBRA Assume that your current

Mathematics

2 answers:

RoseWind [281]2 years ago

7 0

Answer:

10582

Step-by-step explanation:

ASHA 777 [7]2 years ago

3 0

Answer:

48 hours

Step-by-step explanation:

945 per week

18.50x44=814

18.50+13=31.50

945-814=130

130 divided by 31.50 = 4.13

You might be interested in

How do I solve this math question ?

svlad2 [7]

You do the square root of 75
and then with that number you divide it by 5
and that is your answer

3 0

3 years ago

Read 2 more answers

Help ( correct answer ) 16÷(4)(2)-3(2)÷2+1

Bess [88]

16 ÷ (4)(2) - 3(2) <span>÷ 2+1
16 </span>÷ 8 - 6 <span>÷ 3
2 - 2
0

Answer: 0</span>

3 0

2 years ago

Read 2 more answers

I need help with this one.

Mandarinka [93]

Answer:

$T_{n}$ = $\frac{1}{5}$ $T_{n-1}$ with T₁ = 5000

Step-by-step explanation:

There is a common ratio r between consecutive terms in the sequence, that is

r = $\frac{1000}{5000}$ = $\frac{200}{1000}$ = $\frac{40}{200}$ = $\frac{1}{5}$

This indicates the sequence is geometric with recursive rule

$T_{n}$ = r $T_{n-1}$

Here tr = $\frac{1}{5}$ , thus

$T_{n}$ = $\frac{1}{5}$ $T_{n-1}$ , T₁ = 5000

3 0

2 years ago

Which equation is not equivalent to the other equations?

Colt1911 [192]

I think the answer is C. X*4=-12+3

8 0

2 years ago

) f) 1 + cot²a = cosec²a

notsponge [240]

Answer:

It is an identity, proved below.

Step-by-step explanation:

I assume you want to prove the identity. There are several ways to prove the identity but here I will prove using one of method.

First, we have to know what cot and cosec are. They both are the reciprocal of sin (cosec) and tan (cot).

$\displaystyle \large{\cot x=\frac{1}{\tan x}}\\\displaystyle \large{\csc x=\frac{1}{\sin x}}$

csc is mostly written which is cosec, first we have to write in 1/tan and 1/sin form.

$\displaystyle \large{1+(\frac{1}{\tan x})^2=(\frac{1}{\sin x})^2}\\\displaystyle \large{1+\frac{1}{\tan^2x}=\frac{1}{\sin^2x}}$

Another identity is:

$\displaystyle \large{\tan x=\frac{\sin x}{\cos x}}$

Therefore:

$\displaystyle \large{1+\frac{1}{(\frac{\sin x}{\cos x})^2}=\frac{1}{\sin^2x}}\\\displaystyle \large{1+\frac{1}{\frac{\sin^2x}{\cos^2x}}=\frac{1}{\sin^2x}}\\\displaystyle \large{1+\frac{\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}}$

Now this is easier to prove because of same denominator, next step is to multiply 1 by sin^2x with denominator and numerator.

$\displaystyle \large{\frac{\sin^2x}{\sin^2x}+\frac{\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}}\\\displaystyle \large{\frac{\sin^2x+\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}$

Another identity:

$\displaystyle \large{\sin^2x+\cos^2x=1}$

Therefore:

$\displaystyle \large{\frac{\sin^2x+\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}\longrightarrow \boxed{ \frac{1}{\sin^2x}={\frac{1}{\sin^2x}}}$

Hence proved, this is proof by using identity helping to find the specific identity.

6 0

2 years ago