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Natasha_Volkova [10]
3 years ago
8

Written equation for y=-2=1 (0,3)

Mathematics
1 answer:
Alika [10]3 years ago
4 0
Is that supposed to be y=-2x+1?
i cannot solve without the proper equation lol 
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A ________ function specifies the relationship between the demand for a product LaTeX: x and the price of the product LaTeX: p.
Archy [21]

Answer:

Price demand function

Step-by-step explanation:

The price demand function is the function in which there is a relationship between the demand and the price of the product

The Price demand function is

= Price \times demand

= p \times x

hence, the price demand function represents that there is a relationship that lies between the demand and the price

3 0
2 years ago
Can you please help me solve this I need it before 12/21/20 please help
Triss [41]

Answer:

I only know that the first box is and it is counting by twos

Step-by-step explanation:

because 5+2 =7 then 6+4=10 then 7+6=13 then 8+8=16 and lastly 9+10=19

7 0
3 years ago
Three Questions 1) What does 5 7/8 - 2 3/8 equal? and 2)What does 5 7/12 -4 1/12 equal?3) What does 3 5/10 - 1 3/10 equal?
stiks02 [169]
1) 28/8=3 1/2
2) 18/12=1 1/2
3) 22/10=2 1/5
7 0
3 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
Aubree invested $8,300 in an account paying an interest rate of 2.2% compounded quarterly. Assuming no deposits or withdrawals a
Triss [41]

Answer:

$25,735.03

Step-by-step explanation:

I plugged your numbers into the really big formula

7 0
3 years ago
Read 2 more answers
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