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3 years ago

Here is my question

Mathematics

1 answer:

arsen [322]3 years ago

5 0

Answer:

A. 2[24÷ (4+4)]

Step-by-step explanation:

Hope this helps :)

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Let f(x) = 3x+5 and g(x) = x^2. Find g(x)/f(x) and state its domain

lianna [129]

Answer:

1/x-2; domain is the set of all real numbers except 2

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

What is the area of an equilateral triangle that has a perimeter of 36 centimeters? Round to the nearest square centimeter.

Vika [28.1K]

Answer:

(a is size of side of triangle)

the perimeter of equilateral triangle =3a

or, 36=3a

Thus, a = 13cm

area of equilateral triangle

=√3/4*(a^2 )

=√3/4*(13^2)

=73cm^2

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2 years ago

If x+1 and x-1 are the factors of the polynomial ax^3 + x^2 - 2x +b,find the values of a and b

Nadusha1986 [10]

The polynomial remainder theorem says that dividing a polynomial $p(x)$ by $x-c$ leaves a remainder of $p(c)=0$ if $x-c$ is a factor of $p(x)$ . In this case, check $c=-1$ and $c=1$ .

$a(-1)^3+(-1)^2-2(-1)+b=-a+b+3=0$

$a(1)^3+(1)^2-2(1)+b=a+b-1=0$

From the first equation,

$-a+b+3=0\implies a=b+3$

and substituting into the second gives

$(b+3)+b-1=0\implies2b+2=0\implies b=-1\implies a=2$

3 0

3 years ago

Write two numbers so that the numbers are in order _ ;6,010;_ ;5,060

Nastasia [14]

6,050 before 6010 then 5080 after 5060 or basically any numbers that are greater than the one to the left of it :))

5 0

3 years ago

Pls help if pls do most important 15 number 20 number 26 number 28 number 30 number 17 number 25 number 22 number 24 number if u

Tamiku [17]

Answer:

Step-by-step explanation:

15. $\frac{cotx+1}{cotx-1} = \frac{cotx+cotx.tanx}{cotx-cotx.tanx} = \frac{cotx(1+tanx)}{cotx(1-tanx)} = \frac{1+tanx}{1-tanx} \\$

16. $\frac{1+cos\alpha }{sin\alpha } = \frac{sin\alpha }{1-cos\alpha }$

$=> (1+cos\alpha )(1-cos\alpha ) = sin^{2} \alpha \\=> 1-cos^{2}\alpha =sin^{2} \alpha \\=> sin^{2}\alpha +cos^{2}\alpha =1\\$

=> The clause is correct

17. Do the same (16)

18. $\frac{sinx}{1-cosx}+\frac{sinx}{1+cosx} = \frac{sinx(1+cosx) + sinx(1-cosx)}{(1+cosx)(1-cosx)} = \frac{sinx + sinx.cosx + sinx - sinx.cosx}{1-cos^{2}x} = \frac{2sinx}{sin^{2}x }= \frac{2}{sinx} = 2cosecx$

19.

$\frac{sinA}{1+cosA} + \frac{1+cosA}{sinA}=\frac{2}{sinA} \\=> \frac{sinA}{1+cosA} + \frac{1+cosA}{sinA} - \frac{2}{sinA} \\ = 0\\=> \frac{sinA}{1+cosA} + (\frac{1+cosA}{sinA} - \frac{2}{sinA} \\) = 0\\=> \frac{sinA}{1+cosA} + \frac{1+cosA-2}{sinA} = 0\\=> \frac{sinA}{1+cosA} +\frac{cosA-1}{sinA} = 0\\=> \frac{sin^{2} A+(cosA-1)(1+cosA)}{(1+cosA)sinA}=0\\=> \frac{sin^{2} A+cos^{2} A-1}{(1+cosA)sinA}=0\\=> \frac{0}{(1+cosA)sinA} =0\\$

=> The clause is correct

20. Do the same (18)

21. Left = $\frac{1}{cosecA+cotA} = \frac{1}{\frac{1}{sinA}+\frac{cosA}{sinA} } = \frac{1}{\frac{1+cosA}{sinA} }= \frac{sinA}{1+cosA}$

Right = $cosecA-cotA = \frac{1}{sinA}-\frac{cosA}{sinA}= \frac{1-cosA}{sinA} \\$

Same with (16) => Left = Right => The clause is correct

22. Do the same (21)

Too long, i'm so lazy :))))

5 0

3 years ago

Here is my question ​

Here is my question