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2 years ago

These 3 questions pleaseeee!

Mathematics

1 answer:

Sergio039 [100]2 years ago

5 0

Answer:

1. m<T =72 , m<U =54

2. EF= 18 in. , m<F=134'

3. x=4

Step-by-step explanation:

1. <s=<u , there is 180' in a triangle total so 180- <S - <U = <T

180 - 54 - 54 = 72

2. ef = gf

180 - 23 - 23 = 134

3. since bottom 2 angles are = , the 2 sides are also =

see attachment for work

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Decompose x^2+4x-3/x-2

mariarad [96]

Answer:

$\dfrac{x^3-2x^2+4x-3}{x-2}$

Step-by-step explanation:

The given equation is :

$x^2+\dfrac{4x-3}{x-2}$

We need to decompose this equation.

We have,

$x^2+\dfrac{4x-3}{x-2}\\\\=\dfrac{x^2(x-2)+4x-3}{x-2}\\\\=\dfrac{x^3-2x^2+4x-3}{x-2}$

So, $\dfrac{x^3-2x^2+4x-3}{x-2}$ is the decomposed form of the given expression.

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2 years ago

If r(x) = 2 – x2 and w(x) = x – 2, what is the range of ? (w0r)(x)

TiliK225 [7]

Answer:

The range would be all real numbers

Step-by-step explanation:

In order to find (wor)(x), we need to start with the w(x) equation and then input the r(x) equation for every x in the w(x) equation.

w(x) = x - 2

(wor)(x) = (2 - x^2) - 2

(wor)(x) = -x^2

Given this equation, we know that x can be all real numbers

4 0

3 years ago

Read 2 more answers

Find sin(a+b) if tan(a)=7/24 where a is in the third quadrant

ludmilkaskok [199]

The complete question in the attached figure

we have that
tan a=7/24 a----> III quadrant
cos b=-12/13 b----> II quadrant
sin (a+b)=?

we know that
sin(a + b) = sin(a)cos(b) + cos(a)sin(b<span>)
</span>
step 1
find sin b
sin²b+cos²b=1------> sin²b=1-cos²b----> 1-(144/169)---> 25/169
sin b=5/13------> is positive because b belong to the II quadrant

step 2
Find sin a and cos a
tan a=7/24
tan a=sin a /cos a-------> sin a=tan a*cos a-----> sin a=(7/24)*cos a
sin a=(7/24)*cos a------> sin²a=(49/576)*cos²a-----> equation 1
sin²a=1-cos²a------> equation 2
equals 1 and 2
(49/576)*cos²a=1-cos²a---> cos²a*[1+(49/576)]=1----> cos²a*[625/576]=1
cos²a=576/625------> cos a=-24/25----> is negative because a belong to III quadrant
cos a=-24/25
sin²a=1-cos²a-----> 1-(576/625)----> sin²a=49/625
sin a=-7/25-----> is negative because a belong to III quadrant

step 3
find sin (a+b)
sin(a + b) = sin(a)cos(b) + cos(a)sin(b)
sin a=-7/25
cos a=-24/25
sin b=5/13
cos b=-12/13
so
sin (a+b)=[-7/25]*[-12/13]+[-24/25]*[5/13]----> [84/325]+[-120/325]
sin (a+b)=-36/325

the answer is
sin (a+b)=-36/325

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3 years ago

PLEASE HELP TIMED TEST!!!! I WILL REPORT YOU IF YOU GIVE ME A LINK!!

Elenna [48]

Answer:

x=40

Step-by-step explanation:

you have to get a answer of 60

(40)+20=60

and 2(40)-20=60

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2 years ago

ASAP! GIVING BRAINLIEST! Please read the question THEN answer CORRECTLY! NO guessing. I say no guessing because people usually g

mr Goodwill [35]

Answer:

Step-by-step explanation:

It is not B because 7x^2 means multiplying the equation by seven. It isn't C because that would move the graph DOWN seven units. And it's not D because when it is in parenthesis like that, it means that it is a horizontal shift, not vertical.

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2 years ago

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