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3 years ago

13

Help me please guys

2 answers:

ss7ja [257]3 years ago

7 0

-2,-5 because the inverse is the oppisite of any number, so the opposite of 2,5 is -2,-5

Jlenok [28]3 years ago

5 0

Answer:

(9,6)

Step-by-step explanation:

You just take one of the (x,y) pairs and switch the numbers with each other so it's like (y,x)

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Simplify the expression <br> 49<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%7D%20" id="TexFormula1" title=" \frac{1}{

dimaraw [331]

To convert a mixed fraction into an improper fraction:

$a\frac{b}{c} =\frac{(a\cdot c)+b}{c}$

$49 \frac{1}{2} = \frac{(49\cdot 2)+1}{2} = \frac{98 +1}{2} = \boxed{\bf{\frac{99}{2}}}$

8 0

3 years ago

Can someone make up an equation to match this problem plz don’t ignore

Elodia [21]

If it’s about fractions, then the first scale is 3/6 and the second one is
4 1/0 if I’m correct, please message me if I’m wrong.

5 0

3 years ago

---- PLEASE HELP <3 .. with steps :( <3

Zielflug [23.3K]

1) 6, 18, 54

2) 5/3, 14/9, 41/27

3) 1.5, 2.5, 2.5

Step-by-step explanation:

1)

The function that we have in this problem is

$g(x)=3x$

We want to find the first 3 iterations.

The initial value is:

x = 2

To find the value of the 1st iteration, we just substitute this value into the expression of the function, and we get:

$g_1(x)=3x=3\cdot 2 = 6$

The to find the value of the 2nd iteration, we just substitute this value into the expression of the function, and we get:

$g_2(x)=3\cdot g_1(x)=3\cdot 6 = 18$

Finally, the 3rd iteraction is given by:

$g_3(x)=3g_2(x)=3\cdot 18=54$

2)

Here in this problem the function that we have to use is

$g(x)=\frac{1}{3}x+1$

The initial value is

$x=2$

So the first iteration is given by

$g_1(x)=\frac{1}{3}\cdot 2 + 1 = \frac{5}{3}$

To find the 2nd iteration, we substitute this value into g(x) again:

$g_2(x)=\frac{1}{3}g_1+1=\frac{1}{3}\cdot \frac{5}{3}+1=\frac{5}{9}+1=\frac{14}{9}$

Finally, to find the 3rd iteration, we substitute this value into g(x) again:

$g_3(x)=\frac{1}{3}\cdot \frac{14}{9}+1=\frac{14}{27}+1=\frac{41}{27}$

3)

The function in this problem is

$g(x)=-|x-2|+3$

The initial value is

x = 0.5

So, the first iteration is:

$g_1(x)=-1|0.5-2|+3=-1|-1.5|+3=-1\cdot 1.5 +3=-1.5+3=1.5$

The second iteration is given by

$g_2(x)=-|g_1-2|+3=-|1.5-2|+3=--0.5+3=2.5$

Finally, the 3rd iteration is

$g_3(x)=-|g_2-2|+3=-|2.5-2|+3=-0.5+3=2.5$

5 0

3 years ago

Use triangle ABC drawn below & only the sides labeled. Find the side of length AB in terms of side a, side b & angle C o

Brrunno [24]

Answer:

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

Step-by-step explanation:

Given:

The above triangle

Required

Solve for AB in terms of a, b and angle C

Considering right angled triangle BOC where O is the point between b-x and x

From BOC, we have that:

$Sin\ C = \frac{h}{a}$

Make h the subject:

$h = aSin\ C$

Also, in BOC (Using Pythagoras)

$a^2 = h^2 + x^2$

Make $x^2$ the subject

$x^2 = a^2 - h^2$

Substitute $aSin\ C$ for h

$x^2 = a^2 - h^2$ becomes

$x^2 = a^2 - (aSin\ C)^2$

$x^2 = a^2 - a^2Sin^2\ C$

Factorize

$x^2 = a^2 (1 - Sin^2\ C)$

In trigonometry:

$Cos^2C = 1-Sin^2C$

So, we have that:

$x^2 = a^2 Cos^2\ C$

Take square roots of both sides

$x= aCos\ C$

In triangle BOA, applying Pythagoras theorem, we have that:

$AB^2 = h^2 + (b-x)^2$

Open bracket

$AB^2 = h^2 + b^2-2bx+x^2$

Substitute $x= aCos\ C$ and $h = aSin\ C$ in $AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = (aSin\ C)^2 + b^2-2b(aCos\ C)+(aCos\ C)^2$

Open Bracket

$AB^2 = a^2Sin^2\ C + b^2-2abCos\ C+a^2Cos^2\ C$

Reorder

$AB^2 = a^2Sin^2\ C +a^2Cos^2\ C + b^2-2abCos\ C$

Factorize:

$AB^2 = a^2(Sin^2\ C +Cos^2\ C) + b^2-2abCos\ C$

In trigonometry:

$Sin^2C + Cos^2 = 1$

So, we have that:

$AB^2 = a^2 * 1 + b^2-2abCos\ C$

$AB^2 = a^2 + b^2-2abCos\ C$

Take square roots of both sides

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

6 0

3 years ago

What is equivalent to 2x-17

Bad White [126]

Get photo math an app just for math it gives u the answers and how to solve it

8 0

3 years ago