$46 shoes; 2.9% tax how do you solve this question

PLEASE HELP 25 POINTS!!

iris [78.8K]

Answer:

m= masters degree salary and b=bachelors degree salary.

m = 2b - 45000

m + b = 112000

Substitute to get

2b - 45000 + b = 112000

3b = 156000

b = 52000

So m = 59000

Check to see if these numbers work

59000 = 2*52000 - 45000 true

59000 + 52000 = 112000 true

Step-by-step explanation:

7 0

3 years ago

Terrance claims that for the input coordinates (x, y), a rotation of 180° clockwise about the origin, followed by a reflection o

vodka [1.7K]

Answer:

Terrance is incorect.

Correct output coordinates (-y,-x)

Step-by-step explanation:

Let $(x,y)$ be the input coordinates.

First translation is a rotation of 180° clockwise about the origin. This translation has a rule

$(x,y)\rightarrow (-x,-y)$

Second translation is a reflection over the line y = x. The general rule for the reflection across the line y=x has the rule

$(a,b)\rightarrow (b,a)$

When a sequence of two translations are applied to the initial input coordinates, then

$(x,y)\rightarrow (-x,-y)\rightarrow (-y,-x)$

As you can see Terrance made a mistake and these two transformations do not cancel themselves out.

8 0

3 years ago

Write an expression that is equivalent to

finlep [7]

Answer:

15/2r - 40; Distributive Property

Step-by-step explanation:

You can use the Distributive Property to multiply the 5 across the expression in the parentheses.

5(3/2r - 8)

5 × 3/2r = 15/2r

5 × -8 = -40

Combine.

15/2r - 40

Hope this helps!

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

Describe the transformations of f(x) = x3 represented by g(x) = (x - 6) - 2.

devlian [24]

Answer:

Step-by-step explanation:

4 0

3 years ago