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2 years ago

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Mathematics

2 answers:

Gwar [14]2 years ago

8 0

Answer:

4 ther are 4 line symmetery

sukhopar [10]2 years ago

6 0

Answer:

two lines of symmetry

(a vertical and a horizontal)

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Find the point-slope form of the equation of the line that passes through the given point and has the indicated slope. Point: (-

zzz [600]

Answer:

slope : 3

y + 3 = 3(x + 5)

y + 3 = 3x + 15

y = 3x + 12

5 0

3 years ago

What is -3x-5 = 16<br> I'm stuck on a simple problem

mariarad [96]

Answer:

-7

Step-by-step explanation:

-3x-5=16

-3x-5+5=16+5

-3x=21

-3x/-3=21/-3

x=-7

4 0

3 years ago

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Natalie works in a department store selling clothing. She makes a guaranteed salary of $300 per week, but is paid a commission o

Sergio039 [100]

Answer:

The answer is below

Step-by-step explanation:

a) Natalie basic salary is $300 per week. She is paid a commission of 30% of her total sales for the week. Therefore her total weekly salary per week is given by:

Total weekly salary = Basic salary + Commissions

If Natalie makes $225 in sales, her commission = 30% of $225 = 0.3 × $225 = $67.5

Total weekly salary = Basic salary + Commissions = $300 + $67.5

Total weekly salary = $367.5

b) If Natalie makes $x in sales, her commission = 30% of $x = 0.3 × $x = $0.3x

Total weekly salary = Basic salary + Commissions = $300 + $0.3x

Total weekly salary = $(300 + 0.3x)

7 0

3 years ago

| x-3 | < x-3<br><br> can someone give a step by step process on how to do this?

statuscvo [17]

Answer:

There are no solutions to the inequality.

Step-by-step explanation:

|x - 3| < x – 3

1. Separate the inequality into two separate ones.

(1) x – 3 < x – 3

(2) x – 3 < -(x – 3)

2. Solve each equation separately

(a) Equation (1)

$\begin{array}{rcl}x - 3 & < & x - 3\\x & < & x\\\end{array}\\\text{This is impossible. No solutions exist.}$

(b) Equation (2)

$\begin{array}{rcl}x - 3 & < & -(x - 3)\\x - 3 & < & -x + 3\\x &$

For example, if x = 0, we get

|0 - 3| < 0 - 3 or

3 < -3

4 0

3 years ago

Find the x-intercept of the parabola with vertex (1,-13) and y-intercept (0,-11).

Advocard [28]

1. find equation

general solutions are:
$y = a x^{2} +bx +c \\ \\ \frac{dy}{dx} = 2ax +b$

for P(0,-11)

$-11 = 0a + 0b +c$ ⇒ $c = -11$

for p(1,-13)
$-13 = 1a + 1b -11 \\ a+b = -2$

and

$0 = 2a + b \\ b = -2a$

solve for a and b:
$a - 2a = -2 \\ a = 2 \\ b = -4$

the total equation is now:

$y = 2 x^{2} -4x -11$

To find the x-intercept set y=0 and solve for x

$0 = 2 x^{2} -4x-11$

4 0

3 years ago

Read 2 more answers