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3 years ago

The graph below shows the amount of money left in the school’s desk fund, f, after d desks have been purchased.

Mathematics

1 answer:

Scorpion4ik [409]3 years ago

6 0

The answer is $110. Have a good night.

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Please help me solve this for 15 points! -x2 + 4x - 54 = 0

Mazyrski [523]

Answer:

x1 =2-5i*sqrt(2)

x2 =2+5i*sqrt(2)

Step-by-step explanation:

-x^2 +4x-54=0 (quadratic equation)

a=-1, b=4, c=-54

x1=(-b+sqrt(b^2-4ac))/2a

x1=(-4+sqrt(4^2 - 4*(-1)(-54))/2*(-1)

x1=(-4+sqrt(16-216))/(-2)

x1 =(-4+sqrt(-200))/(-2)

x1 =(-4+sqrt(200i^2))/(-2) i^2=-1

x1 =(-4+sqrt(100*2*i^2))/(-2)

x1 =(-4+10i*sqrt(2))/(-2)

x1 =2-5i*sqrt(2)

x2 =(-b-sqrt(b^2-4ac))/2a

x2 =(-4-10i*sqrt(2))/(-2)

x2 =2+5i*sqrt(2)

7 0

3 years ago

Read 2 more answers

Gail is making costumes for a school play. Each rabbit costume needs one and one half yards of white fur fabric, a yard of blue

olga nikolaevna [1]

Answer:

Gail will need 12 yards of white fur, 8 yards of blue striped and 2 yards of the pink felt.

Step-by-step explanation:

We are given that,

The amount of yards for making one costume of different fabrics are,

White fur fabric = $1\frac{1}{2}$ = $\frac{3}{2}$ yards

Blue striped fabric = 1 yard

Pink felt for ears = $\frac{1}{4}$ yards.

So, we have,

The amount required to make 8 costumes are,

White fur fabric = $\frac{3}{2}\times 8$ = 3×4 = 12 yards

Blue striped fabric = 1×8 = 8 yards

Pink felt for ears = $\frac{1}{4}\times 8$ = 2 yards.

Thus, Gail will need 12 yards of white fur, 8 yards of blue striped and 2 yards of the pink felt.

7 0

3 years ago

Read 2 more answers

A car dealership decreased the price of a certain car by 13%. The original price was $45,400.

ira [324]

Answer:

5902

Step-by-step explanation:

Solution: 13% off 45400 is equal to (13 x 13) / 100 = 5902.

5 0

3 years ago

Q5: Identify the graph of the equation and write an equation of the translated or rotated graph in general form.

MAVERICK [17]

Answer:

$d.\:ellipse;\:3x^2+y^2+6x-6y+3=0$ .

Step-by-step explanation:

The given conic has equation;

$3x^2+y^2=9$

Divide through by 9.

$\Rightarrow \frac{3x^2}{9}+\frac{y^2}{9}=\frac{9}{9}$

$\Rightarrow \frac{x^2}{3}+\frac{y^2}{9}=1$ .

This is an ellipse centered at the origin;

This ellipse has been translated so that the center is now at;

$(-1,3)$

The translated ellipse has equation;

$\frac{(x+1)^2}{3}+\frac{(y-3)^2}{9}=1$ .

Clear the fraction;

$3(x+1)^2+(y-3)^2=9$ .

Expand;

$3(x^2+2x+1)+(y^2-6y+9)=9$ .

$3x^2+6x+3+y^2-6y+9=9$ .

Write in general form;

$3x^2+y^2+6x-6y+9-9+3=0$ .

$3x^2+y^2+6x-6y+3=0$ .

The correct choice is D

4 0

3 years ago

Read 2 more answers

Help i dont understand how to complete this

Luda [366]

Answer: Josh = 19.8 hours, Danny = 10.8 hours

Step-by-step explanation:

$Josh: \dfrac{1}{x+9}\\\\\\Danny: \dfrac{1}{x}\\\\\\Together: \dfrac{1}{7}\\\\\\Josh\quad + \quad Danny\quad =\quad Together\\\dfrac{1}{x+9}\quad +\qquad \dfrac{1}{x}\qquad = \qquad \dfrac{1}{7}\\\\\\\underline{\text{Clear the denominator by multiplying by the LCM: (x+9)(x)(7)}}\\\\7(x) + 7(x+9)=x(x+9)\\7x + 7x + 63 = x^2+9x\\14x+63=x^2+9x\\0=x^2-5x-63\\\\\\\underline{\text{Use the quadratic formula to solve for x: }}x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\dfrac{-(-5)\pm \sqrt{(-5)^2-4(1)(-63)}}{2(1)}\\\\\\.\quad =\dfrac{5\pm\sqrt{25+252}}{2}\\\\\\.\quad =\dfrac{5\pm\sqrt{277}}{2}\\\\\\.\quad =\dfrac{5\pm 16.6}{2}\\\\\\x =\dfrac{5+16.6}{2}\qquad x=\dfrac{5-16.6}{2}\\\\\\x=\dfrac{21.6}{2}\qquad \qquad x=\dfrac{-11.6}{2}\\\\\\x=10.8 \qquad \qquad x=-5.8$

Since time cannot be negative, x=-5.8 is an extraneous solution (not valid) so x = 10.8

Josh: x + 9 --> 10.8 + 9 = 19.8

Danny: x --> 10.8

****************************************************************************************

2a) k = 9.45 & 0.55

2b) x = 3/2

2c) x = 2/3 (x = -1 is an extraneous solution so is not valid)

2d) No Solution (x = 1 is an extraneous solution so is not valid)

Here is the work for 2a. Follow this format for b, c, & d

$\dfrac{3}{k^2-8x+12}=\dfrac{k}{k-2}-\dfrac{4}{k-6}\\\\\text{Since the denominator cannot equal zero, then } k \neq2\ and\ k\neq6\\\text{If either of the solutions are 2 or 6, then that solution needs to be crossed out}\\\\\underline{\text{Clear the denominator by multiplying by the LCM: (k-2)(k-6)}}\\\\3 = k(k-6)-4(k-2)\\3=k^2-6k-4k+8\\0=k^2-10k+5\\\\\\\underline{\text{Use the quadratic formula to solve for k}}$

$x=\dfrac{-(-10)\pm \sqrt{(-10)^2-4(1)(5)}}{2(1)}\\\\\\.\quad =\dfrac{10\pm\sqrt{100-20}}{2}\\\\\\.\quad =\dfrac{10\pm\sqrt{80}}{2}\\\\\\.\quad =\dfrac{10\pm8.9}{2}\\\\\\x=\dfrac{10+8.9}{2}\qquad x=\dfrac{10-8.9}{2}\\\\\\x=\dfrac{18.9}{2}\qquad x=\dfrac{1.1}{2}\\\\\\x=9.45\qquad x=0.55\\\\\\\text{Both answers are valid!}$

8 0

3 years ago