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3 years ago

16,000 rolls of paper were purchased and 12,500 were sold at $1.79, what is the total ending inventory?

2 answers:

8 0

12,500 x 1.79 to find the price but 16,000 - 12,500 to get how many were left.
12,500 x 1.79 = $22,375 that they made from selling. 16,000 - 12,500 = 3,500 rolls left in the inventory. This question confused me so I just gave both answers.

asambeis [7]3 years ago

3 0

12500*1.79=22,375 16000*1.79=28,640 Total ending inventory=28640−22,375 =6,265

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The rule which allows us to change the sign of an exponent is called the ____ rule.

Bogdan [553]

Answer:

The rule which allows us to change the sign of an exponent is called the negative exponent rule.

Step-by-step explanation:

Consider the provided information.

The negative exponent rule is the rule which allows to change the sign of an exponent.

$x^{-1}=\frac{1}{x}$

According to negative exponent rule: A negative exponents in the numerator moved to the denominator and become positive exponents.

If A negative exponents in the denominator moved to the numerator and become positive exponents.

Hence, the required statement is: The rule which allows us to change the sign of an exponent is called the negative exponent rule.

6 0

3 years ago

The measures of ∠1, ∠2, and ∠3 are 40%, 12.5%, and 25% of the sum of the angle measures of the quadrilateral. Find the value of

sweet [91]

The value of x is 81

Step-by-step explanation:

The sum of the interior angles of any quadrilateral is 360°

The measure of ∠1 is 40% of the sum of the angle measures of the quadrilateral
The measure of ∠2 is 12.5% of the sum of the angle measures of the quadrilateral
The measure of ∠3 is 25% of the sum of the angle measures of the quadrilateral
We need to find the value of x

∵ The figure have 4 sides

∴ The figure is a quadrilateral

∵ The sum of the measures of the interior angles of a

quadrilateral is 360°

- Add the four angles and equate the sum by 360

∴ m∠1 + m∠2 + m∠3 + x = 360

∵ m∠1 = 40% of the sum of the angle measures of the quadrilateral

∴ m∠1 = 40% × 360 = $\frac{40}{100}$ × 360 = 144°

∵ m∠2 = 12.5% of the sum of the angle measures of the quadrilateral

∴ m∠2 = 12.5% × 360 = $\frac{12.5}{100}$ × 360 = 45°

∵ m∠3 = 25% of the sum of the angle measures of the quadrilateral

∴ m∠3 = 25% × 360 = $\frac{25}{100}$ × 360 = 90°

- Substitute these values in the equation above

∴ 144 + 45 + 90 + x = 360

- Add the like terms in the left hand side

∴ 279 + x = 360

- Subtract 279 from both sides

∴ x = 81°

The value of x is 81

Learn more:

You can learn more about the polygons in brainly.com/question/6281564

#LearnwithBrainly

5 0

3 years ago

Use the figure to find the measure of angle 2.

ale4655 [162]

Answer:

∠ 2 = 70°

Step-by-step explanation:

110° and ∠ 1 are corresponding angles and congruent, thus

∠ 1 = 110°

∠ 1 and ∠ 2 are adjacent angles and are supplementary, thus

∠ 2 = 180° - ∠ 1 = 180° - 110° = 70°

7 0

3 years ago

3. Two cars are parked, 150m apart, and on opposite sides of a building. A camera on the top of the building rotates and can vie

allsm [11]

Answer:

The building is 61.5 m tall

Step-by-step explanation:

The image below is a diagram where all the given distances and angles are shown. We have additionally added some variables:

h = height of the building

a, b = internal angles of each triangle

x = base of each triangle

The angles a and b can be easily found by subtracting the given angles from 90° since they are complementary angles, thus:

a = 90° - 37° = 53°

b = 90° - 42° = 48°

Now we apply the tangent ratio on both triangles separately:

$\displaystyle \tan a=\frac{\text{opposite leg}}{\text{adjacent leg}}$

$\displaystyle \tan 53^\circ=\frac{150-x}{h}$

$\displaystyle \tan 48^\circ=\frac{x}{h}$

From the last equation:

$x=h.\tan 48^\circ$

Substituting into the first equation:

$\displaystyle \tan 53^\circ=\frac{150-h.\tan 48^\circ}{h}$

Operating on the right side:

$\displaystyle \tan 53^\circ=\frac{150}{h}-\tan 48^\circ$

Rearranging:

$\displaystyle \tan 53^\circ+\tan 48^\circ=\frac{150}{h}$

Solving for h:

$\displaystyle h=\frac{150}{\tan 53^\circ+\tan 48^\circ}$

Calculating:

h = 61.5 m

The building is 61.5 m tall

5 0

3 years ago

Match the parabolas represented by the equations with their vertices. y = x2 + 6x + 8 y = 2x2 + 16x + 28 y = -x2 + 5x + 14 y = -

GaryK [48]

Consider all parabolas:

$y = x^2 + 6x + 8,\\y=x^2+6x+9-9+8,\\y=(x^2+6x+9)-1,\\y=(x+3)^2-1.$

When x=-3, y=-1, then the point (-3,-1) is vertex of this first parabola.

$y = 2x^2 + 16x + 28=2(x^2+8x+14),\\y=2(x^2+8x+16-16+14),\\y=2((x^2+8x+16)-16+14),\\y=2((x+4)^2-2)=2(x+4)^2-4.$

When x=-4, y=-4, then the point (-4,-4) is vertex of this second parabola.

$y =-x^2 + 5x + 14=-(x^2-5x-14),\\y=-(x^2-5x+\dfrac{25}{4}-\dfrac{25}{4}-14),\\y=-((x^2-5x+\dfrac{25}{4})-\dfrac{25}{4}-14),\\y=-((x-\dfrac{5}{2})^2-\dfrac{81}{4})=-(x-\dfrac{5}{2})^2+\dfrac{81}{4}.$

When x=2.5, y=20.25, then the point (2.5,20.25) is vertex of this third parabola.

$y =-x^2 + 7x + 7=-(x^2-7x-7),\\y=-(x^2-7x+\dfrac{49}{4}-\dfrac{49}{4}-7),\\y=-((x^2-7x+\dfrac{49}{4})-\dfrac{49}{4}-7),\\y=-((x-\dfrac{7}{2})^2-\dfrac{77}{4})=-(x-\dfrac{7}{2})^2+\dfrac{77}{4}.$

When x=3.5, y=19.25, then the point (3.5,19.25) is vertex of this fourth parabola.

$y =2x^2 + 7x +5=2(x^2+\dfrac{7}{2}x+\dfrac{5}{2}),\\y=2(x^2+\dfrac{7}{2}x+\dfrac{49}{16}-\dfrac{49}{16}+\dfrac{5}{2}),\\y=2((x^2+\dfrac{7}{2}x+\dfrac{49}{16})-\dfrac{49}{16}+\dfrac{5}{2}),\\y=2((x+\dfrac{7}{4})^2-\dfrac{9}{16})=2(x+\dfrac{7}{4})^2-\dfrac{9}{8}.$

When x=-1.75, y=-1.125, then the point (-1.75,-1.125) is vertex of this fifth parabola.

$y =-2x^2 + 8x +5=-2(x^2-4x-\dfrac{5}{2}),\\y=-2(x^2-4x+4-4-\dfrac{5}{2}),\\y=-2((x^2-4x+4)-4-\dfrac{5}{2}),\\y=-2((x-2)^2-\dfrac{13}{2})=-2(x-2)^2+13.$

When x=2, y=13, then the point (2,13) is vertex of this sixth parabola.

3 0

3 years ago