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2 years ago

Solve the inequality 13×-4 < 12x-1

Mathematics

1 answer:

Murrr4er [49]2 years ago

7 0

Answer:

x <3

Step-by-step explanation:

13x-4 < 12x-1

Subtract 12x from each side

13x-12x-4 < 12x-12x-1

x-4 <-1

Add 4 to each side

x-4+4<-1+4

x <3

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How do you do this ?

Maru [420]

Well the first part is 27 but the second part is adding 60 to 21 which is 81. dont forget that!

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3 years ago

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You want to buy a camera. The retail price is $75. The camera is on sale for 15% off . What is the sale price of the camera?

ludmilkaskok [199]

Answer:

$63.75

Step-by-step explanation:

15% off

Get percentage of retail price

1-0.15 = .85

0.85 = 85%

The sale price is 85% of the retail price

Multiply

0.85 * 75

$63.75

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1 year ago

The length of a rectangle is twice the with. If the perimeter of the rectangle is 60 units, find the area of the garden

Darya [45]

w - width

2w - length

60 - perimeter

w + w + 2w + 2w = 6w - perimeter

The equation:

6w = 60 <em>divide both sides by 6</em>

w = 10 → 2w = 2 · 10 = 20

The area: A = width × length

A = (10)(20) = 200

<h3>Answer: The area of the garden is equal 200 square units.</h3>

4 0

2 years ago

Students are planning the backdrop for the school play. The backdrop is 15 feet wide and 10 feet high. Every 16 inches on the sc

alekssr [168]

Answer:6

Step-by-step explanation:

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3 years ago

A disk with radius 3 units is inscribed in a regular hexagon. Find the approximate area of the inscribed disk using the regular

Oksi-84 [34.3K]

Answer:

The approximate are of the inscribed disk using the regular hexagon is $A=18\sqrt{3}\ units^2$

Step-by-step explanation:

we know that

we can divide the regular hexagon into 6 identical equilateral triangles

see the attached figure to better understand the problem

The approximate area of the circle is approximately the area of the six equilateral triangles

Remember that

In an equilateral triangle the interior measurement of each angle is 60 degrees

We take one triangle OAB, with O as the centre of the hexagon or circle, and AB as one side of the regular hexagon

Let

M ----> the mid-point of AB

OM ----> the perpendicular bisector of AB

x ----> the measure of angle AOM

$m\angle AOM =30^o$

In the right triangle OAM

$tan(30^o)=\frac{(a/2)}{r}=\frac{a}{2r}\\\\tan(30^o)=\frac{\sqrt{3}}{3}$

$\frac{a}{2r}=\frac{\sqrt{3}}{3}$

we have

$r=3\ units$

substitute

$\frac{a}{2(3)}=\frac{\sqrt{3}}{3}\\\\a=2\sqrt{3}\ units$

Find the area of six equilateral triangles

$A=6[\frac{1}{2}(r)(a)]$

simplify

$A=3(r)(a)$

we have

$r=3\ units\\a=2\sqrt{3}\ units$

substitute

$A=3(3)(2\sqrt{3})\\A=18\sqrt{3}\ units^2$

Therefore

The approximate are of the inscribed disk using the regular hexagon is $A=18\sqrt{3}\ units^2$

6 0

3 years ago