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4 years ago

Analyze the diagram below and complete the instructions that follow. Solve for x.

Mathematics

1 answer:

Oxana [17]4 years ago

4 0

To solve for x you'll add 6x and 4x and set them equal to 90 since they are complementary.

6x+4x = 90
10x = 90
x = 9

Hope this helps :)

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What is 60 over 40 in a mixed fraction

jeyben [28]

Answer:

1 1/2

Step-by-step explanation:

60/40 = (60 ÷ 20)/(40 ÷ 20) = 3/2; 3 > 2 => improper fraction Rewrite: 3 ÷ 2 = 1 and remainder = 1 => 3/2 = (1 × 2 + 1)/2 = 1 + 1/2 = = 1 1/2, mixed number (mixed fraction)

Hope this helped!

6 0

3 years ago

Read 2 more answers

2000 people are selected randomly from a certain population and it is found that 389 people in the sample are over 6 feet tall.

Airida [17]

Answer: 0.195

Step-by-step explanation:

given data:

population = 2000 people.

people Who are over 6 feet tall = 389.

Solution:

the point estimate of people over 6feet tall

= no of people over 6 feet tall / total population size

= 389/2000

= 0.195

the point estimate of people over 6 feet tall is 0.195

8 0

3 years ago

There are 16 cookies and 4 friends. How many cookies will each friend will have?

Anestetic [448]

Each friend will have 4 cookies.

8 0

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

goldfiish [28.3K]

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

7 0

4 years ago

(-9-11)/(-4)divided by -7^2+(11 times 2^2)

Sergio039 [100]

The answer is -25
Just to make sure the equation is set up as a fraction right?

6 0

3 years ago