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Ray Of Light [21]
3 years ago
10

11. Describe and correct the error in evaluating the expression. –21-7)= -14

Mathematics
1 answer:
likoan [24]3 years ago
8 0

Answer: -28

Step-by-step explanation:

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Find the measure of
lyudmila [28]
The answer is this i’m sure of it
3 0
2 years ago
I need help on questions 48 49 and 50 and u need to find the surface area of the 3 polyhedras please and thank you!
White raven [17]

48. The surface area of a cylinder is given by

... A = 2πr² + 2πrh = 2πr(r+h)

For r = 13 mi and h = 7 mi, this becomes

... A = 2π(13 mi)(13 mi + 7 mi) = 520π mi²

The surface area of the cylinder is 520π mi² ≈ 1634 mi².

49. The surface area of a rectangular prism is

... A = 2(LW + H(L+W))

For L = 20 m, W = 10 m, H = 7 m, this becomes

... A = 2((20 m)(10 m) + (7 m)(20 m + 10 m)) = 2(200 m² + 210 m²) = 820 m²

The surface area of the prism is 820 m².

50. The formula is the same as for problem 48.

For r = 10 m and h = 13 m, this becomes

... A = 2π(10 m)(10 m + 13 m) = 460π m² ≈ 1445 m²

The surface area of the cylinder is 460π m² ≈ 1445 m².

_____

When calculations are repetitive, it is convenient to let a calculator or spreadsheet do them.

7 0
3 years ago
Find the value of x in ⊙Q .
ehidna [41]

Answer:

x= 5

Step-by-step explanation:

both lines are the same length

5x-6=2x+9

-2x -2x

3x-6= +9

+6 +6

3x = 15

÷3 ÷3

x= 5

6 0
3 years ago
I need help with my math homework. The questions is: Find all solutions of the equation in the interval [0,2π).
Aleksandr-060686 [28]

Answer:

\frac{7\pi}{24} and \frac{31\pi}{24}

Step-by-step explanation:

\sqrt{3} \tan(x-\frac{\pi}{8})-1=0

Let's first isolate the trig function.

Add 1 one on both sides:

\sqrt{3} \tan(x-\frac{\pi}{8})=1

Divide both sides by \sqrt{3}:

\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}

Now recall \tan(u)=\frac{\sin(u)}{\cos(u)}.

\frac{1}{\sqrt{3}}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}

or

\frac{1}{\sqrt{3}}=\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}

The first ratio I have can be found using \frac{\pi}{6} in the first rotation of the unit circle.

The second ratio I have can be found using \frac{7\pi}{6} you can see this is on the same line as the \frac{\pi}{6} so you could write \frac{7\pi}{6} as \frac{\pi}{6}+\pi.

So this means the following:

\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}

is true when x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi

where n is integer.

Integers are the set containing {..,-3,-2,-1,0,1,2,3,...}.

So now we have a linear equation to solve:

x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi

Add \frac{\pi}{8} on both sides:

x=\frac{\pi}{6}+\frac{\pi}{8}+n \pi

Find common denominator between the first two terms on the right.

That is 24.

x=\frac{4\pi}{24}+\frac{3\pi}{24}+n \pi

x=\frac{7\pi}{24}+n \pi (So this is for all the solutions.)

Now I just notice that it said find all the solutions in the interval [0,2\pi).

So if \sqrt{3} \tan(x-\frac{\pi}{8})-1=0 and we let u=x-\frac{\pi}{8}, then solving for x gives us:

u+\frac{\pi}{8}=x ( I just added \frac{\pi}{8} on both sides.)

So recall 0\le x.

Then 0 \le u+\frac{\pi}{8}.

Subtract \frac{\pi}{8} on both sides:

-\frac{\pi}{8}\le u

Simplify:

-\frac{\pi}{8}\le u

-\frac{\pi}{8}\le u

So we want to find solutions to:

\tan(u)=\frac{1}{\sqrt{3}} with the condition:

-\frac{\pi}{8}\le u

That's just at \frac{\pi}{6} and \frac{7\pi}{6}

So now adding \frac{\pi}{8} to both gives us the solutions to:

\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}} in the interval:

0\le x.

The solutions we are looking for are:

\frac{\pi}{6}+\frac{\pi}{8} and \frac{7\pi}{6}+\frac{\pi}{8}

Let's simplifying:

(\frac{1}{6}+\frac{1}{8})\pi and (\frac{7}{6}+\frac{1}{8})\pi

\frac{7}{24}\pi and \frac{31}{24}\pi

\frac{7\pi}{24} and \frac{31\pi}{24}

5 0
3 years ago
Look at the picture of a scaffold used to support construction workers. The height of the scaffold can be changed by adjusting t
maria [59]

<u>Part A</u>

Using the Pythagorean on the right triangle PQR, with PQ and QR as the legs and PR as the hypotenuse,

14^2 +6^2 =(PR)^2\\\\(PR)=\sqrt{14^2 +6^2}\\\\PR \approx \boxed{15.23 \text{ ft}}

<u>Part B</u>

(QR)^2 +6^2 =16^2\\\\(QR)^2 =16^2 -6^2\\\\QR=\sqrt{16^2 -6^2}\\\\QR \approx \boxed{14.83 \text{ ft}}

8 0
1 year ago
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