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3 years ago

Someone please help me

Mathematics

2 answers:

____ [38]3 years ago

7 0

b.) 270in2 could you give me a brainlyest answer if it helped thanks have a nice day

lina2011 [118]3 years ago

7 0

Place the point of the compass on one end of this line (A) and draw a semicircle Across the Line. Now place the point of the compass on the other end of the line.

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An elevator is two floors below ground level and goes up 5 floors. Write an addition equation that models the location of the el

makkiz [27]

Answer: d= +5-2 = 3

Step-by-step explanation:

An elevator is two floor below the ground level. Any distance below the ground level is always measured with the negative notation.

So the location of the elevator given by (-2)

Now when elevator goes up by 5 floors, the location of elevator relative to the ground level will be (+5)

d=+5-2

7 0

3 years ago

What is 345 written as a percent? F. 0.0093%. G. 0.93% H. 9.3% I. 93%

zysi [14]

Answer:

34500%

Step-by-step explanation:

5 0

3 years ago

30 POINTS!!!

Nataliya [291]

Answer:

115

Step-by-step explanation:

The measure of x is half the sum of the intercepted arcs. The arc on the left is a semicircle, so 180°.

The arc on the right is the difference between 180° and the two arcs shown, 60° and 70°, so is 180° -60° -70° = 50°.

Then half the sum of the intercepted arcs is ...

x° = (180° +50°)/2 = 230°/2 = 115°

x = 115

5 0

3 years ago

Please I need help!!!!!!!!!!!!!!!

san4es73 [151]

Answer:

flooring crew > plumbers > paint crew > electricians

Step-by-step explanation:

when you turn all of the numbers into percentages, you get:

paint crew: 1/7 = 14%

plumbers: 0.15 = 15%

flooring crew: 1/6 = 16%

electricians = 12% (given)

6 0

3 years ago

Cot^4θ-4cot^2θ-5=0 solve for θ

stich3 [128]

$\cot^4\theta-4\cot^2\theta-5=(\cot^2\theta-5)(\cot^2\theta+1)=0$

$\cot^2\theta-5=0\implies\cot^2\theta=5\implies\cot\theta=\pm\sqrt5$

Recall that $\cot\theta=\cot(\theta+\pi)$ , which is to say $\cot x$ has period $\pi$ . This in turn means that $\cot\theta=0$ will have the same solutions as $\cot(\theta+n\pi)=0$ for any integer $n$ . So the general solution to the first case is

$\cot\theta=\pm\sqrt5\implies\theta=\cot^{-1}(\pm\sqrt5)+n\pi$

where $n$ is any integer.

On the other hand,

$\cot^2\theta+1=0\implies\cot^2\theta=-1$

but $x^2\ge0$ for any value of $x$ , so this equation has no (real) solutions for $\theta$ .

4 0

3 years ago