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DedPeter [7]
3 years ago
9

Fill in the missing number

Mathematics
2 answers:
Fynjy0 [20]3 years ago
7 0
The first on would be 2 and the second would be 1
sladkih [1.3K]3 years ago
3 0
6/ 2 =3
5/ 5 =1
6 divide by 2 is 3
5 divide by 5 is 1
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Two triangles are shown on the graph below. Which is a true statement about the triangles?
andriy [413]

Answer:

where the triangles are........????

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3 years ago
Greatest to least 0.65 0.6 0.5 0.5
Snowcat [4.5K]

Answer:

0.65, 0.6, 0.5, 0.5

Step-by-step explanation:

3 0
3 years ago
Keelah's clothing store bicycles for $50 and then sells them for $80 what is the percent of markup on the price of the coat
kondaur [170]

Answer:

62.5%

Step-by-step explanation:

We can write a proportion. We know that $50 is the original value and $80 is the marked up price. We don't know the marked up percentage. Let's call it x. We write equal ratios of \frac{x}{100} comparing percent and \frac{50}{80} comparing the 50 to the marked up price. We then set them equal.

\frac{x}{100} =\frac{50}{80}

We then cross multiply across the equal sign from numerator to the opposite numerator.

80(x)=50(100)

80x=5000

We divide to find x.

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This means the bicycles are marked up 62.5% and the original value is 37.5%.

8 0
3 years ago
Trig: Which complex number's graph is shown?
Andrei [34K]

ANSWER

2 \sqrt{2} ( \cos( \frac{7\pi}{4}   + i  \sin(  \frac{7\pi}{4} ) )

EXPLANATION

The complex number shown has coordinates (2,-2)

or

z = 2 - 2i

The modulus is

|z|  =  \sqrt{ {2}^{2}  +  {( - 2)}^{2} }

|z|  =  \sqrt{ 8} = 2 \sqrt{2}

The argument is

\theta= \tan^{ - 1} ( \frac{ - 2}{2} )

\theta= \tan^{ - 1} ( -1) =  \frac{7\pi}{4}

The polar form is

2 \sqrt{2} ( \cos( \frac{7\pi}{4}   + i  \sin(  \frac{7\pi}{4} ) )

The first option is correct.

3 0
3 years ago
A mass weighing 3 lb stretches a spring 3 in. If the mass is pushed upward, contracting the spring a distance of 1 in and then s
Bas_tet [7]

Answer:

Step-by-step explanation:

m=\frac{3}{32}\\ \gamma =0\text{ damping constant.}\\ k=\frac{3}{\frac{1}{4}}=12\\ F(t)=0\\ \text{so equation of motion is }\\ mu''+\gamma u'+ku=F(t)\\ \frac{3u''}{32}+12u=0\\ u''+128u=0\\ \text{and initial conditions are}\\ u(0)=-\frac{1}{12}\\ u'(0)=2\\ u''+128u=0\\ \text{characterstic equation is}\\ r^2+128=0\\ \text{roots are }\\ r=\pm 8\sqrt{2}i\\ \text{therefore general solution is}\\ u(t)=A\cos8\sqrt{2}t+Bsin8\sqrt{2}t\\ \text{using initial conditions we get}\\ A=-\frac{1}{12}\\ B=\frac{\sqrt{2}}{8}\\ \text{therefore solution is }\\ u(t)=-\frac{1}{12}\cos8\sqrt{2}t+\frac{\sqrt{2}}{8}\sin8\sqrt{2}t\\
{hence}\\R=\sqrt{\frac{11}{288}}\\\\\sigma=\pi - \tan^{-1}\frac{1}{\sqrt{2}}\\\\\omega_0=8\sqrt{2}\\T=\frac{4}{4\sqrt{2}}

3 0
3 years ago
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