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2 years ago

Someone please help I will give u 1¢

Mathematics

1 answer:

Darya [45]2 years ago

7 0

Answer:

3rd option

y=1/2 x-3

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How do you solve 21r < 7?

quester [9]

21r < 7
Divide by 21
r < 7/21

5 0

3 years ago

Read 2 more answers

What is the value of this expression when c=-4 and d = 10?<br> 1/4(c3+d2)

Mumz [18]

What is the value of this expression when c = -4 and d = 10 ? $\frac{1}{4}$ (c³ + d²)

<h3>Given:-</h3>

$\frac{1}{4}$ (c³ + d²) where c = -4 and d = 10

The value of the expression $\frac{1}{4}$ (c³ + d²)

<h2>Solution:-</h2>

$\frac{1}{4}$ (c³ + d²) [Given expression]

Now, putting the value of c = -4 and d = 10 , we get,

$\frac{1}{4}$ { (-4)³ + (10)² }

$\frac{1}{4}$ ( -64 + 100 )

$\frac{1}{4}$ ( 36 )

$\frac{1}{4}$ × 36

$\bf = \: 9 \: ( Answer )$

3 0

3 years ago

Solve the combined inequality and describe the graph of the solution.

Vikki [24]

I hope this helps you

4x>-4 x>-1

8x <32 x <4

-1 <x <4

3 0

3 years ago

A coin is to be tossed until a head appears twice in a row. What is the sample space for this experiment? If the coin is fair, w

blsea [12.9K]

Answer:

The sample space would be infinite

{HH, TTTHH, TTTTHH, TTTTTTTHH.......................}

1/8

Step-by-step explanation:

The sample space would be infinite

{HH, TTTHH, TTTTHH, TTTTTTTHH.......................}.

This is because the number of times the coin would be flipped is not specified. So, you can keep flipping the coin forever.

If the coin is tossed four times then the sample space would be

{HHHH HTHH THHH HTHT

HHHT HTTH TTHH THTH

HHTT HHTH TTTH THHT

HTTT TTTT TTHT THTT}

HTHH and TTHH are the only two cases where two consecutive tosses will result in two heads

Probability that the coin will be tossed four times is 2/16 = 1/8

6 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