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2 years ago

Write the equation of a line that goes through the points (-7,9) and (-4,3).

Mathematics

1 answer:

solong [7]2 years ago

3 0

Answer:

y= -2x - 5

Step-by-step explanation:

You have use to slope intercept and formula to get the answer.

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-3(-4r - 5)<br> combining like terms

svp [43]

Answer:

Step-by-step explanation:

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3 years ago

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Answer pleassse ???? I'll love you forever

enot [183]

As x increases, y increases. So this data has a positive association. The association looks to be strong, as the values all follow the trend pretty closely.

Answer=Strong Positive

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3 years ago

The circle with center O has a minor arc BSA with a length of

tangare [24]

Answer:

D. $\frac{27\pi}{2}$ inches.

Step-by-step explanation:

We have been given that the circle with center O has a minor arc BSA with a length of $3\pi^{2}$ inches. The central angle is 40°.

To find the circumference of circle we will use formula:

$\frac{\text{Central angle}}{2\pi}=\frac{\text{Arc length}}{2\pi r}$ , where $2\pi$ = measure of 360 degrees in radians and $2\pi r$ = circumference of circle.

Let us convert measure of central angle into radians.

$40^{o}=\frac{40*\pi}{180} =\frac{2\pi}{9}$

Upon substituting our given value in the formula we will get,

$\frac{\frac{2\pi}{9}}{2\pi}=\frac{\frac{3\pi}{2}}{2\pi r}$

$\frac{2\pi}{18\pi}=\frac{3\pi}{4\pi r}$

$\frac{1}{9}=\frac{3}{4r}$

Cross multiplying we will get,

$4r=27$

$r=\frac{27}{4}$

Hence, the radius of our circle is 27/4 inches.

Since the circumference of circle is $2\pi r$ . Upon substituting $r=\frac{27}{4}$ we will get,

$2\pi r=2\pi* \frac{27}{4}=\frac{27\pi}{2}$

Therefore, circumference of our given circle will be $\frac{27\pi}{2}$ inches and option D is the correct choice.

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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:

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

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3 years ago

Given the equation 5 + x -12 = 2x - 7. Part A. Solve the equation 5 + x - 12 = 2x - 7. In your final answer, be sure to state th

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Answer:

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3 years ago

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