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3 years ago

If 4x^2 - 100 = 0, the roots of the equation are

1 answer:

5 0

$4x^2-100=0 \ \ \ \ \ \ |+100 \\
4x^2=100 \ \ \ \ \ \ \ \ \ \ \ |\div 4 \\
x^2=25 \\
x=\pm \sqrt{25} \\
x=\pm 5 \\
x=-5 \hbox{ or } x=5$

The answer is (3) -5 and 5.

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Write an equation of a line in slope-intercept form that passes through the point (-2, 4) and has a slope of -3. (no spaces in y

mrs_skeptik [129]

Answer:

y = - 3x - 2

Step-by-step explanation:

Step 1:

y = mx + b Slope Intercept Form

Step 2:

y = - 3x + b Input Slope

Step 3:

4 = - 3 ( - 2 ) + b Input x and y

Step 4:

4 = 6 + b Multiply

Step 5:

- 2 = b Subtract 6 on both sides

Answer:

y = - 3x - 2

Hope This Helps :)

8 0

2 years ago

First-order linear differential equations

kkurt [141]

Answer:

$(1)\ logy\ =\ -sint\ +\ c$

$(2)\ log(y+\dfrac{1}{2})\ =\ t^2\ +\ c$

Step-by-step explanation:

1. Given differential equation is

$\dfrac{dy}{dt}+ycost = 0$

$=>\ \dfrac{dy}{dt}\ =\ -ycost$

$=>\ \dfrac{dy}{y}\ =\ -cost dt$

On integrating both sides, we will have

$\int{\dfrac{dy}{y}}\ =\ \int{-cost\ dt}$

$=>\ logy\ =\ -sint\ +\ c$

Hence, the solution of given differential equation can be given by

$logy\ =\ -sint\ +\ c.$

2. Given differential equation,

$\dfrac{dy}{dt}\ -\ 2ty\ =\ t$

$=>\ \dfrac{dy}{dt}\ =\ t\ +\ 2ty$

$=>\ \dfrac{dy}{dt}\ =\ 2t(y+\dfrac{1}{2})$

$=>\ \dfrac{dy}{(y+\dfrac{1}{2})}\ =\ 2t dt$

On integrating both sides, we will have

$\int{\dfrac{dy}{(y+\dfrac{1}{2})}}\ =\ \int{2t dt}$

$=>\ log(y+\dfrac{1}{2})\ =\ 2.\dfrac{t^2}{2}\ + c$

$=>\ log(y+\dfrac{1}{2})\ =\ t^2\ +\ c$

Hence, the solution of given differential equation is

$log(y+\dfrac{1}{2})\ =\ t^2\ +\ c$

8 0

3 years ago

Mr. Anderson drove 168 miles in 3 1/2 hours. He then drove the next 2 1/4 hours at a rate of 5 miles per hour faster than the fi

BlackZzzverrR [31]

First we need to know how many miles per hour they drove the first time. This would be found by division. 168/3.5=48
Then we need to add 5 miles per hour for the average.48+5= 53
Next we need to multiply 2 1/4 by the new miles per hour 53*2.25=119.25
After that we need to know how many miles they drove 119.25+168=287.25 miles.
Finally to get an improper fraction we need to change 287.25 miles to it.
We already have 1/4 for .25 so we multiply 287 by 4 and add it to the 1/4 we already have and it makes it out to be 1149/4

Hope this helps!

8 0

2 years ago

Ryan flew from Wiley Post to Ponca City and back. Ryan maintained an average rate of 450 mph going to Ponca City and an average

Reil [10]

The answer is B, and here's why. Set up a table for "there" and "back" and use the distance = rate * time formula, like this:
               d             r            t
there       d           450         t
back        d           400        1-t

Let me explain this table to you. The distance is d, we don't know what it is, that's what we are actually looking for. We only know that if we go somewhere from point A to point B, then back again to point A, the distance there is the same as the distance back. Hence, the d in both spaces. There he flew 450 mph, back he flew 400 mph. If the total distance was 1 hour, he flew an unknown time there and one hour minus that unknown time back. For example, if he flew for 20 minutes there, one hour minus 20 minutes means that he flew 60 minutes - 20 minutes = 40 minutes back. See? Now, because the distance there = the distance back, we can set the rt in both equal to each other. If d = rt there and d = rt back and the d's are the same, then we can set the rt's equal to each other. 450t = 400(1-t) and
450t = 400 - 400t and 850t = 400. Solve for t to get t = .47058. Now, t is time, not the distance and we are looking for distance. So multiply that t value by the rate (cuz d = r*t) to get that the distance one way is
d = 450(.470580 and d = 211. 76 or, rounded like you need, 212.

4 0

2 years ago

Name two triangles that are congruent by the ASA Postulate.

lubasha [3.4K]

Is there a picture to this question?

8 0

3 years ago