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2 years ago

11

15 subtracted from the square of an integer the result is equal to three less than four times the integer find the negative inte

ger

1 answer:

dangina [55]2 years ago

5 0

Answer:

Let the integer be called "n"

So we have:

n^2-15=4n-3

n^2=4n+12

n^2-4n-12=0

(n+2)(n-6)=0

So, the negative integer is -2.

Let me know if this helps!

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Find the volume (Don't forget your UNITS) *

Radda [10]

Answer:

135 cm³

Step-by-step explanation:

To find volume: you multiply length*width*height

2. $l*w*h$

3. 6*5*4 1/2

4. 6*5=30

5. 30*4 1/2=135

6. 135 cm³

Hope this helped!! :)

Stay safe and have a wonderful day/night!!!!!

Brainliest?!?!

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2 years ago

Donald bought a box of golf balls for $9.27. There were 18 golf balls in the box.

wlad13 [49]

Answer:

Each golf ball costs about <u>$2</u>.

Step-by-step explanation:

Given:

Donald bought a box of golf balls for $9.27. There were 18 golf balls in the box.

Rename the decimal dividend as a whole number that is compatible with the divisor to estimate the quotient.

Now, to find the cost of each golf ball.

<u>As given:</u>

<em>Rename the decimal dividend as a whole number that is compatible with the divisor to estimate the quotient.</em>

As, the cost of golf ball box = $9.27.

So, 9.27 nearest to whole number is 9.

Thus to get the cost of each ball:

18 golf balls cost = $9.

So, 1 golf ball cost = $18\div 9=2.$

Therefore, each golf ball costs about $2.

3 0

3 years ago

Find a linear second-order differential equation f(x, y, y', y'') = 0 for which y = c1x + c2x3 is a two-parameter family of solu

Alisiya [41]

Let $y=C_1x+C_2x^3=C_1y_1+C_2y_2$ . Then $y_1$ and $y_2$ are two fundamental, linearly independent solution that satisfy

$f(x,y_1,{y_1}',{y_1}'')=0$
$f(x,y_2,{y_2}',{y_2}'')=0$

Note that ${y_1}'=1$ , so that $x{y_1}'-y_1=0$ . Adding $y''$ doesn't change this, since ${y_1}''=0$ .

So if we suppose

$f(x,y,y',y'')=y''+xy'-y=0$

then substituting $y=y_2$ would give

$6x+x(3x^2)-x^3=6x+2x^3\neq0$

To make sure everything cancels out, multiply the second degree term by $-\dfrac{x^2}3$ , so that

$f(x,y,y',y'')=-\dfrac{x^2}3y''+xy'-y$

Then if $y=y_1+y_2$ , we get

$-\dfrac{x^2}3(0+6x)+x(1+3x^2)-(x+x^3)=-2x^3+x+3x^3-x-x^3=0$

as desired. So one possible ODE would be

$-\dfrac{x^2}3y''+xy'-y=0\iff x^2y''-3xy'+3y=0$

(See "Euler-Cauchy equation" for more info)

6 0

3 years ago

State how many imaginary and real zeros the function has.

sammy [17]

Answer:

Step-by-step explanation:

If you'd graph this function on a graphing calculator or graphing utility, you'd see quickly that the graph never touches or crosses the x-axis. This tells us immediately that there are no real roots; all four roots are complex or imaginary.

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2 years ago

What is2+2x45+6-7 i cant figure it out

Vaselesa [24]

The answer is 179
yup

6 0

3 years ago

Read 2 more answers