What is a math problem that solves for 6????

A bakery sells a pack of peanut butter cookies for $3 and a

Westkost [7]

F(x,y) =3x + 5y (hope this helps )

5 0

2 years ago

A force of 60 N is used to stretch two springs that are initially the same length. Spring A has a spring constant of 4 N/m, and

goblinko [34]

Answer:

D:Spring A is 3 m longer than spring B because 15 – 12 = 3.

Step-by-step explanation:

In this question, you should remember the Hooke's Law in physics.

The Hooke's Law simply explains that the extension that occurs on a spring is directly proportional to the load applied on it.

The mathematical expression for this law is

$F=-kx$

where;

F= force applied on the spring

x = the extension on the spring

k= the spring constant which varies in spring.

The question will need you to calculate the extension on the springs A and B then compare the values obtained.

<u>In spring A</u>

Force, F=60N and spring constant ,k=4 N/m

To find the extension x apply the expression;

$F=-kx\\\\60=-4*x\\\\60=-4x\\\\\frac{60}{-4} =\frac{-4x}{-4} \\\\\\-15=x$

Here the spring extension is 15 m

<u>In spring B</u>

Force, F=60N and spring constant , k=5N/m

To find the extension x apply the same expression

$F=-kx\\\\60=-5*x\\\\60=-5x\\\\\\\frac{60}{-5} =\frac{-5x}{-5} \\\\\\-12=x$

Here the extension on the spring is 12 m

<u>Compare</u>

The extension on spring A is 3 m longer than that in spring B because when you subtract the value of spring B from that in spring A you get 3m

$=15m-12m=3m$

5 0

3 years ago

PLEASE HELP!!!

AfilCa [17]

Answer:

$h=\dfrac{10\sqrt{3}}{2}=5\sqrt{3}\ cm.$

$JP=5\sqrt{3}\ cm$

Step-by-step explanation:

Connecting points O and E and points O and J, we get triangle EOJ. This triangle is equilateral triangle, because OJ=OE=JE=r=10 cm.

Since EP⊥IJ, then segment JP is the height of the triangle EOJ.

The height of the equilateral triangle can be found using formula

$h=\dfrac{a\sqrt{3}}{2},$

where a is the side length.

So,

$h=\dfrac{10\sqrt{3}}{2}=5\sqrt{3}\ cm.$

Therefore JP is 5√3 cm

3 0

3 years ago

Find a factorization of x² + 2x³ + 7x² - 6x + 44, given that<br> −2+i√√7 and 1 - i√/3 are roots.

Levart [38]

A factorization of $x^4+2x^3+7x^2-6x+44$ is $(x^2+4x+11)(x^2-2x+4)$ .

<h3>What are the properties of roots of a polynomial?</h3>

The maximum number of roots of a polynomial of degree $n$ is $n$ .
For a polynomial with real coefficients, the roots can be real or complex.
The complex roots of a polynomial with real coefficients always exist in a pair of conjugate numbers i.e., if $a+ib$ is a root, then $a-ib$ is also a root.

If the roots of the polynomial $p(x)=ax^4+bx^3+cx^2+dx+e$ are $r_1,r_2,r_3,r_4$ , then it can be factorized as $p(x)=(x-r_1)(x-r_2)(x-r_3)(x-r_4)$ .

Here, we are to find a factorization of $p(x)=x^4+2x^3+7x^2-6x+44$ . Also, given that $-2+i\sqrt{7}$ and $1-i\sqrt{3}$ are roots of the polynomial.

Since $p(x)=x^4+2x^3+7x^2-6x+44$ is a polynomial with real coefficients, so each complex root exists in a pair of conjugates.

Hence, $-2-i\sqrt{7}$ and $1+i\sqrt{3}$ are also roots of the given polynomial.

Thus, all the four roots of the polynomial $p(x)=x^4+2x^3+7x^2-6x+44$ , are: $r_1=-2+i\sqrt{7}, r_2=-2-i\sqrt{7}, r_3=1-i\sqrt{3}, r_4=1+i\sqrt{3}$ .

So, the polynomial $p(x)=x^4+2x^3+7x^2-6x+44$ can be factorized as follows:

$\{x-(-2+i\sqrt{7})\}\{x-(-2-i\sqrt{7})\}\{x-(1-i\sqrt{3})\}\{x-(1+i\sqrt{3})\}\\=(x+2-i\sqrt{7})(x+2+i\sqrt{7})(x-1+i\sqrt{3})(x-1-i\sqrt{3})\\=\{(x+2)^2+7\}\{(x-1)^2+3\}\hspace{1cm} [\because (a+b)(a-b)=a^2-b^2]\\=(x^2+4x+4+7)(x^2-2x+1+3)\\=(x^2+4x+11)(x^2-2x+4)$

Therefore, a factorization of $x^4+2x^3+7x^2-6x+44$ is $(x^2+4x+11)(x^2-2x+4)$ .

To know more about factorization, refer: brainly.com/question/25829061

#SPJ9

3 0

1 year ago

Read 2 more answers

Find the equation of the line which passes through (4, -1) and (2,-1).

-BARSIC- [3]

I think it’s D maybe

8 0

3 years ago