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3 years ago

How many solutions can a quadratic-linear system have?

Mathematics

2 answers:

Contact [7]3 years ago

4 0

Answer:

Step-by-step explanation:

Paul [167]3 years ago

3 0

Answer:

Step-by-step explanation:

This type of system can have one solution, two solutions, or no solutions. Graph both equations on the same coordinate plane.

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Mrs.Flores’ rectangular garden has a length of 20 meters and width of 15 meters. Her neighbor, Mr.Sosa, has a similar garden in

natima [27]

180 m

Step-by-step explanation:

Information given:

length of Mrs. Flores' garden= 20 m
width of Mrs. Flores' garden = 15 m
Scale factor = 3

Since scale factor is 3, to get measurements of sides Mr. Sosa's garden, sides of Mrs. Flores's have to be multiplied by 3. Hence,

length of Mr. Sosa's garden= 3×20= 60 m

width of Mr. Sosa's garden= 3×15= 45 m

Diagonal of a rectangle = √length²+width²

Let Diagonal = D,

D= √60²+45²

=√3600+2025

=√5625

=75 m

If Mr. Sosa cut his garden diagonally,

perimeter= length+width+diagonal

=60+45+75=180 m

7 0

3 years ago

What is five percent of 150

3241004551 [841]

7.5 .. multiply .05 times 150

6 0

3 years ago

Read 2 more answers

Find the polynomial of minimum degree, with real coefficients, zeros at

drek231 [11]

Answer:

$\huge\boxed{p(x)=4x^3-20x^2+4x+300}$

Step-by-step explanation:

$\text{If}\ x=4\pm3i\ \text{and}\ x=-3\ \text{are the zeros of a polynomial, then it has a form:}\\\\p(x)=\bigg(x-(4-3i)\bigg)\bigg(x-(4+3i)\bigg)\bigg(x-(-3)\bigg)\bigg(r(x)\bigg)\\\\p(x)=(x-4+3i)(x-4-3i)(x+3)\bigg(r(x)\bigg)\\\\p(x)=\underbrace{\bigg((x-4)+3i\bigg)\bigg((x-4)-3i\bigg)}_{\text{use}\ (a+b)(a-b)=a^2-b^2}(x+3)\bigg(r(x)\bigg)\\\\p(x)=\bigg((x-4)^2-(3i)^2\bigg)(x+3)\bigg(r(x)\bigg)\qquad\text{use}\ (a-b)^2=a^2-2ab+b^2$

$p(x)=(x^2-2(x)(4)+4^2-3^2i^2)(x+3)\bigg(r(x)\bigg)\qquad\text{use}\ i^2=-1\\\\p(x)=(x^2-8x+16-9(-1))(x+3)\bigg(r(x)\bigg)\\\\p(x)=(x^2-8x+16+9)(x+3)\bigg(r(x)\bigg)\\\\p(x)=(x^2-8x+25)(x+3)\bigg(r(x)\bigg)\qquad\text{use FOIL}:\ (a+b)(c+d)=ac+ad+bc+bd\\\\p(x)=\bigg((x^2)(x)+(x^2)(3)+(-8x)(x)+(-8x)(3)+(25)(x)+(25)(3)\bigg)\bigg(r(x)\bigg)\\\\p(x)=(x^3+3x^2-8x^2-24x+25x+75)\bigg(r(x)\bigg)\qquad\text{combine like terms}\\\\p(x)=(x^3-5x^2+x+75)\bigg(r(x)\bigg)$

$\text{The y-intercept is at 300}.\\\\\text{For}\ w(x)=a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+...+a_1x+a_0\\\\\text{y-intercept is}\ a_0\\\\\text{Therefore for}\ p(x)=(x^3-5x^2+x+75)\bigg(r(x)\bigg)\\\\\text{y-intercet is}\ 75\bigg(r(x)\bigg)\\\\75\bigg(r(x)\bigg)=300\qquad\text{divide both sides by 75}\\\\r(x)=4\\\\\text{Finally:}\\\\p(x)=(x^3-5x^2+x+75)(4)\qquad\text{use the distributive property}\\\\p(x)=(x^3)(4)+(-5x^2)(4)+(x)(4)+(75)(4)\\\\p(x)=4x^3-20x^2+4x+300$

7 0

3 years ago

What needs to be corrected in this construction of a line parallel to AB passing through C?

monitta

Answer;

C) The second arc should be centered at C.

Explanation;

Assuming the goal is to construct a line parallel to AB that passes through given point C.

-Draw a line through C and across AB at an angle creating D.

- With the compass width about half of DC, and center D, draw the first arc to cross both lines.

-Using the same compass width , draw the second arc with center C.

-Then set the compass width to the lower arc (the first arc)

- Move the compass to the second arc. Mark off an arc to make point E

-Draw a straight line through C and E

Thus the line CE will be parallel to line AB

3 0

3 years ago

Read 2 more answers

What are the answers

kolezko [41]

Answer:

I use app called math

way

Step-by-step explanation:

7 0

2 years ago