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2 years ago

8

Suppose you cut along a diagonal of a rhombus, a rectangle, or a square. Would you get two triangles that have the same shape an

d size?

1 answer:

stich3 [128]2 years ago

3 0

I believe the answer to your question is

Yes, you would get two triangles that have the same shape and size

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A Robin comes to Tonyas birdfeeder every five days and the Bluejay comes by every three days. Today, The Robin and Bluejay both

fredd [130]

The answer is 4 days

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3 years ago

What is the midpoint of 2,6 and 10,12

-BARSIC- [3]

Answer:

= 6, 9

Step-by-step explanation:

To find the midpoint , we use the formula

midpoint = (x1+x1)/2, (y1+y2)/2

= (2+10)/2, (6+12)/2

= 12/2, 18/2

= 6, 9

5 0

2 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

Please help I need this ASAP!

yanalaym [24]

Answer:

nope

Step-by-step explanation:

3 0

2 years ago

Dovator [93]

12^2=144

144 is divided by b which also is an odd interger, there only 2 numbers : 1 and 3

if b=3, a^2=48, then there will be no a available

if b=1 , a^2=144, then a is 12 and 12 can not be divided by 9

The answer is D

3 0

3 years ago