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3 years ago

What conditional statement is represented by the Venn diagram below

Mathematics

2 answers:

andre [41]3 years ago

7 0

Answer:

Option C. is the correct answer.

Step-by-step explanation:

Since in the given Venn diagram "Blue circle" is contained in a "purple circle".

Blue circle represents Equilateral triangles and purple color circle represents isosceles triangle.

Since by this Venn diagram, equilateral triangles are contained in isosceles triangle then we can conclude that all equilateral triangles are isosceles triangle.

Therefore, option C. if shape is an equilateral triangle, then it is an isosceles triangle is true.

ipn [44]3 years ago

6 0

Answer:

Step-by-step explanation:

It's one of each or not both tho:)

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A potato sorting machine produces 5 pounds of potatoes an average for the standard aviation is a half a pound assuming the weigh

mixas84 [53]

For a 95% confidence interval, the corresponding z-score is 1.96. Therefore the deviation will by 1.96*0.5 lbs = 0.98 lbs. Therefore, the confidence interval will be (5 - 0.98, 5 + 0.98), which is (4.02, 5.98). The weight range is from 4.02 lbs to 5.98 lbs.

5 0

3 years ago

Pablo graphs a system of equations. One equation is quadratic and the other equation is linear. What is the greatest number of p

Nady [450]

For this case we have the following type of equations:
Quadratic equation:
$y = ax ^ 2 + bx + c
$
Linear equation:
$y = mx + b
$
We observe that when equating the equations we have:
$ax ^ 2 + bx + c = mx + b
$
Rewriting we have:
$ax ^ 2 + (b-m) x + (c-b) = 0
$
We obtain a polynomial of second degree, therefore, the maximum number of solutions that we can obtain is 2.
Answer:
The greatest number of possible solutions to this system is:
c.2

7 0

3 years ago

Read 2 more answers

Use the drop-down menus to choose steps in order to correctly solve 5(z + 6) – 2 = 9z for z.

shusha [124]

Solve for z:
5 (z + 6) - 2 = 9 z

5 (z + 6) = 5 z + 30:
5 z + 30 - 2 = 9 z

Add like terms. 30 - 2 = 28:
5 z + 28 = 9 z

Subtract 9 z from both sides:
(5 z - 9 z) + 28 = 9 z - 9 z

5 z - 9 z = -4 z:
-4 z + 28 = 9 z - 9 z

9 z - 9 z = 0:
28 - 4 z = 0

Subtract 28 from both sides:
(28 - 28) - 4 z = -28

28 - 28 = 0:
-4 z = -28

Divide both sides of -4 z = -28 by -4:
(-4 z)/(-4) = (-28)/(-4)

(-4)/(-4) = 1:
z = (-28)/(-4)

The gcd of 28 and -4 is 4, so (-28)/(-4) = (-(4×7))/(4 (-1)) = 4/4×(-7)/(-1) = (-7)/(-1):
z = (-7)/(-1)

(-7)/(-1) = (-1)/(-1)×7 = 7:
Answer: z = 7

5 0

3 years ago

Read 2 more answers

Subtract the given function and indicate the domain of the difference

maxonik [38]

Answer:

The domain is $x\in (-\infty,\infty).$

Step-by-step explanation:

Given functions $f(x)=x^2+3x+1$ and $g(x)=2x^2-4x-1$

Subtract these two functions:

$f(x)-g(x)\\ \\=(x^2+3x+1)-(2x^2-4x-1)\\ \\=x^2+3x+1-2x^2+4x+1\\ \\=(x^2-2x^2)+(3x+4x)+(1+1)\\ \\=-x^2+7x+2$

Plot these difference on the coordinate plane (see attached diagram). This function is defined for all vlues of x, so the domain is $x\in (-\infty,\infty).$

6 0

3 years ago

Define the double factorial of n, denoted n!!, as follows:n!!={1⋅3⋅5⋅⋅⋅⋅(n−2)⋅n} if n is odd{2⋅4⋅6⋅⋅⋅⋅(n−2)⋅n} if n is evenand (

tekilochka [14]

Answer:

Radius of convergence of power series is $\lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}=\frac{1}{108}$

Step-by-step explanation:

Given that:

n!! = 1⋅3⋅5⋅⋅⋅⋅(n−2)⋅n n is odd

n!! = 2⋅4⋅6⋅⋅⋅⋅(n−2)⋅n n is even

(-1)!! = 0!! = 1

We have to find the radius of convergence of power series:

$\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}](8x+6)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}]2^{n}(4x+3)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}](x+\frac{3}{4})^{n}\\$

Power series centered at x = a is:

$\sum_{n=1}^{\infty}c_{n}(x-a)^{n}$

$\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}](8x+6)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}]2^{n}(4x+3)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}4^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}](x+\frac{3}{4})^{n}\\$

$a_{n}=[\frac{8^{n}4^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}n!(3(n+1)+3)!(2(n+1))!!}{[(n+1+9)!]^{3}(4(n+1)+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}(n+1)!(3n+6)!(2n+2)!!}{[(n+10)!]^{3}(4n+7)!!}]$

Applying the ratio test:

$\frac{a_{n}}{a_{n+1}}=\frac{[\frac{32^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}]}{[\frac{32^{n+1}(n+1)!(3n+6)!(2n+2)!!}{[(n+10)!]^{3}(4n+7)!!}]}$

$\frac{a_{n}}{a_{n+1}}=\frac{(n+10)^{3}(4n+7)(4n+5)}{32(n+1)(3n+4)(3n+5)(3n+6)+(2n+2)}$

Applying n → ∞

$\lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}= \lim_{n \to \infty}\frac{(n+10)^{3}(4n+7)(4n+5)}{32(n+1)(3n+4)(3n+5)(3n+6)+(2n+2)}$

The numerator as well denominator of $\frac{a_{n}}{a_{n+1}}$ are polynomials of fifth degree with leading coefficients:

$(1^{3})(4)(4)=16\\(32)(1)(3)(3)(3)(2)=1728\\ \lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}=\frac{16}{1728}=\frac{1}{108}$

4 0

2 years ago