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zalisa [80]
3 years ago
10

There are 4 veggie pizzas for every 5 cheese pizzas. How many cheese pizzas are there if 108 pizzas total are ordered? please sh

ow work
Mathematics
2 answers:
Andre45 [30]3 years ago
7 0

Answer:

6.25

Step-by-step explanation:

Katarina [22]3 years ago
5 0

Answer:

60 because [total of pizzas]/(veggie + cheese)= 12 ((108/(4+5) = 12)) so you take 5 (the amount of cheese pizzas) and multiply it by 12 to get 60

sorry if i explained it bad im not too good at explaining

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Since 35 is an angle at the circumference
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3 years ago
Write all the terms of the following X+y+z​
lutik1710 [3]

Answer:

  • x
  • y
  • z

Step-by-step explanation:

The terms are the addends in the sum:

  x, y, z

4 0
3 years ago
Math help 20 points<br> help
astra-53 [7]

Answer:

1. x =8

2.  x=9

Step-by-step explanation:

Since these figures are parallelograms, the opposite sides are equal.

1.     6 = 2x-10

Add 10 to each side

   6+10 = 2x-10+10

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Divide  each side by 2

16/2 = 2x/2

 8 =x


2.  x+14 =23

    Subtract 14 from each side

x+14-14 = 23-14

x = 9



6 0
3 years ago
Find the taylor series centered at the given value of a and find the associated radius of convergence. (1) f(x) = 1 x , a = 1 (2
Kitty [74]

a) The radius of convergence is calculated as

R=1.

b) Due to the fact that it converges in every direction, the radius of convergence is either infinity or zero.

<h3>What is the associated radius of convergence.?</h3>

(a)

Take into consideration the function f with respect to the number a,

f(x)=\frac{1}{x}, \quad a=1

In case you forgot, the Taylor series for the function $f$ at the number a looks like this:

\begin{aligned}f(x) &=\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n !}(x-a)^{n} \\&=f(a)+\frac{f^{\prime}(a)}{1 !}(x-a)+\frac{f^{*}(a)}{2 !}(x-a)^{2}+\ldots\end{aligned}

Determine the function f as well as any derivatives of the function $f by setting a=1 and working backward from there.

\begin{aligned}f(x) &=\frac{1}{x} & f(1)=\frac{1}{1}=1 \\\\f^{\prime}(x) &=-\frac{1}{x^{2}} &  f^{\prime}(1)=-\frac{1}{(1)^{2}}=-1 \\\\f^{\prime \prime}(x) &=\frac{2}{x^{3}} &  f^{\prime \prime}(1)=\frac{2}{(1)^{3}}=2 \\\\f^{\prime \prime}(x) &=-\frac{2 \cdot 3}{x^{4}} & f^{\prime \prime}(1)=-\frac{2 \cdot 3}{(1)^{4}}=-2 \cdot 3 \\\\f^{(*)}(x) &=\frac{2 \cdot 3 \cdot 4}{x^{5}} & f^{(n)}(1)=\frac{2 \cdot 3 \cdot 4}{(1)^{5}}=2 \cdot 3 \cdot 4\end{aligned}

At the point when a = 1, the Taylor series for the function f looks like this:

f(x) &=f(a)+\frac{f^{\prime}(a)}{1 !}(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2}+\frac{f^{\prime \prime \prime}(a)}{3 !}(x-a)^{3}+\cdots \\\\&=f(1)+\frac{f^{\prime}(1)}{1 !}(x-1)+\frac{f^{\prime}(1)}{2 !}(x-1)^{2}+\frac{f^{\prime \prime}(1)}{3 !}(x-1)^{3}+\cdots \\

&=1+\frac{-1}{1 !}(x-1)+\frac{2}{2 !}(x-1)^{2}+\frac{-2 \cdot 3}{3 !}(x-1)^{3}+\frac{2 \cdot 3 \cdot 4}{4 !}(x-1)^{4}+\cdots \\\\&=1-(x-1)+(x-1)^{2}-(x-1)^{3}+(x-1)^{4}+\cdots \\\\&=\sum_{1=0}^{\infty}(-1)^{n}(x-1)^{n}

In conclusion,

&=\sum_{1=0}^{\infty}(-1)^{n}(x-1)^{n}

Find the radius of convergence by using the Ratio Test in the following manner:

\begin{aligned}L &=\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right| \\&=\lim _{n \rightarrow \infty} \frac{(-1)^{n+1}(x-1)^{n+1}}{(-1)^{n}(x-1)^{n}} \mid \\&=\lim _{n \rightarrow \infty}|x-1| \\&=|x-1|\end{aligned}

The convergence of the series when L<1, that is, |x-1|<1.

The radius of convergence is calculated as

R=1.

For B

Take into consideration the function f with respect to the number a,

a_{n}=(-1)^{n}(x-1)^{n}

f(x)=\left(x^{2}+2 x\right) e^{x},  a=0 The Taylor series for f(x)=e^{x} at a=0 is,

e^{2}=1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\frac{x^{4}}{4 !}+\ldots

f(x) &=\left(x^{2}+2 x\right) e^{x} \\&=\left(x^{2}+2 x\right)\left(1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\frac{x^{4}}{4 !}+\ldots\right)+2 x\left(1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\frac{x^{4}}{4 !}+\ldots\right) \\&=x^{2}\left(1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\frac{x^{4}}{4 !}+\ldots\right)+\left(\frac{x^{4}}{2 !}+\frac{x^{5}}{3 !}+\frac{x^{6}}{4 !}+\ldots\right)+\left(2 x+2 x^{2}+\frac{2 x^{3}}{2 !}+\frac{2 x^{4}}{3 !}+\frac{2 x^{5}}{4 !}+\ldots\right) \\

&=\left(x^{2}+x^{3}+\frac{x^{4}}{2 !}\right) \\&=2 x+3 x^{2}+\left(1+\frac{2}{2 !}\right) x^{3}+\left(\frac{1}{2 !}+\frac{2}{3 !}\right) x^{4}+\left(\frac{1}{4 !}\right) x^{5}+\ldots \\&=2 x+3 x^{2}+2 x^{3}+\frac{5}{6} x^{4}+\frac{1}{4} x^{5}+\ldots

Due to the fact that it converges in every direction, the radius of convergence is either infinity or zero.

Read more about convergence

brainly.com/question/15415793

#SPJ4

The complete question is attached below

3 0
2 years ago
The shaded part of the diagram shows what tina had left from a yard of fabric.she now use 1/3 yard yard of fabric for one projec
ioda

Answer: Left over yard after the two projects is \frac{1}{2}\ yards

Step-by-step explanation:

Since we have given that

Number of yard of fabric for one project is given by

\frac{1}{3}

Number of yard of fabric for second project is given by

\frac{1}{6}

Total number of yards of first and second project is given by

\frac{1}{3}+\frac{1}{6}\\\\=\frac{2+1}{6}\\\\=\frac{3}{6}\\\\=\frac{1}{2}\ yards

Number of original of fabric Tina have left after the two project is given by

1-\frac{1}{2}\\\\=\frac{2-1}{2}\\\\=\frac{1}{2}\ yards

Hence, Left over yard after the two projects is \frac{1}{2}\ yards

8 0
3 years ago
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