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castortr0y [4]
3 years ago
11

Mario's pizza has two choices of crust: deep dish and thin-and-crispy. the restaurant also has a choice of 5 toppings: tomatoes,

sausage, peppers, onions, and pepperoni. finally, mario's offers every pizza in extracheese as well as ‘regular’. if linda's volleyball team decides to order a pizza with four toppings, how many different choices do the teammates have at mario's pizza?
Mathematics
1 answer:
marusya05 [52]3 years ago
6 0
They have a choice of four toppings
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Match each function with its inverse
Brut [27]

Ok. This is an incomplete question. I don't want to report. Please tell me the answer choices so I can match them for you.

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3 years ago
Select the correct answer.
Elina [12.6K]

Answer:

  A)  6.86×10⁴

Step-by-step explanation:

You want to write 68600 in scientific notation.

<h3>Expanded form</h3>

The number 68600 can be written in expanded form with exponents as ...

  68600 = 6×10⁴ +8×10³ +6×10² +0×10¹ +0×10⁰

The left-most term of this sum tells you the exponent in scientific notation:

  68600 = 6.86×10⁴

5 0
1 year ago
PLEASE HELP!! Question with picture is below
Alenkinab [10]

Answer:

It honestly does not matter, but I'd choose the ring because I feel like it is more pleasant to look at.  

Step-by-step explanation:

We need to find the area of both shapes to compare:

3/4 of a circle:

A = 6^2pi

= 36pi

36pi(0.75) = 27pi

Ring:

A = 36pi

A2 = 3^2pi = 9pi

36pi-9pi = 27pi

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Hope this helped! :)

8 0
3 years ago
Read 2 more answers
A circle with a radius of 5 inches would have an area of about what?
Marrrta [24]
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Hence here the area is A=5^2*pi=25pi~78.5 square inches
6 0
3 years ago
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Prove that an = 4^n + 2(-1)^nis the solution to
olga nikolaevna [1]

Answer:

See proof below

Step-by-step explanation:

We have to verify that if we substitute a_n=4^n+2(-1)^n in the equation a_n=3a_{n-1}+4a_{n-2} the equality is true.

Let's substitute first in the right hand side:

3a_{n-1}+4a_{n-2}=3(4^{n-1}+2(-1)^{n-1})+4(4^{n-2}+2(-1)^{n-2})

Now we use the distributive laws. Also, note that (-1)^{n-1}=\frac{1}{-1}(-1)^n=(-1)(-1)^{n} (this also works when the power is n-2).

=3(4^{n-1})+6(-1)^{n-1}+4(4^{n-2})+8(-1)^{n-2}

=3(4^{n-1})+(-1)(6)(-1)^{n}+4^{n-1}+(-1)^2(8)(-1)^{n}

=4(4^{n-1})-6(-1)^{n}+8(-1)^{n}=4^n+2(-1)^n=a_n

then the sequence solves the recurrence relation.

4 0
3 years ago
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