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Sophie [7]
3 years ago
13

?????????????????????????

Mathematics
2 answers:
Inga [223]3 years ago
7 0
Answer= 1

In an equilateral triangle all sides are equivalent and all angles are equivalent. Since all sides are equivalent, they all equal 60°.

180°/3=60°

Since we know all side lengths and all angle measurements of the triangle, their is only one possibility of how the triangle will look.


frez [133]3 years ago
3 0
I'm pretty sure just 1, i don't really know what the question is asking
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Please help with this question.
horsena [70]

Answer:

Apples and Peaches

Step-by-step explanation:

It is convenient to compare the prices for 3 lbs of each of the fruits.

Multiply the price and quantity of Apples by 4:

... 4×(3/4 lb) = 4×$1.59

... 3 lb = $6.36 . . . . for apples

Multiply the price and quantity of Plums by 6:

... 6×(0.5 lb) = 6×$1.15

... 3 lb = $6.90 . . . . for plums

The two lowest-priced fruits are apples (6.36/(3 lb)) and peaches (5.37/(3 lb)).

_____

<em>Comment on</em> the best deal

No criteria are provided for determining the best deal. Though plums are higher-priced, they have somewhat higher polyphenol content than apples or peaches, so may provide the best deal in terms of nutrition per dollar.

4 0
3 years ago
Use the following table to find the Rate of Change.
ELEN [110]

Answer:

x=1 and y= 6

Step-by-step explanation:

and for y its a positive cause their both negitives

3 0
3 years ago
Ashley earn 60 points every time she shops at a grocery store she needs a total of 2580 points to receive a prize so far she has
lutik1710 [3]
Answer: 35 more times
How: (if she gets 60 points for each grocery store visit then first do 2580-480=2100. she still needs 2100 points so then 2100/60 = 35 grocery visits!)
4 0
3 years ago
Mathematical induction, prove the following two statements are true
adelina 88 [10]
Prove:
1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+n\left(\frac12\right)^{n-1}=4-\dfrac{n+2}{2^{n-1}}
____________________________________________

Base Step: For n=1:
n\left(\frac12\right)^{n-1}=1\left(\frac12\right)^{0}=1
and
4-\dfrac{n+2}{2^{n-1}}=4-3=1
--------------------------------------------------------------------------

Induction Hypothesis: Assume true for n=k. Meaning:
1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+k\left(\frac12\right)^{k-1}=4-\dfrac{k+2}{2^{k-1}}
assumed to be true.

--------------------------------------------------------------------------

Induction Step: For n=k+1:
1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+k\left(\frac12\right)^{k-1}+(k+1)\left(\frac12\right)^{k}

by our Induction Hypothesis, we can replace every term in this summation (except the last term) with the right hand side of our assumption.
=4-\dfrac{k+2}{2^{k-1}}+(k+1)\left(\frac12\right)^{k}

From here, think about what you are trying to end up with.
For n=k+1, we WANT the formula to look like this:
1+2\left(\frac12\right)+...+k\left(\frac12\right)^{k-1}+(k+1)\left(\frac12\right)^{k}=4-\dfrac{(k+1)+2}{2^{(k+1)-1}}

That thing on the right hand side is what we're trying to end up with. So we need to do some clever Algebra.

Combine the (k+1) and 1/2, put the 2 in the bottom,
=4-\dfrac{k+2}{2^{k-1}}+\dfrac{(k+1)}{2^{k}}

We want to end up with a 2^k as our final denominator, so our middle term is missing a power of 2. Let's multiply top and bottom by 2,
=4+\dfrac{-2(k+2)}{2^{k}}+\dfrac{(k+1)}{2^{k}}

Distribute the -2 and combine the fractions together,
=4+\dfrac{-2k-4+(k+1)}{2^{k}}

Combine like-terms,
=4+\dfrac{-k-3}{2^{k}}

pull the negative back out,
=4-\dfrac{k+3}{2^{k}}

And ta-da! We've done it!
We can break apart the +3 into +1 and +2,
and the +0 in the bottom can be written as -1 and +1,
=4-\dfrac{(k+1)+2}{2^{(k-1)+1}}
3 0
3 years ago
Which is equivalent to (-1+2i)(5i)
Naya [18.7K]

Answer:

.

Step-by-step explanation:

3 0
3 years ago
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