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2 years ago

Margo is practicing her free throws for basketball

Mathematics

1 answer:

kipiarov [429]2 years ago

5 0

Answer:

Step-by-step explanation:

Well if you think about it, if she has a total of 10 attempts she has to make 8 of them to make it on the team. Now the total attempts is double so we double the number of throws she needs to make in order to make the team so to do this we would do 8*2 which is 16. She needs to make 16 throws to make it on the team.

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The larger of two numbers is 10 less than twice the smaller number. if the sum of the two numbers is 38, find the tqo nubers

Iteru [2.4K]

Let two number A and B, and A >B

as we know A+10=2B① && A+B =38②

①-② solve that
B=16 so A=22

so, they are 16 and 22

3 0

3 years ago

What is the missing numbers<br><br> 9 tenths + 7 hundredths<br><br> 6 tenths + _ hundredths = 0.66

dolphi86 [110]

The first one is 0.97 the second one is 6

7 0

2 years ago

Read 2 more answers

The table shows the number of high school employees by gender and position, what percent of the

miss Akunina [59]

Answer: 22%

Step-by-step explanation:

6 0

3 years ago

Prove or disprove (from i=0 to n) sum([2i]^4) <= (4n)^4. If true use induction, else give the smallest value of n that it doe

ddd [48]

Answer:

The statement is true for every n between 0 and 77 and it is false for $n\geq 78$

Step-by-step explanation:

First, observe that, for n=0 and n=1 the statement is true:

For n=0: $\sum^{n}_{i=0} (2i)^4=0 \leq 0=(4n)^4$

For n=1: $\sum^{n}_{i=0} (2i)^4=16 \leq 256=(4n)^4$

From this point we will assume that $n\geq 2$

As we can see, $\sum^{n}_{i=0} (2i)^4=\sum^{n}_{i=0} 16i^4=16\sum^{n}_{i=0} i^4$ and $(4n)^4=256n^4$ . Then,

$\sum^{n}_{i=0} (2i)^4 \leq(4n)^4 \iff \sum^{n}_{i=0} i^4 \leq 16n^4$

Now, we will use the formula for the sum of the first 4th powers:

$\sum^{n}_{i=0} i^4=\frac{n^5}{5} +\frac{n^4}{2} +\frac{n^3}{3}-\frac{n}{30}=\frac{6n^5+15n^4+10n^3-n}{30}$

Therefore:

$\sum^{n}_{i=0} i^4 \leq 16n^4 \iff \frac{6n^5+15n^4+10n^3-n}{30} \leq 16n^4 \\\\ \iff 6n^5+10n^3-n \leq 465n^4 \iff 465n^4-6n^5-10n^3+n\geq 0$

and, because $n \geq 0$ ,

$465n^4-6n^5-10n^3+n\geq 0 \iff n(465n^3-6n^4-10n^2+1)\geq 0 \\\iff 465n^3-6n^4-10n^2+1\geq 0 \iff 465n^3-6n^4-10n^2\geq -1\\\iff n^2(465n-6n^2-10)\geq -1$

Observe that, because $n \geq 2$ and is an integer,

$n^2(465n-6n^2-10)\geq -1 \iff 465n-6n^2-10 \geq 0 \iff n(465-6n) \geq 10\\\iff 465-6n \geq 0 \iff n \leq \frac{465}{6}=\frac{155}{2}=77.5$

In concusion, the statement is true if and only if n is a non negative integer such that $n\leq 77$

So, 78 is the smallest value of n that does not satisfy the inequality.

Note: If you compute $(4n)^4- \sum^{n}_{i=0} (2i)^4$ for 77 and 78 you will obtain:

$(4n)^4- \sum^{n}_{i=0} (2i)^4=53810064$

$(4n)^4- \sum^{n}_{i=0} (2i)^4=-61754992$

7 0

3 years ago

Whats the cost of each vegetable if 5 squash and 2 zucchini is $1.32

alex41 [277]

Answer:

Step-by-step explanation:

We first need to define a couple of variables. Let s = the cost of 1 squash and z = the cost of 1 zucchini.

Now lets translate the words into algebra:

"The cost of 5 squash and 2 zucchini is $1.32" ===> 5s + 2z = 1.32

"Three squash and 1 zucchini cost $0.75" ===> 3s + z = 0.75

There are several ways to solve systems of equations. Let's use substitution. We can find what z equals in terms of s by manipulating the second equation:

3s + z = 0.75

-3s -3s

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

z = -3s +0.75

Now lets substitute (-3s + 0.75) into the first equation for z, then solve for s:

5s + 2(-3s + 0.75) = 1.32

Can you handle it from here?

(Hint: Once you have solved for s, you can substitute that value back into either of the equations and solve for z.)

3 0

3 years ago