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3 years ago

Can someone help pleassssssseeeeee!!!!!!!!!????????? ILL MARK BRAINLIEST!!!!

Mathematics

1 answer:

Lelu [443]3 years ago

4 0

Answer:

$b-17$

Step-by-step explanation:

So, Tanvi started with b buttons.

And she gave away 17.

Since she "gave away," this means that we should use subtraction.

Therefore, she now has b-17 buttons.

Our expression is:

$b-17$

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The value of y varies jointly with x and z. If y = 7 when z = 196 and x = 2, find the value of y when x = 3 and z = 336. I will

eimsori [14]

Answer:

Step-by-step explanation:

Given that:

y∞ xz

y=kxz. Where k is constant

When z=196 and x= 2 then y= 7

7=(196)(2)k

7=392k

k=1/56

There fore y=(1/56)xz

When x=3 and z =336

y=(1/56)xz

y=(1/56)(336)(3)

y=18

8 0

3 years ago

Complete the function table for y=2.5x−8 .

Black_prince [1.1K]

Answer:

x=3.2

Step-by-step explanation:

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2 years ago

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Leatherbound Stores, Inc., rejects a shipment of goods that does not conform to its contract with Cowhide Corporation, but is un

Dimas [21]

Answer:

resell, return, or store the goods

Step-by-step explanation:

8 0

3 years ago

The future value of a certain money market account with a fixed interest rate of

rosijanka [135]

Answer:

D = $8637.45

Step-by-step explanation:

Rate = 3.65% = 0.0365

Principal = 5000

Time (t) = 15 years

N = 12 (since its compounded monthly)

Compound interest (A) = P(1 + r/n)^nt

A = 5000(1 + 0.0365 / 12)^15*12

A = 5000(1 + 0.00304)¹⁸⁰

A = 5000(1.00304)¹⁸⁰

A = 5000 * 1.7269

A = 8634.86

The investment would worth $8634.86

Note: the final answer may vary slightly from the answer in the options due to ± from approximation

3 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