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3 years ago

Solve the equation using the order of operations. 5+{2×[(5−1)+6]}÷4

Mathematics

1 answer:

creativ13 [48]3 years ago

5 0

Answer:

The order of operations is PEMDAS

First step is to do everything is parenthesis or brackets

5+{2*[(5-1)+6]}/4

{2*[(5-1)+6]}

First you would go 5-1=4, because that is the first equation that is in parenthesis by itself.

{2*[4+6]}

Next you would go 4+6=10, because that is your next smallest bracket

{2*10}

Last you would go 2*10=20

Your equation now looks like this

5+{20}/4

In PEMDAS your next step is exponents, but we don't have any so we go on to the next one which is multiply/division

5+5

You would go 20/4=5

Last step is to add/subtract

5+5=10

Your final answer would be 10

Hope this helps ;)

Step-by-step explanation:

You might be interested in

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

4 years ago

Use the diagram below to solve. The m∠ 1 = 29x + 12 and the m∠ 5 = 15x - 8. All lines that appear parallel are parallel.

Kamila [148]

Answer:

Equation: (29x + 12)° + (15x - 8)° = 180°

x = 4

m∠ 5 = 52°

Step-by-step explanation:

Given:

m∠ 1 = 29x + 12,

m∠ 5 = 15x - 8

Since all lines that appear are parallel, therefore, opposite angles of the quadrilateral are congruent.

Therefore, m<1 = m<4 = 29x + 12

m<4 + m<5 = 180° (linear pair)

(29x + 12) + (15x - 8) = 180° (substitution)

The equation above is what we would use in solving for x. Let's solve for x:

29x + 12 + 15x - 8 = 180

Collect like terms

44x + 4 = 180

Subtract 4 from both sides

44x = 180 - 4

44x = 176

Divide both sides by 44

x = 176/44

x = 4

m∠ 5 = 15x - 8

Plug in the value of x

m∠ 5 = 15(4) - 8 = 60 - 8

m∠ 5 = 52°

7 0

3 years ago

Find the vertex of the parabola. y = -4x2 - 16x - 11

emmasim [6.3K]

The standard equation is y=ax^2+bx+k and the vertex way is y=a(x-h)(x-h)+x
so the coordinate for x is -b/2a, for your problem a=-4 b=-16 so the coordinate for x is -(-16)/(2*(-4))=-2. Now you have to replace x=-2 in your equation y=-4(-2)^2-16(-2)-11= 5.
The vertex is (-2, 5)

5 0

4 years ago

Pls help I’m rlly desperate

LenaWriter [7]

Answer: nah

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

I need help with - The product of two integers is -21. The difference between the integers is -10. The sum of the two integers i

Soloha48 [4]

The two integers are +7 and -3.

We didn't need all three hints.
The answer can be found with
any two of them.

4 0

3 years ago