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Naya [18.7K]
3 years ago
7

Need helpp fastt plss will give brainliest to who is correct plss helpp

Mathematics
2 answers:
malfutka [58]3 years ago
8 0

Answer:

308, 77, large

Step-by-step explanation:

simple multiplication, and the larger one is more per cent.

antoniya [11.8K]3 years ago
6 0

Answer:

$308, $294, large

Step-by-step explanation:

5.5 * 56 = 308

10.5*28=294

294 is less than 308, so the large is the better deal

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To rent a certain meeting room, a college charges a reservation fee of $31 and an additional fee of $7.40 per hour. The chemistr
Mamont248 [21]
T= # of rental hours
Reservation Fee= $31
Per Hour Fee= $7.40


Multiply number of hours rented by the cost per hour, add that total to the reservation fee. The total needs to be less than $97.60.


Res Fee + ($ per hr * # hrs) < $97.60

$31 + $7.40t < $97.60
subtract 31 from both sides

7.40t < 66.60
divide both sides by 7.40

t < 9


ANSWER: To spend less than $97.60, they need to rent the room less than 9 hours.

Hope this helps! :)
7 0
3 years ago
Find the next term is in the explicit and recursive rule for the in term of the sequence. NO LINKS!!!
BARSIC [14]

9514 1404 393

Answer:

  5) 729, an=3^n, a[1]=3; a[n]=3·a[n-1]

  6) 1792, an=7(4^(n-1)), a[1]=7; a[n]=4·a[n-1]

Step-by-step explanation:

The next term of a geometric sequence is the last term multiplied by the common ratio. (This is the basis of the recursive formula.)

The Explicit Rule is ...

  a_n=a_1\cdot r^{n-1}

for first term a₁ and common ratio r.

The Recursive Rule is ...

  a[1] = a₁

  a[n] = r·a[n-1]

__

5. First term is a₁ = 3; common ratio is r = 9/3 = 3.

Next term: 243×3 = 729

Explicit rule: an = 3·3^(n-1) = 3^n

Recursive rule: a[1] = 3; a[n] = 3·a[n-1]

__

6. First term is a₁ = 7; common ratio is r = 28/7 = 4.

Next term: 448×4 = 1792

Explicit rule: an = 7·4^(n-1)

Recursive rule: a[1] = 7; a[n] = 4·a[n-1]

7 0
3 years ago
D, h, and p are coplanar ?
Allushta [10]

Answer:

Yes

Step-by-step explanation:

But if you stuck a pencil through the center of the picture the line that it makes would not be coplanar.

6 0
3 years ago
Read 2 more answers
Can anyone help me integrate :
worty [1.4K]
Rewrite the second factor in the numerator as

2x^2+6x+1=2(x+2)^2-2(x+2)-3

Then in the entire integrand, set x+2=\sqrt3\sec t, so that \mathrm dx=\sqrt3\sec t\tan t\,\mathrm dt. The integral is then equivalent to

\displaystyle\int\frac{(\sqrt3\sec t-2)(6\sec^2t-2\sqrt3\sec t-3)}{\sqrt{(\sqrt3\sec t)^2-3}}(\sqrt3\sec t)\,\mathrm dt
=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\sec^2t-1}}\,\mathrm dt
=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\tan^2t}}\,\mathrm dt
=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{|\tan t|}\,\mathrm dt

Note that by letting x+2=\sqrt3\sec t, we are enforcing an invertible substitution which would make it so that t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3} requires 0\le t or \dfrac\pi2. However, \tan t is positive over this first interval and negative over the second, so we can't ignore the absolute value.

So let's just assume the integral is being taken over a domain on which \tan t>0 so that |\tan t|=\tan t. This allows us to write

=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\tan t}\,\mathrm dt
=\displaystyle\int(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\csc t\,\mathrm dt

We can show pretty easily that

\displaystyle\int\csc t\,\mathrm dt=-\ln|\csc t+\cot t|+C
\displaystyle\int\sec t\csc t\,\mathrm dt=-\ln|\csc2t+\cot2t|+C
\displaystyle\int\sec^2t\csc t\,\mathrm dt=\sec t-\ln|\csc t+\cot t|+C
\displaystyle\int\sec^3t\csc t\,\mathrm dt=\frac12\sec^2t+\ln|\tan t|+C

which means the integral above becomes

=3\sqrt3\sec^2t+6\sqrt3\ln|\tan t|-18\sec t+18\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|-6\ln|\csc t+\cot t|+C
=3\sqrt3\sec^2t-18\sec t+6\sqrt3\ln|\tan t|+12\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|+C

Back-substituting to get this in terms of x is a bit of a nightmare, but you'll find that, since t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}, we get

\sec t=\dfrac{x+2}{\sqrt3}
\sec^2t=\dfrac{(x+2)^2}3
\tan t=\sqrt{\dfrac{x^2+4x+1}3}
\cot t=\sqrt{\dfrac3{x^2+4x+1}}
\csc t=\dfrac{x+2}{\sqrt{x^2+4x+1}}
\csc2t=\dfrac{(x+2)^2}{2\sqrt3\sqrt{x^2+4x+1}}

etc.
3 0
3 years ago
What is the value of x in the eqation 5x+3=4x
Reika [66]

Answer :x=-3

Step-by-step explanation:Solve for  x by simplifying bothh sides of the equation then isoolating the variable

7 0
2 years ago
Read 2 more answers
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