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3 years ago

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