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2 years ago

What is another way to write the expression t⋅(14−5) ?

Mathematics

2 answers:

grigory [225]2 years ago

8 0

t⋅(14−5)

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14t -5t

Choice A

Nastasia [14]2 years ago

4 0

The answer A..i hope it helped

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Help please asap!!<br> A- z=15<br> B- z= 20<br> C- z=23<br> D- z=41

grigory [225]

Answer:

z = 15

Step-by-step explanation:

The sum S of the interior angles of a regular polygon is given by the formula

S = (n-2) x 180 where n is the number sides

Here n = 9

So S = (9-2) x 180 = 7 x 180 = 1,260

There are 9 interior angles and each angle is (5z + 65)

So the sum of all 9 interior angles = 9 (5z + 65)

= 45z + 585

Set these equal to each other and solve for z

45z + 585 = 1260

45z = 675

z = 675/45 = 15

4 0

5 months ago

Can anyone help me with this? I'll mark brainliest for best answer

soldier1979 [14.2K]

Answer:

The answer is k = -12

4 0

3 years ago

How do you do this problem

Sliva [168]

Answer:

Step-by-step explanation:

sec= x

cross multiply

60 mb x 2

120 mb/29 sec

6 0

3 years ago

PLEASEEE SOMEONE HELP ME ON THIS!!!! It’s due in a couple hours

zimovet [89]

Answer:

x = 8.0 m

Step-by-step explanation:

With the angle presented, I'm assuming you know basic trigonometry. We know cosθ = adj/hyp from SOH CAH TOA. We have angle and hypotenuse, so plugging in those values we get cos(20°) = x / 8.5. Solve for x to get 8.5cos(20°) = 7.987387277 m = 8.0 m.

4 0

2 years ago

Read 2 more answers

Integrate this two questions

tankabanditka [31]

Simplify the integrands by polynomial division.

$\dfrac{t^2}{1 - 3t} = -\dfrac19 \left(3t + 1 - \dfrac1{1 - 3t}\right)$

$\dfrac t{1 + 4t} = \dfrac14 \left(1 - \dfrac1{1 + 4t}\right)$

Now computing the integrals is trivial.

$\displaystyle \int \frac{t^2}{1 - 3t} \, dt = -\frac19 \int \left(3t + 1 - \frac1{1-3t}\right) \, dt \\\\ = -\frac19 \left(\frac32 t^2 + t + \frac13 \ln|1 - 3t|\right) + C \\\\ = \boxed{-\frac{t^2}6 - \frac t9 - \frac{\ln|1-3t|}{27} + C}$

where we use the power rule,

$\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C ~~~~ (n\neq-1)$

and a substitution to integrate the last term,

$\displaystyle \int \frac{dt}{1-3t} = -\frac13 \int \frac{du}u \\\\ = -\frac13 \ln|u| + C \\\\ = -\frac13 \ln|1-3t| + C ~~~ (u=1-3t \text{ and } du = -3\,dt)$

$\displaystyle \int \frac t{1+4t} \, dt = \frac14 \int \left(1 - \frac1{1+4t}\right) \, dt \\\\ = \frac14 \left(t - \frac14 \ln|1 + 4t|\right) + C \\\\ = \boxed{\frac t4 - \frac{\ln|1+4t|}{16} + C}$

using the same approach as above.

5 0

1 year ago

What is another way to write the expression ​ t⋅(14−5) ​?

What is another way to write the expression t⋅(14−5) ?