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2 years ago

Help please (I have to write something else to send this question)

Mathematics

1 answer:

sp2606 [1]2 years ago

4 0

Answer:

You could do -2×1 which is -2

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Simplify (1+tan^2y)(cos2y-1)

Reil [10]

1+tan^2(y)=1/cos^2(y)
cos 2y -1=cos^2(y) -sin^2(y)-cos^2(y) -sin^2(y)

8 0

3 years ago

A rectangle on a coordinate plane has vertices at (7,5),(-7,5),(-7,-2),and (7,-2) what is the perimeter of the rectangle?

Nadya [2.5K]

Answer:

Your Perimiter is 42

Step-by-step explanation:

Distance from 7,5 to -7,5 is 14

Distance from -7,5 to -7,-2 is 7

Then we double both those numbers as it is a rectangle

2l + 2w = Per...

28 + 14 = 42

MARK AS BRAINLIEST

5 0

3 years ago

But his teacher that he needed to measure the height of a building in which his classroom is located he backs away from the buil

Ivenika [448]

Answer:

<h2>IT'S 22 JUST FOUND OUT </h2>

Step-by-step explanation:

3 0

2 years ago

An alternating current E(t)=120sin(12t) has been running through a simple circuit for a long time. The circuit has an inductance

Dvinal [7]

Answer:

14.488 amperes

Step-by-step explanation:

The amplitude I of the current is given by

$\large I=\displaystyle\frac{E_m}{Z}$

where

$\large E_m$ = amplitude of the energy source E(t).

Z = Total impedance.

The amplitude of the energy source is 120, the maximum value of E(t)

The total impedance is given by

$\large Z=\sqrt{R^2+(X_L-X_C)^2}$

where

<em> R= Resistance </em>

<em>L = Inductance </em>

<em>C = Capacitance </em>

<em>w = Angular frequency </em>

$\large X_L=wL$ = inductive reactance

$\large X_C=\displaystyle\frac{1}{wC}$ = capacitive reactance

As E(t) = 120sin(12t), the angular frequency w=12

$\large X_L=12*0.37=4.44\\\\X_C=1/(12*7)=0.012$

and

$\large Z=\sqrt{7^2+(4.44-0.012)^2}=8.283$

Finally

$\large I=\displaystyle\frac{E_m}{Z}=\frac{120}{8.283}=14.488\;amperes$

7 0

2 years ago

Evaluate the series

dalvyx [7]

Answer:

the value of the series;

$\sum_{k=1}^{6}(25-k^2) = 59$

C) 59

Step-by-step explanation:

Recall that;

$\sum_{1}^{n}a_n = a_1+a_2+...+a_n\\$

Therefore, we can evaluate the series;

$\sum_{k=1}^{6}(25-k^2)$

by summing the values of the series within that interval.

the values of the series are evaluated by substituting the corresponding values of k into the equation.

$\sum_{k=1}^{6}(25-k^2) =(25-1^2)+(25-2^2)+(25-3^2)+(25-4^2)+(25-5^2)+(25-6^2)\\\sum_{k=1}^{6}(25-k^2) =(25-1)+(25-4)+(25-9)+(25-16)+(25-25)+(25-36)\\\sum_{k=1}^{6}(25-k^2) =24+21+16+9+0+(-11)\\\sum_{k=1}^{6}(25-k^2) = 59\\$

So, the value of the series;

$\sum_{k=1}^{6}(25-k^2) = 59$

6 0

3 years ago

Help please (I have to write something else to send this question)​

Help please (I have to write something else to send this question)