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1 year ago

can you please give me the answer to this question please hurry I need this answer for my homework 3¹. (3³)³ =

Mathematics

1 answer:

Otrada [13]1 year ago

5 0

Answer:

59,049

Step-by-step explanation:

3^3 = 27

27^3 = 19683

19683*3 = 59,049

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16th term in the geometric sequence 1024, -512, 256, ....

tresset_1 [31]

Answer:-1/32

Step-by-step explanation:

First term=a=1024

Common ratio=r=-512/1024

r=-1/2

Using them formula

Tn=a x r^(n-1)

T16=1024 x (-1/2)^(16-1)

T16=1024 x (-1/2)^15

T16=1024 x -1/32768

T16=-1024/32768

T16=-1/32

5 0

2 years ago

Please help me I’m so lost

Svetllana [295]

Answer:

759.88 square inches

Step-by-step explanation:

$r = \frac{1}{2}d = \frac{1}{2} \times 44 = 22 \\ \\ area = \frac{1}{2} \pi \: {r}^{2} \\ area = \frac{1}{2} \times 3.14 \times 22 \times 22 \\ area = 759.88 \: square \: inches$

4 0

2 years ago

Read 2 more answers

Which are the solutions of x2 = –7x – 8?

vovangra [49]

Answer:

The solutions are:

$x_1=\frac{-7+\sqrt{17}}{2}$ $x_2=\frac{-7-\sqrt{17}}{2}$

Step-by-step explanation:

We have the following quadratic equation

$x^2 = -7x - 8$

We can rewrite the equation as follows

$x^2+7x + 8=0$

Now we use the quadratic formula to solve the equation

For an equation of the form $ax ^ 2 + bx + c = 0$ the quadratic formula is:

$x=\frac{-b\±\sqrt{b^2-4ac}}{2a}$

In this case:

$a=1,\ b=7,\ c=8$

Then:

$x=\frac{-7\±\sqrt{7^2-4(1)(8)}}{2(1)}$

$x=\frac{-7\±\sqrt{49-32}}{2}$

$x=\frac{-7\±\sqrt{17}}{2}$

$x_1=\frac{-7+\sqrt{17}}{2}$

$x_2=\frac{-7-\sqrt{17}}{2}$

7 0

2 years ago

Read 2 more answers

Two equations are given below:

ch4aika [34]

Rewrite the equations;
m+4n=8
m-n=-2

Eliminating m gives;
5n=10
n=2
Replacing for n in the first equation;
m+4(2)=8
m=8
Answer;
(0, 2)

6 0

2 years ago

A projectile is launched into the air. The function h(t) = –16t2 + 32t + 128 gives the height, h, in feet, of the projectile t s

Zinaida [17]

Answer:

t = 4 seconds

Step-by-step explanation:

The height of the projectile after it is launched is given by the function :

$h(t)=-16t^2+32t+128$

t is time in seconds

We need to find after how many seconds will the projectile land back on the ground. When it land, h(t)=0

So,

$-16t^2+32t+128=0$

The above is a quadratic equation. It can be solved by the formula as follows :

$t=\dfrac{-b\pm \sqrt{b^2-4ac} }{2a}$

Here, a = -16, b = 32 and c = 128

$t=\dfrac{-32\pm \sqrt{(32)^2-4\times (-16)(128)} }{2\times (-16)}\\\\t=\dfrac{-32+ \sqrt{(32)^2-4\times (-16)(128)} }{2\times (-16)}, \dfrac{-32\- \sqrt{(32)^2-4\times (-16)(128)} }{2\times (-16)}\\\\t=-2\ s\ \text{and}\ 4\ s$

Neglecting negative value, the projectile will land after 4 seconds.

4 0

3 years ago