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3 years ago

15

A number increased by 27 is 233. Find the number.

1 answer:

miss Akunina [59]3 years ago

8 0

Answer:

206

Step-by-step explanation:

233 - 27 = 206

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What is the true solution to the equation below?

Naddik [55]

Yeeee

assuming your equaiton is
$2ln(e^{ln(2x)})-ln(e^{ln(10x)})=ln(30)$

remember some nice log rules
$log_a(b)=c$ translates to $a^c=b$
and
$a^{log_a(b)}=b$
and
$xlog_c(b)=log_c(b^x)$
and
$ln(x)=log_e(x)$
and
$log(a)-log(b)=log(\frac{a}{b})$
and
if $log(a)=log(b)$ then a=b

so

we can simplify a bit of stuff here

the $e^{ln(2x)} \space\ and \space\ the \space\ e^{ln(10x)}$ can be simplified to $2x \space\ and \space\ 10x$

so we gots now

$2ln(2x)-ln(10x)=ln(30)$
$ln((2x)^2)-ln(10x)=ln(30)$
$ln(4x^2)-ln(10x)=ln(30)$
$ln(\frac{4x^2}{10x})=ln(30)$
same base so
$\frac{4x^2}{10x}=30$
$\frac{2x}{5}=30$
times both sides by 5
$2x=150$
divide both sides by 2
$x=75$
answer is x=75

5 0

3 years ago

Helpppp me plz i dont get it

SOVA2 [1]

Try 3.14x2= 6.28 im so sorry if its wrong

6 0

3 years ago

Read 2 more answers

I need the answer written as a base and exponent not decimal. please and thanks

Dmitry [639]

I think the answer would be -8^4

4 0

3 years ago

Can someone help me pls if u do u will be the best and on the first one and the second one

likoan [24]

Answer:

#1

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5 0

3 years ago

The graph h = −16t^2 + 25t + 5 models the height and time of a ball that was thrown off of a building where h is the height in f

Thepotemich [5.8K]

Answer:

part 1) 0.78 seconds

part 2) 1.74 seconds

Step-by-step explanation:

step 1

At about what time did the ball reach the maximum?

Let

h ----> the height of a ball in feet

t ---> the time in seconds

we have

$h(t)=-16t^{2}+25t+5$

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

so

The x-coordinate of the vertex represent the time when the ball reach the maximum

Find the vertex

Convert the equation in vertex form

Factor -16

$h(t)=-16(t^{2}-\frac{25}{16}t)+5$

Complete the square

$h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+5+\frac{625}{64}$

$h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}\\h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}$

Rewrite as perfect squares

$h(t)=-16(t-\frac{25}{32})^{2}+\frac{945}{64}$

The vertex is the point $(\frac{25}{32},\frac{945}{64})$

therefore

The time when the ball reach the maximum is 25/32 sec or 0.78 sec

step 2

At about what time did the ball reach the minimum?

we know that

The ball reach the minimum when the the ball reach the ground (h=0)

For h=0

$0=-16(t-\frac{25}{32})^{2}+\frac{945}{64}$

$16(t-\frac{25}{32})^{2}=\frac{945}{64}$

$(t-\frac{25}{32})^{2}=\frac{945}{1,024}$

square root both sides

$(t-\frac{25}{32})=\pm\frac{\sqrt{945}}{32}$

$t=\pm\frac{\sqrt{945}}{32}+\frac{25}{32}$

the positive value is

$t=\frac{\sqrt{945}}{32}+\frac{25}{32}=1.74\ sec$

8 0

3 years ago