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Maslowich
3 years ago
15

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

Mathematics
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
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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

158 students voted for toby

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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
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