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poizon [28]
2 years ago
14

Find the distance between (4,-3),(3,-8)

Mathematics
1 answer:
WARRIOR [948]2 years ago
8 0
So you just have to use distance formula

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Ming throws a stone off a bridge into a river below.
AURORKA [14]

Answer:

1 seconds after being thrown, the stone reaches its max height

Step-by-step explanation:

The parabolic (quadratic) equation is:

h(x)=-5(x-1)^2+45

Lets expand this in the form ax^2+bx+c, so we have:

h(x)=-5(x-1)^2+45\\h(x)=-5(x^2-2x+1)+45\\h(x)=-5x^2+10x-5+45\\h(x)=-5x^2+10x+40

We can say the values of a,b, and c, now to be:

a = -5

b = 10

c = 40

The number of seconds at which the max would occur is given by the point, x, at:

x=-\frac{b}{2a}

We know a and b, let's find the seconds, x,

x=-\frac{b}{2a}=-\frac{10}{2(-5)}=-\frac{10}{-10}=--1=1

Hence,

1 seconds after being thrown, the stone reach its max height

6 0
3 years ago
Read 2 more answers
kali has a 6 centimeter ladder and puts the base of the ladder 2 meters away from the tree how high into the tree will the lader
weqwewe [10]
I am going to assume ‘centimetre’ was a typo.

A^2 + B^2 = C^2

C^2 - A^2 = B^2

32 = B^2

Square root both sides

B = 5.7m high

The ladder will reach 5.7 meters high
5 0
3 years ago
Who can do 20 assignments in math for me rn needs to b finish in 5 hours?
GREYUIT [131]

u need to finish 4 assignments in 1 hour

5 0
3 years ago
Which sequence could be described by the explicit definition: t_n=2^n+1t n = 2 n + 1
yuradex [85]

Answer:

2. 27

3. Option C. 3, 5, 9, 17, 33, ...

Step-by-step explanation:

2. The sequence is defined by the explicit function t_{n} = 2^{n}  - n

Therefore, the 5th term i.e. t_{5} of the sequence.

t_{5} = 2^{5}  - 5 = 32 - 5 = 27 (Answer)

3. The explicit definition is t_{n} = 2^{n}  + 1

Hence, t_{1} = 2^{1}  + 1 = 3

t_{2} = 2^{2}  + 1 = 5

t_{3} = 2^{3}  + 1 = 9

t_{4} = 2^{4}  + 1 = 17

Therefore, the option C is the right sequence. (Answer)

4 0
3 years ago
Read 2 more answers
What is the similarity ratio of the smaller to the larger similar cylinders?
lbvjy [14]
\bf \qquad \qquad \textit{ratio relations}
\\\\
\begin{array}{ccccllll}
&Sides&Area&Volume\\
&-----&-----&-----\\
\cfrac{\textit{similar shape}}{\textit{similar shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3}
\end{array} \\\\
-----------------------------\\\\
\cfrac{\textit{similar shape}}{\textit{similar shape}}\qquad \cfrac{s}{s}=\cfrac{\sqrt{s^2}}{\sqrt{s^2}}=\cfrac{\sqrt[3]{s^3}}{\sqrt[3]{s^3}}\\\\
-------------------------------\\\\

\bf \cfrac{smaller}{larger}\qquad \cfrac{s}{s}=\cfrac{\sqrt{48\pi }}{\sqrt{75\pi }}\implies \cfrac{s}{s}=\cfrac{\sqrt{(2^2)^2\cdot 3}}{\sqrt{5^2\cdot 3}}\implies \cfrac{s}{s}=\cfrac{4\sqrt{3}}{5\sqrt{3}}
\\\\\\
\cfrac{s}{s}=\cfrac{4}{5}
8 0
2 years ago
Read 2 more answers
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