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Rudiy27
2 years ago
13

PLZ Help!!! Look at the pic.

Mathematics
2 answers:
stealth61 [152]2 years ago
8 0
435 being the constant means that is what they started with

Hope this helps :)
strojnjashka [21]2 years ago
4 0
The constant 435 represents how much they started with

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Find the x- and y- intercepts of parabola y=5x^2-16x+10
____ [38]

Y-INTERCEPT

y = 5x^2 - 16x + 10

The y-intercept is where the equation/curve/parabola cosses the y-axis.

The y-axis is where x = 0. (The x-axis is where y = 0)

To find the y-intercept:

\text{y-axis} \rightarrow \text{x = 0} \rightarrow y = 5(0)^2 -16(0) + 10 = 10

The y-intercept must be at (0, 10)

X-INTERCEPT (ROOTS/SOLUTIONS)

y = 5x^2 - 16x + 10\\\text{make it equal 0}\\y = 0\\\therefore 5x^2 - 16x + 10 = 0

We need to use the quadratic formula

The quadratic formula helps us find what values of x make the equation = 0

Quadratic formula: x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

x=\frac{-(-16) + \sqrt{(-16)^2-4(5)(10)}}{2(5)}\\\\x = \frac{16 + \sqrt{256-200}}{10}\\x = \frac{16 + \sqrt{56}}{10}\\x = \frac{16 + 2\sqrt{14}}{10}\\x = \frac{8 + \sqrt{14}}{5}\\\\\\x=\frac{-(-16) - \sqrt{(-16)^2-4(5)(10)}}{2(5)}\\\text{doing the same thing...}\\\text{end up with...}\\x = \frac{8 - \sqrt{14}}{5}\\

The x-intercepts are at:

(\frac{8 + \sqrt{14}}{5}, 0)\\(\frac{8 - \sqrt{14}}{5}, 0)

5 0
2 years ago
Need Help! 10 points!
larisa86 [58]

The answer is 1680. To get this you do 8*7*6*5 because there are 8 options for the first character, and then you can repeat that for the second one with 7, 6, and 5

6 0
3 years ago
Pls help thank you jdxjdjcjdjjjdjddjiwjdodidsidjjsjd
nadezda [96]

Answer:

The answer is B.

Step-by-step explanation:

3< x _<5

3 0
2 years ago
2/3=1.2/x solve for x
Anvisha [2.4K]

First isolate x by multiplying by x on both sides.  It will look like this:

(2/3)x=1.2

Then devise by (2/3) on both sides to get:

x= 1.2/(2/3)

The easiest way to get this is plug that into a calculator and get the answer:

x=1.8

5 0
3 years ago
A certain freely falling object, released from rest, requires 1.50 s to travel the last 30.0 m before it hit the ground. a.Find
Lyrx [107]
A.)
<span>s= 30m
u = ? ( initial velocity of the object )
a = 9.81 m/s^2 ( accn of free fall )
t = 1.5 s
s = ut + 1/2 at^2 
\[u = \frac{ S - 1/2 a t^2 }{ t }\]
\[u = \frac{ 30 - ( 0.5 \times 9.81 \times 1.5^2) }{ 1.5 } \]
\[u = 12.6 m/s\]
</span>
b.)
<span>s = ut + 1/2 a t^2
u = 0 ,
s = 1/2 a t^2
\[s = \frac{ 1 }{ 2 } \times a \times t ^{2}\]
\[s = \frac{ 1 }{ 2 } \times 9.81 \times \left( \frac{ 12.6 }{ 9.81 } \right)^{2}\]
\[s = 8.0917...\]
\[therfore total distance = 8.0917 + 30 = 38.0917.. = 38.1 m \] </span>
4 0
3 years ago
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