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IRISSAK [1]
3 years ago
12

Please help!

Mathematics
2 answers:
DedPeter [7]3 years ago
6 0
(7-(-1))^2 +(-2)^2=60 Answer should be B
irakobra [83]3 years ago
6 0
Actually if you do the math correct using PEMDAS the answer would be a 68
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What is the slope of the line that goes through points (−3,−4) and (−7,16)?
Oliga [24]
I need the same question as this
4 0
3 years ago
HELP PLS ASAP!!!!!!!!!!!!!!!!!!!!!!!! MATH! HELP! (BRAINLISYT) PLEASE HELP IM BEGGING YOU!!
inn [45]

Answer:

Step-by-step explanation:

360(50/360)=50

5 0
3 years ago
HELP PLEASE NEED ANSWERS​
r-ruslan [8.4K]

Answer:

<u>Answer</u><u>:</u><u> </u><u>-</u><u>3</u><u>0</u>

Step-by-step explanation:

{(8 - 10)}^{2}  - 30 \div 5 \times 6 + 2

BODMAS

let's first solve the bracket:

=  {( - 2)}^{2}  - 30 \div 5 \times 6 + 2 \\  = 4 - 30 \div 5 \times 6 + 2

then let's divide:

= 4 - (30 \div 5) \times 6 + 2 \\  = 4 - 6 \times 6 + 2

then let's multiply:

= 4 - (6 \times 6) + 2 \\  = 4 - 36 + 2

simplify:

=  - 30

6 0
3 years ago
Newton's Method: Calculus I need assistance and a clear explanation so I can understand. Please be clear
marishachu [46]
Newtons method is a way to approximate the zero of a function. 
It is a recursive method such that the output becomes the new input, the goal is to approach the zero by having the inputs change by a smaller amount each iteration until that change is nearly 0.

First, let me explain where the method comes from.
The formula is really just a manipulation of the slope formula using 2 points; the point tangent to the curve, and the point where the tangent line crosses the x-axis.
If (x,y) is point on curve, then (x*,0) is point on tangent line at x-axis.
m = \frac{dy}{dx} = \frac{y - 0}{x - x^*}

Rearranging for x*:
x^* = x - \frac{y}{dy/dx}

This is the formula for newtons method. y = f(x), dy/dx = f'(x)
Now what happens is (x*,y*) becomes new point on curve with new tangent line with different slope.
This new line will cross x-axis at different point x** and so on until eventually f(x) gets really really close to 0.

Now for the example:
you need to take derivative f'(x) using product rule:
f(x) = x tan(x) - 1 \\  \\ f'(x) = tan(x) + x sec^2 (x)

Then its just a matter of plugging in values for x, and repeating until we get close to a zero.

First plug in x = 1
x_1 = 1 - \frac{(1)(tan 1) - 1}{tan 1 + (1)sec^2 (1)} = 0.888136 \\  \\ x_2 = .888136 - \frac{f(.888136)}{f'(.888136)} = 0.861465 \\  \\ x_3 = .861465 - \frac{f(.861465)}{f'(.861465)} = 0.860335 \\  \\ x_4 = .860335 - \frac{f(.860335)}{f'(.860335)} = 0.860334 \\  \\ x_5 = .860334 - \frac{f(.860334)}{f'(.860334)} = 0.860334

Now we can stop because x5 = x4 to 6 decimals, this means f(x) is very close to 0 and will serve as a good approximation for a solution.

Final Answer:
x = 0.860334
8 0
3 years ago
Solve for x: X^2 = 4/16
Rainbow [258]

We know that

x = \sqrt{\frac{4}{16}}

So:

x = \frac{2}{4}

6 0
4 years ago
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