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4 years ago

2 + 2 = ? idk the answer bro i need help

Mathematics

2 answers:

Leto [7]4 years ago

8 0

Answer:

That would be 4 man

Lera25 [3.4K]4 years ago

5 0

Answer:

ima say 4 but I'm really clueless very hard

Step-by-step explanation:

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Use f’( x ) = lim With h ---> 0 [f( x + h ) - f ( x )]/h to find the derivative at x for the given function. 5-x²

beks73 [17]

<h2>Answer:</h2>

The derivative of the function f(x) is:

$f'(x)=-2x$

<h2>Step-by-step explanation:</h2>

We are given a function f(x) as:

$f(x)=5-x^2$

We have:

$f(x+h)=5-(x+h)^2\\\\i.e.\\\\f(x+h)=5-(x^2+h^2+2xh)$

( Since,

$(a+b)^2=a^2+b^2+2ab$ )

Hence, we get:

$f(x+h)=5-x^2-h^2-2xh$

Also, by using the definition of f'(x) i.e.

$f'(x)= \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$

Hence, on putting the value in the formula:

$f'(x)= \lim_{h \to 0} \dfrac{5-x^2-h^2-2xh-(5-x^2)}{h}\\\\\\f'(x)=\lim_{h \to 0} \dfrac{5-x^2-h^2-2xh-5+x^2}{h}\\\\i.e.\\\\f'(x)=\lim_{h \to 0} \dfrac{-h^2-2xh}{h}\\\\f'(x)=\lim_{h \to 0} \dfrac{-h^2}{h}+\dfrac{-2xh}{h}\\\\f'(x)=\lim_{h \to 0} -h-2x\\\\i.e.\ on\ putting\ the\ limit\ we\ obtain:\\\\f'(x)=-2x$

Hence, the derivative of the function f(x) is:

$f'(x)=-2x$

3 0

3 years ago

Read 2 more answers

F(x) = 5x3 + 2x2 + 7x – 3

REY [17]

Answer:

Step-by-step explanation:

f= 7x+16 over x

4 0

4 years ago

Pls no links, I really need help

mylen [45]

Answer:

The answer should be b. (0,2) because that the order pair for y

4 0

3 years ago

Read 2 more answers

Triangle PQR is transformed to triangle P'Q'R'. Triangle PQR has vertices P(4, 0), Q(0, −4), and R(−8, −4). Triangle P'Q'R' has

Whitepunk [10]

Answer:

Please find attached the required plot accomplished with an online tool

Part A:

1/4

Part B:

P''(-1, 0), Q''(0, -1), and R''(2, -1)

Part C:

Triangle PQR is similar to triangle P''Q''R'' but they are not congruent

Step-by-step explanation:

Part A:

Triangle ΔPQR has vertices P(4, 0), Q(0, -4), R(-8, -4)

Triangle ΔP'Q'R' has vertices P'(1, 0), Q'(0, -1), R'(-2, -1)

The dimensions of the sides of the triangle are given by the relation;

$l = \sqrt{\left (y_{2}-y_{1} \right )^{2}+\left (x_{2}-x_{1} \right )^{2}}$

Where;

(x₁, y₁) and (x₂, y₂) are the coordinates on the ends of the segment

For segment PQ, we place (x₁, y₁) = (4, 0) and (x₂, y₂) = (0, -4);

By substitution into the length equation, we get;

The length of segment PQ = 4·√2

The length of segment PR = 4·√10

The length of segment RQ = 8

The length of segment P'Q' = √2

The length of segment P'R' = √10

The length of segment R'Q' = 2

Therefore, the scale factor of the dilation of ΔPQR to ΔP'Q'R' is 1/4

Part B:

Reflection of (x, y) across the y-axis gives;

(x, y) image after reflection across the y-axis = (-x, y)

The coordinates after reflection of P'(1, 0), Q'(0, -1), R'(-2, -1) across the y-axis is given as follows;

P'(1, 0) image after reflection across the y-axis = P''(-1, 0)

Q'(0, -1) image after reflection across the y-axis = Q''(0, -1)

R'(-2, -1) image after reflection across the y-axis = R''(2, -1)

Part C:

Triangle PQR is similar to triangle P''Q''R'' but they are not congruent as the dimensions of the sides of triangle PQR and P''Q''R'' are not the same.