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3 years ago

Find the value of x if Sin80 = Cos( 90 -x)

Mathematics

1 answer:

natita [175]3 years ago

5 0

Answer:

Step-by-step explanation:

$\sin(80) = \cos(10) \\ 90 - x = 10 \\ x = 80$

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What's the slope of the graph

kotykmax [81]

Answer:

1/-2

Step-by-step explanation:

To find the slope:

rise/run

From point to point

So....

From (-1,0)

you go up 2(2)

then left 4(-4)

to get to (-5,2)

so 2/-4 reduced is 1/-2

6 0

3 years ago

What is 2^4+5*6*4+4/2*50 plz help

Lelechka [254]

It is 236. Happy to help (:

6 0

3 years ago

Read 2 more answers

If two 12-sided dice are rolled, what is the probability that both numbers will be even?

geniusboy [140]

1/4 I think, I’m not sure

7 0

3 years ago

What are the transformations to this quadratic function y= (x+4) power of 2 - 8

I am Lyosha [343]

Answer:

Horizontal shift of 4 units to the left.

Vertical translation of 8 units downward.

Step-by-step explanation:

Given the quadratic function, y = (x + 4)² - 8, which represents the horizontal and vertical translations of the parent graph, y = x²:

The vertex form of the quadratic function is y = a(x - h)² + k

Where:

The vertex is (h , k), which is either the minimum (upward facing graph) or maximum (downward-facing graph).

The axis of symmetry occurs at x = h.

a = determines whether the graph opens up or down, and makes the graph wider or narrower.

h = determines how far left or right the parent function is translated.

k = determines how far up or down the parent function is translated.

Going back to your quadratic function,

y = (x + 4)² - 8

The vertex occrs at (-4, -8)
a is assumed to have a value of 1.
Given the value of h = -4, then it means that the graph shifted horizontally by 4 units to the left.
Since k = -8, then it implies that the graph translated vertically at 8 units downward.

Please mark my answers as the Brainliest, if you find this helpful :)

4 0

2 years ago

Let f (x) = 3x − 1 and ε > 0. Find a δ > 0 such that 0 < ∣x − 5∣ < δ implies ∣f (x) − 14∣ < ε. (Find the largest

s344n2d4d5 [400]

Answer:

This proves that f is continous at x=5.

Step-by-step explanation:

Taking f(x) = 3x-1 and $\varepsilon>0$ , we want to find a $\delta$ such that $|f(x)-14|$

At first, we will assume that this delta exists and we will try to figure out its value.

Suppose that $|x-5|$ . Then

$|f(x)-14| = |3x-1-14| = |3x-15|=|3(x-5)| = 3|x-5|< 3\delta$ .

Then, if $|x-5|$ , then $|f(x)-14|$ . So, in this case, if $3\delta \leq \varepsilon$ we get that $|f(x)-14|$ . The maximum value of delta is $\frac{\varepsilon}{3}$ .

By definition, this procedure proves that $\lim_{x\to 5}f(x) = 14$ . Note that f(5)=14, so this proves that f is continous at x=5.

3 0

3 years ago