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aksik [14]
3 years ago
15

And why is it a function

Mathematics
1 answer:
andreyandreev [35.5K]3 years ago
8 0
It's not a function because the x value 2 has two y values
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Find the value of the trigonometric ratio. Make sure to simplify the fraction if needed. plss help
Natalija [7]

Answer:

3/5

Step-by-step explanation:

sinA = opp/hyp

= 21/35

= 3/5

8 0
3 years ago
If AE=6x-55 and EC=3x-16, find DB. (Hint: Find x first and then substitute.)
erica [24]

<u>Given</u>:

Given that ABCD is a rectangle.

The diagonals of the rectangle are AC and DB.

The length of AE is (6x -55)

The length of EC is (3x - 16)

We need to determine the length of the diagonal DB.

<u>Value of x:</u>

The value of x can be determined by equating AE and EC

Thus, we have;

AE=EC

Substituting the values, we get;

6x-55=3x-16

3x-55=-16

       3x=39

         x=13

Thus, the value of x is 13.

<u>Length of AC:</u>

Length of AE = 6(13)-55=78-55=23

Length of EC = 3(13)-16=39-16=23

Thus, the length of AC can be determined by adding the lengths of AE and EC.

Thus, we have;

AC=AE+EC

AC=23+23

AC=46

Thus, the length of AC is 46.

<u>Length of DB:</u>

Since, the diagonals AC and DB are perpendicular to each other, then their lengths are congruent.

Hence, we have;

AC=DB

 46=DB

Thus, the length of DB is 46.

6 0
3 years ago
-2(4a+4b)+5a&gt;-35<br> Solve the inequality
Amiraneli [1.4K]

Step-by-step explanation:

-2 (4a + 4b) + 5a > -35

-8a -8b + 5a > -35

-3a -8b > -35

for a

-3a > -35 + 8b

a < -35 + 8b ÷ -3

a < 35 - 8b ÷ 3

for b

-3a - 8b > -35

-8b > -35 + 3a

b < -35 + 3a ÷ -8

b < 35 - 3a ÷ 8

5 0
3 years ago
Since at t=0, n(t)=n0, and at t=∞, n(t)=0, there must be some time between zero and infinity at which exactly half of the origin
Airida [17]
Answer: t-half = ln(2) / λ ≈ 0.693 / λ

Explanation:

The question is incomplete, so I did some research and found the complete question in internet.

The complete question is:

Suppose a radioactive sample initially contains N0unstable nuclei. These nuclei will decay into stable nuclei, and as they do, the number of unstable nuclei that remain, N(t), will decrease with time. Although there is no way for us to predict exactly when any one nucleus will decay, we can write down an expression for the total number of unstable nuclei that remain after a time t:

N(t)=No e−λt,

where λ is known as the decay constant. Note that at t=0, N(t)=No, the original number of unstable nuclei. N(t) decreases exponentially with time, and as t approaches infinity, the number of unstable nuclei that remain approaches zero.

Part (A) Since at t=0, N(t)=No, and at t=∞, N(t)=0, there must be some time between zero and infinity at which exactly half of the original number of nuclei remain. Find an expression for this time, t half.

Express your answer in terms of N0 and/or λ.

Answer:

1) Equation given:

N(t)=N _{0} e^{-  \alpha  t} ← I used α instead of λ just for editing facility..

Where No is the initial number of nuclei.

2) Half of the initial number of nuclei: N (t-half) =  No / 2

So, replace in the given equation:

N_{t-half} =  N_{0} /2 =  N_{0}  e^{- \alpha t}

3) Solving for α (remember α is λ)

\frac{1}{2} =  e^{- \alpha t} &#10;&#10;2 =   e^{ \alpha t} &#10;&#10; \alpha t = ln(2)

αt ≈ 0.693

⇒ t = ln (2) / α ≈ 0.693 / α ← final answer when you change α for λ




4 0
3 years ago
Drag each tile to the correct box.
Julli [10]
For f(x), which has a vertex at (2,0), the y-intercept at (0,4) is above this vertex, so the parabola opens upward. This means that the vertex is the only point that touches the x-axis, so there is only 1 x-intercept.
For h(x), the graph does not have any x-intercepts.
For g(x) = x^2 + x - 2 = (x+2)(x-1), this intersects the x-axis at x = -2 and x = 1, so there are 2 x-intercepts.
From least to greatest: h(x), f(x), g(x).
7 0
3 years ago
Read 2 more answers
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