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deff fn [24]
2 years ago
12

A couple quick algebra 1 questions for 50 points!

Mathematics
1 answer:
kirill [66]2 years ago
6 0

All the relevant transformations are as shown in the explanations below.

<h3>How to Interpret Transformations?</h3>

We are given the original function as y = x and told the transformation is as follows;

1) y = x + 5; This means that the function was shifted 5 units to the left.

2) y = ³/₈x; This means that the function was vertically stretched by a scale factor of  ³/₈.

3) y = -x - 2 or y = -(x + 2); This means that the function was first shifted by 2 units to the left before being reflected over the x-axis.

4) y = 6x + 1; This means that the function was stretched by a factor of 6 and then shifted 1  unit to the left.

5) y = x - 11; This means that the function was shifted 11 units to the right.

6) y = 8x; This means that the function was vertically stretched by a scale factor of  8.

7) y = -1/3x; This means that the function was horizontally stretched by a scale factor of  1/3.

Read more about Transformations at; brainly.com/question/4289712

#SPJ1

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Is the relationship between the 7s in 7,742 and the 7s in 7,785 different in any way
Usimov [2.4K]
No, because they're in the same place value.
4 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
5. Given: BCDF is a parallelogram; line AB is congruent to line BF
solong [7]
A and E are corresponding angles so A=E
3 0
2 years ago
Identify the relative maximum value of g(x) for the function shown below g(x)=7/x^2+5
Liula [17]
To find t<span>he relative maximum value of the function we need to find where the function has its first derivative equal to 0.

Its first derivative is -7*(2x)/(x^2+5)^2
</span>7*(2x)/(x^2+5)^2 =0 the numerator needs to be eqaul to 0
2x=0
x=0

g(0) = 7/5

The <span>relative maximum value is at the point (0, 7/5).</span>
8 0
2 years ago
Read 2 more answers
Please help me to solve this sum<br>please gI've correct answer​
disa [49]

This question requires creating a few equations and working through them step-by-step. Now, first let's give each of the shapes a variable: let's say that the blue shape is a, the orange shape is b and the green shape is c.

1. We can technically create six formulas for the magic square, with three for sum of the rows and three for the sum of the columns, however the smartest way to approach this is to observe whether there are any obvious answers that we can get.

We can see in row 2 that there are three of the same shape (a) that add to 57. This makes it very simple to calculate the value of the shape.

Since 3a = 57

a = 57/3 = 19

2. Now we need to find a row or column that includes a and one other shape; we could choose either column 2 or 3, so let's go with column 2. Remembering that the blue shape is a and the orange shape is b:

2a + b = 50

Now, given that a = 19:

2(19) + b = 50

38 + b = 50

b = 12

3. We can now take any of the rows or columns that include the third shape (c) since we already know the values of the other two shapes. Let's take column 1:

a + b + c = 38

19 + 12 + c = 38

31 + c = 38

c = 38 - 31

c = 7

Thus, the value of the blue shape is 19, the value of the orange shape is 12 and the value of the green shape is 7.

3 0
3 years ago
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