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marishachu [46]
3 years ago
14

What is the answer!!!!!!

Mathematics
1 answer:
weqwewe [10]3 years ago
8 0

Answer:

The answer is c: the lendth of the photo frame is between 11 and 12 inches.

I hope this helps you!

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The table on the left is that of a linear function, and the one on the right is that of an exponential function. Can you tell wh
Brut [27]

The correct answer is D.

As you can see, the exponential function grows by doubling the previous output with each increment of the input: start with 1, you double it to get 2, then you double it to get 4, 8 and so on.

On the other hand, the linear function adds 7 with each step. This means that the exponential function will eventually reach and pass the linear one, and will definitely be grater from that point on. In fact, if we continue the table, we get

\begin{array}{c|c|c}\text{x value}&\text{linear}&\text{exponential}\\4&28&8\\5&35&16\\6&42&32\\7&49&64\\8&56&128\\9&63&256\end{array}

and you can see how the exponential growth is much faster than the linear one.

6 0
3 years ago
Question 6 of 10
Greeley [361]
I believe the answer is V = 216^3
7 0
3 years ago
PLZ HELP ME
Likurg_2 [28]

Answer:

x ≈ 14.87

Step-by-step explanation:

Using Pythagoras' identity in the right triangle.

The square on the hypotenuse is equal to the sum of the squares on the other 2 sides, that is

x² = 10² + 11² = 100 + 121 = 221 ( take the square root of both sides )

x = \sqrt{221} ≈ 14.87 ( to the nearest hundredth )

8 0
3 years ago
A survey of a random sample of voters shows that 29% plan to vote for Martin and 71% plan to vote for
Iteru [2.4K]

Range=±7%

Martin=71%, Lang=29%

Range for Martin=71%±7%

So, highest %=71%+7%=78%

Lowest %=71%-7%=64%

Therefore,range for Martin is between 64%-78% which is 14%

3 0
3 years ago
If the function h is defined by h(x)=<img src="https://tex.z-dn.net/?f=x%5E%7B2%7D" id="TexFormula1" title="x^{2}" alt="x^{2}" a
harkovskaia [24]

Given:

The function is:

h(x)=x^2-3x+5

To find:

The value of h(2x+1).

Solution:

We have,

h(x)=x^2-3x+5

Putting x=2x+1, we get

h(2x+1)=(2x+1)^2-3(2x+1)+5

h(2x+1)=(2x)^2+2(2x)(1)+(1)^2-3(2x)-3(1)+5

h(2x+1)=4x^2+4x+1-6x-3+5

On combining like terms, we get

h(2x+1)=4x^2+(4x-6x)+(1-3+5)

h(2x+1)=4x^2-2x+3

Therefore, the required function is h(2x+1)=4x^2-2x+3.

3 0
3 years ago
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