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Crazy boy [7]
2 years ago
7

Answer the precalculus question from khan academy pls! I will give brainliest and 5 stars. 30 points.

Mathematics
1 answer:
rosijanka [135]2 years ago
7 0

The value of f(h(2)) =2 and h(f(16))= 14

<h3>What is function?</h3>

Functions are the fundamental part of the calculus in mathematics. The functions are the special types of relations. A function in math is visualized as a rule, which gives a unique output for every input . Mapping or transformation is used to denote a function in math. These functions are usually denoted by letters . The domain is defined as the set of all the values that the function can input while it can be defined. The range is all the values that come out as the output of the function involved.

given:

f(x)= √x-1 , h(x)= x² + 5

Now,

f(h(2))= f( (2)² +5 )

=f(4+5)

=f(9)

=√9-1

= 3-1

=2

h(f(16)) = h( √16-1)

=h( 4-1)

=h(3)

=3² + 5

=9+5

=14

Learn more about function here:

brainly.com/question/12431044

#SPJ1

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The graph of y ≤-2x+4 is shown. Which set contains only points that satisfy the inequality?
Alecsey [184]
The answer is A
(0,0) (1,2) (3,-3)
5 0
2 years ago
A rectangular field has an area of 1764 square meters. The width of the field is 13 meters more than the length. What is the per
sveticcg [70]

Answer:

170m

Step-by-step explanation:

The answer to the above question is letter d which is 170 m. To get the 170 m, kindly check the below solution: 

x^2 + 13x = 1764 so x = -49 and 36, we take 36 as its the positive value. And the other side is 49. Now use 2(l+b) to find perimeter. You get (36+49)*2 = 170

4 0
3 years ago
Find the distance between the points
Tasya [4]

Answer:

5 units

Step-by-step explanation:

Calculate the distance d using the distance formula

d = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2    }

with (x₁, y₁ ) = (2, 1) and (x₂, y₂ ) = (6, 4)

d = \sqrt{(6-2)^2+(4-1)^2}

   = \sqrt{4^2+3^2}

   = \sqrt{16+9}

    = \sqrt{25}

    = 5

3 0
3 years ago
Read 2 more answers
a. Fill in the midpoint of each class in the column provided. b. Enter the midpoints in L1 and the frequencies in L2, and use 1-
Tresset [83]

Answer:

\begin{array}{ccc}{Midpoint} & {Class} & {Frequency} & {64} & {63-65} & {1}  & {67} & {66-68} & {11} & {70} & {69-71} & {8} &{73} & {72-74} & {7}  & {76} & {75-77} & {3} & {79} & {78-80} & {1}\ \end{array}

Using the frequency distribution, I found the mean height to be 70.2903 with a standard deviation of 3.5795

Step-by-step explanation:

Given

See attachment for class

Solving (a): Fill the midpoint of each class.

Midpoint (M) is calculated as:

M = \frac{1}{2}(Lower + Upper)

Where

Lower \to Lower class interval

Upper \to Upper class interval

So, we have:

Class 63-65:

M = \frac{1}{2}(63 + 65) = 64

Class 66 - 68:

M = \frac{1}{2}(66 + 68) = 67

When the computation is completed, the frequency distribution will be:

\begin{array}{ccc}{Midpoint} & {Class} & {Frequency} & {64} & {63-65} & {1}  & {67} & {66-68} & {11} & {70} & {69-71} & {8} &{73} & {72-74} & {7}  & {76} & {75-77} & {3} & {79} & {78-80} & {1}\ \end{array}

Solving (b): Mean and standard deviation using 1-VarStats

Using 1-VarStats, the solution is:

\bar x = 70.2903

\sigma = 3.5795

<em>See attachment for result of 1-VarStats</em>

8 0
3 years ago
In ATUV, the measure of V=90°, TU = 5.4 feet, and VT = 1.6 feet. Find the measure
luda_lava [24]

Answer:73

Step-by-step explanation:

7 0
3 years ago
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