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Whitepunk [10]
3 years ago
8

B is the midpoint of segment AC. Find the value of x.

Mathematics
1 answer:
tester [92]3 years ago
3 0

Answer:

x = 2

Step-by-step explanation:

Since B is the midpoint, AB and BC are congruent in length.

   5x-6 = 2x

      5x = 2x + 6

5x - 2x = 6

       3x = 6

         x = 2

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If f(x) = 4x - 3, evaluate f(2).
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Answer: f(2) = 5

Explanation: If f(x) - 4x - 3, then f(2) = 4(2) - 3 which simplifies to 5.

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3 years ago
Layla arked a group of students to chooue their favorite type of exerche from the choices of jogging, biling, and wwimming. The
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50 students

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7 0
3 years ago
Writing Trig Ratios.
notsponge [240]

Answer:

The ratios are as follows for the given angle:

sineX=opposite/hypotenuse

cosineX=adjacent/hypotenuse

tangentX=opposite/adjacent

Insert the appropriate values for each:

7.) sinZ=24/40

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3 years ago
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7 0
3 years ago
F $1 \le a \le 10$ and $1 \le b \le 36$, for how many ordered pairs of integers $(a, b)$ is $\sqrt{a + \sqrt{b}}$ an integer?
svetlana [45]
You have such entry data: 
1\ \textless \ a\ \textless \ 10 \\ 1\ \textless \ b\ \textless \ 36

Consider expression \sqrt{a+ \sqrt{b} }. If this expression becomes an integer, then b=4,9,16,25, because then \sqrt{b} = 2,3,4,5, respectively. In other cases \sqrt{b} is not integer and thus the expression \sqrt{a+ \sqrt{b} } also is not integer.

1. b=4, then \sqrt{a+ \sqrt{b} }=\sqrt{a+2}. Here a=2,7 (in other cases \sqrt{a+2} is not integer). When a=2, \sqrt{a+2}=\sqrt{2+2}=2 and when a=7, \sqrt{a+2}=\sqrt{7+2}=3.

2. b=9, then a=6 and \sqrt{a+ \sqrt{b} }= \sqrt{a+3}=\sqrt{6+3}=3.

3. b=16, then a=5 and \sqrt{a+ \sqrt{b} }= \sqrt{a+4}=\sqrt{5+4}=3.

4. b=25, then a=4 and \sqrt{a+ \sqrt{b} }=\sqrt{a+5}=\sqrt{4+5}=3.

Answer: (2,4), (7,4), (6,9), (5,16), (4,25).







3 0
3 years ago
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