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arsen [322]
3 years ago
8

What is f(3) for f(x) = 3x + 2

Mathematics
2 answers:
Blizzard [7]3 years ago
5 0

Answer:

<h2>f(3) = 11</h2>

Step-by-step explanation:

f(x) = 3x + 2

In order to find f(3) substitute the value of x that's 3 into f (x) that's replace every x in f (x) by 3

That's

f(3) = 3(3) + 2

= 9 + 2

We have the final answer as

<h3>f(3) = 11</h3>

Hope this helps you

lesya [120]3 years ago
4 0

Answer:

11

Step-by-step explanation:

f(x) = 3x + 2

Let x = 3

f(3) = 3*3 +2

f(3) = 9+2

      = 11

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PLZ SOLVE PIC ATTACHED BELOW
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\tan(51°)  =  \frac{10}{x}  \\x =  \frac{10}{ \tan(51°) }  \\ x = \frac{10}{1.235}  \\ \boxed{ x = 8.1}

<h3>★ Value of x is <u>8.1</u>.</h3>

\sin(51°)=  \frac{10}{H}  \\ h =  \frac{10}{ \sin(51°) }  \\ H=  \frac{10}{0.777}  \\  \boxed{H= 12.8676}

<h3>★ Value of H is <u>12.8676</u>.</h3>

Remember:

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4 0
3 years ago
determine the area of a trapezoid the length of the midline of 14 centimeters and the length of the height of 8 centimeters
emmainna [20.7K]

Answer:

Area = 112cm^2

Step-by-step explanation:

Area = (a + b)*h / 2

M = (a + b) / 2

14 = (a + b) /

2*14 = a + b

28 = a + b

Area = 28*8 / 2

Area = 112 cm^2

5 0
2 years ago
Select all the correct answers. Which expressions are equal to 10^5?
Vlada [557]

Answer:

b, d, and E

Step-by-step explanation:

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7 0
3 years ago
Points M, N, and P are respectively the midpoints of sides AC , BC , and AB of △ABC. Prove that the area of △MNP is on fourth of
Hunter-Best [27]

Answer:

The area of △MNP is one fourth of the area of △ABC.

Step-by-step explanation:

It is given that the points M, N, and P are the midpoints of sides AC, BC and AB respectively. It means AC, BC and AB are median of the triangle ABC.

Median divides the area of a triangle in two equal parts.

Since the points M, N, and P are the midpoints of sides AC, BC and AB respectively, therefore MN, NP and MP are midsegments of the triangle.

Midsegments are the line segment which are connecting the midpoints of tro sides and parallel to third side. According to midpoint theorem the length of midsegment is half of length of third side.

Since MN, NP and MP are midsegments of the triangle, therefore the length of these sides are half of AB, AC and BC respectively. In triangle ABC and MNP corresponding side are proportional.

\triangle ABC \sim \triangle NMP

MP\parallel BC

MP=\frac{BC}{2}

By the property of similar triangles,

\frac{\text{Area of }\triangle MNP}{\text{Area of }\triangle ABC}=\frac{PM^2}{BC^2}

\frac{\text{Area of }\triangle MNP}{\text{Area of }\triangle ABC}=\frac{(\frac{BC}{2})^2}{BC^2}

\frac{\text{Area of }\triangle MNP}{\text{Area of }\triangle ABC}=\frac{1}{4}

Hence proved.

5 0
3 years ago
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