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Zinaida [17]
2 years ago
7

What is 342 in fraction form

Mathematics
1 answer:
oee [108]2 years ago
7 0

Answer:

342 / 1

Step-by-step explanation:

342 / 1 = 342

342 in fraction form is 342 / 1.

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(9 x 10 ^(-5)) - (3 x 10^(-16))

Your subtracting to find the difference in size

9 x 10^(-5)

7 0
3 years ago
Write a simplified expression for the perimeter of the triangle.
Anit [1.1K]

Answer:

<em>15m+13</em>

Step-by-step explanation:

<em>The perimeter of a triangle is the sum of all sides.</em>

<em />=(7m+4)+(8m-3)+(12)\\\\=7m+4+8m-3+12\\\\=15m+13

8 0
2 years ago
help here plz ! …..A taxi company charges a flat fee of $6.00, plus $1.10 per mile after the first 3 miles. Sarah’s taxi cab rid
Igoryamba

Answer:

2.5 miles

Step-by-step explanation:

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Read 2 more answers
In right ABC, AN is the altitude to the hypotenuse. FindBN, AN, and AC,if AB =2 5 in, and NC= 1 in.
Rama09 [41]

From the statement of the problem, we have:

• a right triangle △ABC,

,

• the altitude to the hypotenuse is denoted AN,

,

• AB = 2√5 in,

,

• NC = 1 in.

Using the data above, we draw the following diagram:

We must compute BN, AN and AC.

To solve this problem, we will use Pitagoras Theorem, which states that:

h^2=a^2+b^2\text{.}

Where h is the hypotenuse, a and b the sides of a right triangle.

(I) From the picture, we see that we have two sub right triangles:

1) △ANC with sides:

• h = AC,

,

• a = ,NC = 1,,

,

• b = NA.

2) △ANB with sides:

• h = ,AB = 2√5,,

,

• a = BN,

,

• b = NA,

Replacing the data of the triangles in Pitagoras, Theorem, we get the following equations:

\begin{cases}AC^2=1^2+NA^2, \\ (2\sqrt[]{5})^2=BN^2+NA^2\text{.}\end{cases}\Rightarrow\begin{cases}NA^2=AC^2-1, \\ NA^2=20-BN^2\text{.}\end{cases}

Equalling the last two equations, we have:

\begin{gathered} AC^2-1=20-BN^2.^{} \\ AC^2=21-BN^2\text{.} \end{gathered}

(II) To find the values of AC and BN we need another equation. We find that equation applying the Pigatoras Theorem to the sides of the bigger right triangle:

3) △ABC has sides:

• h = BC = ,BN + 1,,

,

• a = AC,

,

• b = ,AB = 2√5,,

Replacing these data in Pitagoras Theorem, we have:

\begin{gathered} \mleft(BN+1\mright)^2=(2\sqrt[]{5})^2+AC^2 \\ (BN+1)^2=20+AC^2, \\ AC^2=(BN+1)^2-20. \end{gathered}

Equalling the last equation to the one from (I), we have:

\begin{gathered} 21-BN^2=(BN+1)^2-20, \\ 21-BN^2=BN^2+2BN+1-20 \\ 2BN^2+2BN-40=0, \\ BN^2+BN-20=0. \end{gathered}

(III) Solving for BN the last quadratic equation, we get two values:

\begin{gathered} BN=4, \\ BN=-5. \end{gathered}

Because BN is a length, we must discard the negative value. So we have:

BN=4.

Replacing this value in the equation for AC, we get:

\begin{gathered} AC^2=21-4^2, \\ AC^2=5, \\ AC=\sqrt[]{5}. \end{gathered}

Finally, replacing the value of AC in the equation of NA, we get:

\begin{gathered} NA^2=(\sqrt[]{5})^2-1, \\ NA^2=5-1, \\ NA=\sqrt[]{4}, \\ AN=NA=2. \end{gathered}

Answers

The lengths of the sides are:

• BN = 4 in,

,

• AN = 2 in,

,

• AC = √5 in.

7 0
1 year ago
How do I find the area of a trapezoid?​
konstantin123 [22]

Answer:

Step-by-step explanation:

Area of a trapezoid is found with the formula, A=(a+b)/2 x h. Learn how to use the formula to find area of trapezoids.

The two parallel sides are the bases, and height, as always, is the perpendicular distance from one base to the opposite. The area of this parallelogram is its height (half-height of the trapezoid) times its base (sum of the bases of the trapezoid), so its area is half-height × (base1 + base2).

3 0
2 years ago
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