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fiasKO [112]
3 years ago
6

Dakota wants to build 8 birdhouses, he needs to point 6 square feet of plywood, for each birdhouse is the plywood cause $0.56 pe

r square foot, how much will it cost to go to to make his bird houses
Mathematics
1 answer:
Westkost [7]3 years ago
3 0

Answer:

$26.88

Step-by-step explanation:

Cost of a ply wood = $0.56 per square foot

If Dakota wants to build 8 birdhouses and he needs to paint 6 square feet of plywood for each birdhouse, then the total area fir the 8 bird house will be expressed as:

8 * 6

= 48 sq. feet

Hence the area of 8 bird houses is 46sq. feet

Next is to calculate the amount it will cost to make the 8 bird houses.

Since 1 sq. foot = $0.56

48 sq. feet = y

cross multiply

y * 1 = 48 * 0.56

y =  26.88

Hence it will cost Dakota $26.88 to make hid bird houses

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Help!!! i don’t understand how i would be able to find rt
liq [111]

Given:

M is the mid-point of RS

N is the mid-point of ST

MN = 18.4

To find:

The length of RT.

Solution:

The reference image is attached below.

Joining mid-point M and N, we get mid-segment MN.

MN is parallel to RT.

Triangle mid-segment theorem:

If a segments joins the mid point of a two sides of triangle, then the segment is parallel to the third side and is half of that side.

$\Rightarrow MN=\frac{1}{2} RT

Substitute MN = 18.4

$\Rightarrow 18.4 =\frac{1}{2} RT

Multiply by 2 on both sides.

$\Rightarrow 2\times 18.4 =2\times \frac{1}{2} RT

$\Rightarrow 36.8=RT

The length of RT is 36.8.

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2 years ago
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kozerog [31]
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6 0
2 years ago
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The average temperature of the week is 80 degrees. The first 3 days have an average of 78 and the average of the last 3 days is
Tju [1.3M]

Answer:

80

Step-by-step explanation:Solution:

step 1 Address the formula, input parameters & values.

Input parameters & values:

The given numbers are 78 & 80

step 2 Find the sum of the given two numbers.

sum = 78 + 80 = 158

step 3 Divide the sum by 2 to get the average.

average = 158/2

= 79

Thus, 79 is an average of positive integers 78 and 80.

3 0
3 years ago
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bagirrra123 [75]
A right angle triangle normally relates to Pythagoras Theorem , and this is related to the hypotenuse, the answer maybe the hypotenuse. As the other sides are the opposite and alternative  


8 0
3 years ago
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zysi [14]

Given: The following functions

A)cos^2\theta=sin^2\theta-1B)sin\theta=\frac{1}{csc\theta}\begin{gathered} C)sec\theta=\frac{1}{cot\theta} \\ D)cot\theta=\frac{cos\theta}{sin\theta} \\ E)1+cot^2\theta=csc^2\theta \end{gathered}

To Determine: The trigonometry identities given in the functions

Solution

Verify each of the given function

\begin{gathered} cos^2\theta=sin^2\theta-1 \\ Note\text{ that} \\ sin^2\theta+cos^2\theta=1 \\ cos^2\theta=1-sin^2\theta \\ Therefore \\ cos^2\theta sin^2\theta-1,NOT\text{ }IDENTITIES \end{gathered}

B

\begin{gathered} sin\theta=\frac{1}{csc\theta} \\ Note\text{ that} \\ csc\theta=\frac{1}{sin\theta} \\ sin\theta\times csc\theta=1 \\ sin\theta=\frac{1}{csc\theta} \\ Therefore \\ sin\theta=\frac{1}{csc\theta},is\text{ an identities} \end{gathered}

C

\begin{gathered} sec\theta=\frac{1}{cot\theta} \\ note\text{ that} \\ cot\theta=\frac{1}{tan\theta} \\ tan\theta cot\theta=1 \\ tan\theta=\frac{1}{cot\theta} \\ Therefore, \\ sec\theta\ne\frac{1}{cot\theta},NOT\text{ IDENTITY} \end{gathered}

D

\begin{gathered} cot\theta=\frac{cos\theta}{sin\theta} \\ Note\text{ that} \\ cot\theta=\frac{1}{tan\theta} \\ cot\theta=1\div tan\theta \\ tan\theta=\frac{sin\theta}{cos\theta} \\ So, \\ cot\theta=1\div\frac{sin\theta}{cos\theta} \\ cot\theta=1\times\frac{cos\theta}{sin\theta} \\ cot\theta=\frac{cos\theta}{sin\theta} \\ Therefore \\ cot\theta=\frac{cos\theta}{sin\theta},is\text{ an Identity} \end{gathered}

E

\begin{gathered} 1+cot^2\theta=csc^2\theta \\ csc^2\theta-cot^2\theta=1 \\ csc^2\theta=\frac{1}{sin^2\theta} \\ cot^2\theta=\frac{cos^2\theta}{sin^2\theta} \\ So, \\ \frac{1}{sin^2\theta}-\frac{cos^2\theta}{sin^2\theta} \\ \frac{1-cos^2\theta}{sin^2\theta} \\ Note, \\ cos^2\theta+sin^2\theta=1 \\ sin^2\theta=1-cos^2\theta \\ So, \\ \frac{1-cos^2\theta}{sin^2\theta}=\frac{sin^2\theta}{sin^2\theta}=1 \\ Therefore \\ 1+cot^2\theta=csc^2\theta,\text{ is an Identity} \end{gathered}

Hence, the following are identities

\begin{gathered} B)sin\theta=\frac{1}{csc\theta} \\ D)cot\theta=\frac{cos\theta}{sin\theta} \\ E)1+cot^2\theta=csc^2\theta \end{gathered}

The marked are the trigonometric identities

3 0
1 year ago
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