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sergij07 [2.7K]

3 years ago

14

You have just planted a new tree. To support the tree, you attach a wire from the trunk of th tree to a stake in the ground. Eac

h wire has a length of 7 feet. Suppose you put the stakes in the ground 5 feet from the base of the trunk. Approximately how far up the trunk should you attach the wires?

1 answer:

Tomtit [17]3 years ago

5 0

Answer: 4.9 ft

Step-by-step explanation:

Given

Length of wire $l=7\ ft$

Stake is at a distance of $5\ ft$ from the base of tree

suppose wire is attached y ft above the base of tree

from the figure, we can write

$\Rightarrow 7^2=5^2+y^2\\\Rightarrow y^2=49-25\\\Rightarrow y=\sqrt{24}\\\Rightarrow y=4.89\approx 4.9\ ft$

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mr chu is slicing 4 cantaloupes for his class . There are 24 students in the class. If he cuts the cantal to make an equal porti

HACTEHA [7]

1/6 because 4/24 simplifies to 1/6

4 0

4 years ago

-3/8 - 7/8 = <br>explicar el procedimiento por favor

pogonyaev

Answer:

-1/4

Step-by-step explanation:

-3/8 - 7/8= 4/8=1/2

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3 years ago

PLEASE HELP ME WITH THIS QUESTION!

RSB [31]

Peta has 10 plums. The second sentence is unnecessary.

8 0

3 years ago

Read 2 more answers

The vertices of ∆ABC are A(-2, 2), B(6, 2), and C(0, 8). The perimeter of ∆ABC is units is? What is the area?

11111nata11111 [884]

we have

$A(-2, 2),B(6, 2),C(0, 8)$

see the attached figure to better understand the problem

we know that

The perimeter of the triangle is equal to

$P=AB+BC+AC$

and

the area of the triangle is equal to

$A=\frac{1}{2}*base *heigth$

in this problem

$base=AB\\heigth=DC$

we know that

The distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Step $1$

<u>Find the distance AB</u>

$A(-2, 2),B(6, 2)$

Substitute the values in the formula

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

$d=\sqrt{(0)^{2}+(8)^{2}}$

$dAB=8\ units$

Step $2$

<u>Find the distance BC</u>

$B(6, 2),C(0, 8)$

Substitute the values in the formula

$d=\sqrt{(8-2)^{2}+(0-6)^{2}}$

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

$dBC=6\sqrt{2}\ units$

Step $3$

<u>Find the distance AC</u>

$A(-2, 2),C(0, 8)$

Substitute the values in the formula

$d=\sqrt{(8-2)^{2}+(0+2)^{2}}$

$d=\sqrt{(6)^{2}+(2)^{2}}$

$dAC=2\sqrt{10}\ units$

Step $4$

<u>Find the distance DC</u>

$D(0, 2),C(0, 8)$

Substitute the values in the formula

$d=\sqrt{(8-2)^{2}+(0-0)^{2}}$

$d=\sqrt{(6)^{2}+(0)^{2}}$

$dDC=6\ units$

Step $5$

<u>Find the perimeter of the triangle</u>

$P=AB+BC+AC$

substitute the values

$P=8\ units+6\sqrt{2}\ units+2\sqrt{10}\ units$

$P=22.81\ units$

therefore

The perimeter of the triangle is equal to $22.81\ units$

Step $6$

<u>Find the area of the triangle</u>

$A=\frac{1}{2}*base *heigth$

in this problem

$base=AB=8\ units\\heigth=DC=6\ units$

substitute the values

$A=\frac{1}{2}*8*6$

$A=24\ units^{2}$

therefore

the area of the triangle is $24\ units^{2}$

4 0

3 years ago

Read 2 more answers

A cheetah ran 300 feet in 2.92 seconds. What was the cheetah’s average speed in miles per hour? Show your work.

alexdok [17]

Answer:

about 70.0 mph

Step-by-step explanation:

speed = distance/time

(300 ft)/(2.92 s)·(1 mi)/(5280 ft)·(3600 s)/(1 h) = (300·3600/(2.92·5280)) mi/h

≈ 70.0498 mi/h

8 0

3 years ago