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3 years ago

If a 5 ft tall man cast an 8 ft long shadow at the same time a tree cast a 24 ft long shadow, how tall is the tree?

Mathematics

2 answers:

Taya2010 [7]3 years ago

5 0

Answer:

15 feet

Step-by-step explanation:

We have 2 similar right triangles with legs height and length of shadows.

height of men : length of shadows of the man = height of tree : length of shadows of the tree

5 : 8 = x : 24

8x = 5* 24

x = 5*24/8 = 15 (feet)

erma4kov [3.2K]3 years ago

3 0

Answer:

15ft

Step-by-step explanation:

5 ft is to 8 ft

A ft is to 24 ft

A = 24*5/8

A = 15ft

15ft

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Tomtit [17]

Answer:

$Point A = (-\frac{1}{2}, -\frac{1}{4})$

$Point B = (-2\frac{1}{2},1\frac{1}{2} )$

$Point C = (\frac{1}{4},\frac{1}{2})$

8 0

3 years ago

Bob neff, owner of an automotive dealership, pays one of his salesmen, mike, a $1,300 draw per week plus 6% on all commission sa

Sedaia [141]

Draw per week = $1,300

Draw for four weeks = 1300 x 4 = $5,200

Four week sales amount = $186,900

6% commission on sales = 0.06 x 186,900 = $11,214

<span>Thus the amount equal to mike’s commission minus the draw = 11,214 – 5,200 = $6,014</span>

4 0

4 years ago

[5y]^4 when y= 2 Evaluate the expression

Novay_Z [31]

Answer:

10000

Step-by-step explanation:

Well substitute it

(5*2)^4

now follow order of operations (Parenthesis, exponents, multipication-division left to right, etc.)

5*2=10

10^4=10000

so the answer would be

10000 if y is 2

5 0

3 years ago

Read 2 more answers

In general, the point

Zanzabum

Answer:

we conclude that when we put the ordered pair (0, a), both sides of the function equation becomes the same.

Therefore, the point (0, a) is on the graph of the function f(x) = abˣ

Hence, option (D) is correct.

Step-by-step explanation:

Given the function

f(x) = abˣ

Let us substitute all the points one by one

FOR (b, 0)

y = abˣ

putting x = b, y = 0

0 = abᵇ

FOR (a, b)

y = abˣ

putting x = a, y = b

b = abᵃ

FOR (0, 0)

y = abˣ

putting x = 0, y = 0

0 = ab⁰

0 = a ∵b⁰ = 1

FOR (0, a)

y = abˣ

putting x = 0, y = a

a = ab⁰

a = a ∵b⁰ = 1

TRUE

Thus, we conclude that when we put the ordered pair (0, a), both sides of the function equation becomes the same.

Therefore, the point (0, a) is on the graph of the function f(x) = abˣ

Hence, option (D) is correct.

8 0

3 years ago

Use a triple integral to find the volume of the tetrahedron T bounded by the planes x+2y+z=2, x=2y, x=0 and z=0

Tanzania [10]

Answer:

Volume of the Tetrahedron T = $\frac{1}{3}$

Step-by-step explanation:

As given, The tetrahedron T is bounded by the planes x + 2y + z = 2, x = 2y, x = 0, and z = 0

We have,

z = 0 and x + 2y + z = 2

⇒ z = 2 - x - 2y

∴ The limits of z are :

0 ≤ z ≤ 2 - x - 2y

Now, in the xy- plane , the equations becomes

x + 2y = 2 , x = 2y , x = 0 ( As in xy- plane , z = 0)

Firstly , we find the intersection between the lines x = 2y and x + 2y = 2

∴ we get

2y + 2y = 2

⇒4y = 2

⇒y = $\frac{2}{4} = \frac{1}{2}$ = 0.5

⇒x = 2( $\frac{1}{2}$ ) = 1

So, the intersection point is ( 1, 0.5)

As we have x = 0 and x = 1

∴ The limits of x are :

0 ≤ x ≤ 1

Also,

x = 2y

⇒y = $\frac{x}{2}$

and x + 2y = 2

⇒2y = 2 - x

⇒y = 1 - $\frac{x}{2}$

∴ The limits of y are :

$\frac{x}{2}$ ≤ y ≤ 1 - $\frac{x}{2}$

So, we get

Volume = $\int\limits^1_0 {\int\limits^{1-\frac{x}{2}}_{y = \frac{x}{2}}\int\limits^{2-x-2y}_{z=0} {dz} \, dy \, dx$

= $\int\limits^1_0 {\int\limits^{1-\frac{x}{2}}_{y = \frac{x}{2}}{[z]}\limits^{2-x-2y}_0 {} \, \, dy \, dx$

= $\int\limits^1_0 {\int\limits^{1-\frac{x}{2}}_{y = \frac{x}{2}}{(2-x-2y)} \, \, dy \, dx$

= $\int\limits^1_0 {[2y-xy-y^{2} ]}\limits^{1-\frac{x}{2}} _{\frac{x}{2} } {} \, \, dx$

= $\int\limits^1_0 {[2(1-\frac{x}{2} - \frac{x}{2}) -x(1-\frac{x}{2} - \frac{x}{2}) -(1-\frac{x}{2}) ^{2} + (\frac{x}{2} )^{2} ] {} \, \, dx$

= $\int\limits^1_0 {(1 - 2x + x^{2} )} \, \, dx$

= ${(x - x^{2} + \frac{x^{3}}{3} )}\limits^1_0$

= 1 - 1² + $\frac{1^{3} }{3}$ - 0 + 0 - 0

= 1 - 1 + $\frac{1 }{3}$ = $\frac{1}{3}$

So, we get

Volume = $\frac{1}{3}$

7 0

3 years ago