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Nata [24]
3 years ago
12

Can you please help me with this I will mark the BRAINLIEST who ever can help thanks​

Mathematics
2 answers:
son4ous [18]3 years ago
5 0

QUESTION:

WORK OUT THE  VALUE OF 2^{6}

Or 2 times 2 times 2 times 2 times 2 times 2 times 2

ANSWER:

So, 2^6 is:

64

Now, 7^4 is:

2,401

Hope it helps!

Grace Rosalia

Tanya [424]3 years ago
4 0

Answer:

a) 2x2x2x2x2x2=64

b)7x7x7x7=2401

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A 12 foot ladder is leaning against a building. If the bottom of the ladder is sliding along the pavement directly away from the
ira [324]

Using Pythagoras theorem, the top of the ladder moving down when the foot of the ladder is 3 feet from the wall is of -0.518 feet/sec.                      

Let distance from the wall to the foot of the ladder is 'x' feet and the height of the top of the ladder is 'y' feet.

Pythagoras theorem, x^{2} + y^{2} = (12)^{2}       --->(1)

Given,\frac{dx}{dt}= 2feet/second   at x=3

Put x=3 in Pythagoras theorem equation (1)

(3)^{2} + y^{2} = 144

         y^{2} = 144 - 9

        y^{2}  =  135

        y = 11.61

Derive equation (1) w.r.t to 't'

2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0                ---->(2)

substitute the value of 'x', 'dx/dt' and 'y' in equation (2), we get the fast of the top of the ladder moving down when the foot of the ladder is 3 feet from the wall

2(3)(2) + 2 (11.61)\frac{dy}{dt}  = 0

12 + 23.22 \frac{dy}{dt}  = 0

                  \frac{dy}{dt}= \frac{-12}{23.22}

                  \frac{dy}{dt} = -0.518

Hence,  using Pythagoras theorem the top of the ladder moving down when the foot of the ladder is 3 feet from the wall is of -0.518 feet/sec.  

Learn more about Pythagoras theorem here

brainly.com/question/21511305  

#SPJ4    

     

5 0
2 years ago
Can someone plz help me with this one problem
crimeas [40]

Answer:

24 1/2

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
The length of a rectangle is 5 in longer than its width. if the perimeter of the rectangle is 50 in , find its area.
GrogVix [38]
The area is 150 in^2

6 0
4 years ago
A flat rectangular piece of aluminum has a perimeter of 62 inches. The length is 15
Salsk061 [2.6K]

Answer:

So, the required width of rectangular piece of aluminium is 8 inches

Step-by-step explanation:

We are given:

Perimeter of rectangular piece of aluminium = 62 inches

Let width of rectangular piece of aluminium = w

and length of rectangular piece of aluminium  = w+15

We need to find width i.e value of x

The formula for finding perimeter of rectangle is: Perimeter=2(Length+Width)\\

Now, Putting values in formula for finding Width w:

Perimeter=2(Length+Width)\\\\62=2(w+15+w)\\62=2(2w+15)\\62=4w+30\\62-30=4w\\4w=32\\w=\frac{32}{4}\\w=8

After solving we get the width of rectangular piece :w = 8

So, the required width of rectangular piece of aluminium is 8 inches

6 0
3 years ago
Is it possible to place another box of volume 420 cubic inches inside of the box shown?
frozen [14]
Yes, the volume it is 8³ = 512 
The volume it is 420
512 >>> 420 
The side the cube is ³√420 = 7.48 in
8 >> 7.44

6 0
4 years ago
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