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3 years ago

What are the dimensions of the base of the pyramid?

Mathematics

1 answer:

aleksley [76]3 years ago

4 0

Answer:

Step-by-step explanation:

its a wild guess because you didn't add a picture or anything to show the pyramid

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Use the given conditions to write an equation in slope-intercept form. passing through (-2, 0) & (0, 2)

Crazy boy [7]

Answer:

Step-by-step explanation:

(2 - 0)/(0 + 2)= 2/2= 1

y - 0 = x + 2

y = x + 2

6 0

3 years ago

Please help with 40-42 please if you could

likoan [24]

40.) 10.5^2 and the area is 110.25

41.) 126

42.) 81 millimeters

7 0

3 years ago

What is the value of the expression<br> 9r-5s+3 to the 3rd power if r=2/3 and s= -2

kramer

Answer:

Step-by-step explanation:

(9r-5s+3^3)

Let r = 2/3 and s = -2

(9 * 2/3-5* -2+3^3)

(6+ 10+27)

5 0

3 years ago

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How far can you walk half an hour?

lys-0071 [83]

depends on how fast you are going

3 0

3 years ago

A car leaves an intersection traveling west. Its position 4 sec later is 18 ft from the intersection. At the same time, another

Evgen [1.6K]

Answer:

The rate at which the distance between the two cars is changing is;

15.53 ft/sec.

Step-by-step explanation:

To solve the question, we note that

Position of car A 4 s after start of motion, w = 18 ft west,

Position of car B 4 s after start of motion, n = 27 ft north

Therefore

The distance between the two cars at the 4 s instance is

d² = w² + n²

d² = 18² + 27² = 1053 ft² and

d = 32.45 ft

The rate at which the distance between the two cars is changing is given by;

Differentiating both sides of the equation, d² = w² + n², with respect to t as follows.

$\frac{dd^2}{dt} = \frac{dw^2}{dt} + \frac{dn^2}{dt} \Longrightarrow 2 d\frac{dd}{dt} = 2w\frac{dw}{dt} + 2n\frac{dn}{dt}$

It is given that the speeds of car A and car B at the 4 second instant are 7 ft/sec and 14 ft/sec, respectively

That is;

$\frac{dw}{dt} = 7\frac{ft}{sec}$ and $\frac{dn}{dt} = 14\frac{ft}{sec}$

Substituting the values of speed in the equation of rate of change gives

$2 d\frac{dd}{dt} = 2w\frac{dw}{dt} + 2n\frac{dn}{dt}\Longrightarrow d\frac{dd}{dt} = w\frac{dw}{dt} + n\frac{dn}{dt}$

$d\frac{dd}{dt} = w\frac{dw}{dt} + n\frac{dn}{dt} \Longrightarrow 32.45\frac{dd}{dt} = 18\times 7 + 27\times 14 = 504$

$32.45\frac{dd}{dt} = 504$

$\frac{dd}{dt} = \frac{504}{32.45} = 15.53 \frac{ft}{sec}$

The rate at which the distance between the two cars is changing = dd/dt = 15.53 ft/sec.

7 0

3 years ago

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