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3 years ago

How many inches are in 4 feet 3 inches?

Mathematics

2 answers:

mina [271]3 years ago

7 0

51 in.
4(12)+3=51
12 inches in a foot

V125BC [204]3 years ago

4 0

52 because 4 feet is 48 inches 9 (which is like 1 feet is 12 inches) and 48 plus 4 equals 52

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In ABC, the measure of sides c is 3.9 cm If DEF is a dilation ABC with a scale factor of 2.5 what is the measure of side f

White raven [17]

Answer:

$F = 9.75cm$

Step-by-step explanation:

Given

In ABC;

$C = 3.9cm$

$Dilation = 2.5$

Required

Determine side F

In the given question;

DEF is a dilation of ABC

Comparing both triangles side by side,

Side F is a direct dilation of side C and this is calculated as follows

$F = Dilation * C$

$F = 2.5 * 3.9cm$

$F = 9.75cm$

3 0

3 years ago

given a point A is on a coordinate plane in quadrant I. if it is rotated 90° counter clockwise then reflecred across the y axis.

den301095 [7]

Answer:

it would be -90⁰ because its clockwise its the opposite direction

7 0

3 years ago

Given the point (3,4 ) and the slope of 6 ,find y when x=27.

Irina-Kira [14]

Work shown above! y = 148

6 0

3 years ago

Rich measured the height of a desk to be 80.7 cm. The actual height of the desk is 80 cm. What is Rich's percent error in this c

Hitman42 [59]

Answer:

0.9%

Step-by-step explanation:

We have been given that Rich measured the height of a desk to be 80.7 cm. The actual height of the desk is 80 cm.

We will use percentage error formula to solve our given problem.

$\text{Percentage error}=\frac{\text{Experimental value-Actual value }}{\text{Actual actual}}\times 100$

$\text{Percentage error}=\frac{80.7-80}{80}\times 100$

$\text{Percentage error}=\frac{0.7}{80}\times 100$

$\text{Percentage error}=0.00875\times 100$

$\text{Percentage error}=0.875\approx 0.9$

Therefore, Rich's percent error in calculation is 0.9%.

5 0

3 years ago

An object is acted upon by the forces F1=<10,6,4> and F2=<0,3,9>. Find the force F3 that must act on the object

murzikaleks [220]

$\vec F_3=\langle x,y,z\rangle$ is a force vector such that

$\vec F_1+\vec F_2+\vec F_3=\vec0$

$\implies\langle10,6,4\rangle+\langle0,3,9\rangle+\langle x,y,z\rangle=\langle0,0,0\rangle$

$\implies\begin{cases}10+x=0\\9+y=0\\13+z=0\end{cases}\implies x=-10,y=-9,z=-13$

So the missing force is

$\boxed{\vec F_3=\langle-10,-9,-13\rangle}$

8 0

2 years ago