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sweet-ann [11.9K]
3 years ago
8

Simplify -4 1/6 - -6 1/3 Which of the following is correct?

Mathematics
1 answer:
leonid [27]3 years ago
3 0

Answer:

its 2 1/6 (B)

Step-by-step explanation:

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Find the missing side length of the right triangle. show work.
motikmotik

Answer:

6.32

Step-by-step explanation:

To find the side length on a right angled triangle we follow The Pythagorean Theorem:

{a}^{2}  +  {b}^{2}  =  {c}^{2}

c being the hypothesis (longest side: the hypotenuse is always opposite the right angle)

so..

{6}^{2}  +   {2}^{2}  =  {x}^{2}  \\ 36 + 4 =  {x}^{2}  \\ 40 =  {x}^{2} \\  \sqrt{40}   =  \sqrt{ {x}^{2} } \\

6.32 = x

8 0
2 years ago
Read 2 more answers
Someone please help me!! I don’t know the answer!!
coldgirl [10]

Answer:

Step-by-step explanation:

Remark

The measurement of the arc is 3Pi

that represents 108 / 360 of the circle.

Equation

108/350 2 * pi * R = 3* pi

Solution                            

Pi is on both sides of the equation. They both cancel.

108/360 = 0.3

0.3 * 2 * R = 3

0.6 * R = 3

R = 3/0.6

R = 5

8 0
3 years ago
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Larinda cooked a 4-kilogram roast. The roast left over after the meal weighed 3 kilograms. How many grams of roast were eaten du
blsea [12.9K]

Answer:

1 kilogram was eaten

Step-by-step explanation:

7 0
4 years ago
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What is the solution to the system x+ y =4 and 2x - y = -1
Lostsunrise [7]

Answer:

Step-by-step explanation:

hello :

the system is :  x+ y =4 ...(*)

                       2x - y = -1...(**)

add(*) and (**) : 3x=3    so : x=1

put this value into (*) : 1+y=4     so : y=3

one solution : (1 , 3)

4 0
3 years ago
The position of an object along a vertical line is given by s(t) = −t3 + 3t2 + 7t + 4, where s is measured in feet and t is meas
saw5 [17]

Answer:

The maximum velocity of the object in the time interval [0, 4] is 10 ft/s.

Step-by-step explanation:

Given : The position of an object along a vertical line is given by s(t) = -t^3+3t^2+7t +4, where s is measured in feet and t is measured in seconds.

To find : What is the maximum velocity of the object in the time interval [0, 4]?

Solution :

The velocity is rate of change of distance w.r.t time.

Distance in terms of t is given by,

s(t) = -t^3+3t^2+7t +4

Derivate w.r.t. time,

v(t)=s'(t) = -3t^2+6t+7

It is a quadratic function so its maximum is at vertex of the function.

The x point of the function is given by,

x=-\frac{b}{2a}

Where, a=-3, b=6 and c=7

t=-\frac{6}{2(-3)}

t=-\frac{6}{-6}

t=1

As 1 lie between interval [0,4]

Substitute t=1 in the function,

v(t)= -3(1)^2+6(1)+7

v(t)= -3+6+7

v(1)=10

Th maximum velocity is 10 ft/s.

Therefore, the maximum velocity of the object in the time interval [0, 4] is 10 ft/s.

8 0
3 years ago
Read 2 more answers
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