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2 years ago

How do you write 14.8 as a mixed fraction Step by step method please?

1 answer:

lisabon 2012 [21]2 years ago

5 0

$14.8=14+0.8\\\\0.8=\dfrac{8}{10}=\dfrac{8:2}{10:2}=\dfrac{4}{5}\\\\therefore\\\\14.8=14+\dfrac{4}{5}=\boxed{14\frac{4}{5}}$

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in the following equation A & B are both integers. What is the value of a? What is the value of b?

MariettaO [177]

3ax-8a=b-18x
3ax+18x=b+8a
3x(a+6)=b+8a

3x=(b+8a)/(a+6)
x=(b+8a)/3(a+6)
3a*(b+8a)/3(a+6)=b-18*(b+8a)/3(a+6)
a(b+8a)/(a+6)=b-6(b+8a)/(a+6)
a(b+8a)=b(a+6)-6(b+8a)
ab+8a^2=ab+6b-6b-48a
ab+8a^2=ab-48a
8a^2=-48a
8a^2+48a=0
8a(a+6)=0 so a=0,-6
0=b-18x or -18x+48=b-18x
-b=-18x or 48=b
let b=18 if a =-6

8 0

2 years ago

The lengths of two sides of a triangle are 15 inches each the third side measures 10 inches what type of triangle is this

Effectus [21]

It's an isosceles triangle. an isosceles triangle has two congruent sides that have the same measures and the third side can be whatever measure. Those 2 congruent sides are called "legs"

6 0

2 years ago

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The position function of a particle in rectilinear motion is given by s(t) = 2t3 – 21t2 + 60t + 3 for t ≥ 0 with t measured in s

Norma-Jean [14]

The positions when the particle reverses direction are:

$s(t_1)=55ft\\\\s(t_2)=28ft$

The acceleraton of the paticle when reverses direction is:

$a(t_1)=-18\frac{ft}{s^{2}}\\ \\a(t_2)=a(5s)=18\frac{ft}{s^{2}}$

Why?

To solve the problem, we need to remember that if we derivate the position function, we will get the velocity function, and if we derivate the velocity function, we will get the acceleration function. So, we will need to derivate two times.

Also, when the particle reverses its direction, the velocity is equal to 0.

We are given the following function:

$s(t)=2t^{3}-21t^{2}+60t+3$

So,

- Derivating to get the velocity function, we have:

$v(t)=\frac{ds}{dt}=(2t^{3}-21t^{2}+60t+3)\\\\v(t)=3*2t^{2}-2*21t+60*1+0\\\\v(t)=6t^{2}-42t+60$

Now, making the function equal to 0, to find the times when the particle reversed its direction, we have:

$v(t)=6t^{2}-42t+60\\\\0=6t^{2}-42t+60\\\\0=t^{2}-7t+10\\(t-5)*(t-2)=0\\\\t_{1}=5s\\t_{2}=2s$

We know that the particle reversed its direction two times.

- Derivating the velocity function to find the acceleration function, we have:

$a(t)=\frac{dv}{dt}=6t^{2}-42t+60\\\\a(t)=12t-42$

Now, substituting the times to calculate the accelerations, we have:

$a(t_1)=a(2s)=12*2-42=-18\frac{ft}{s^{2}}\\ \\a(t_2)=a(5s)=12*5-42=18\frac{ft}{s^{2}}$

Now, substitutitng the times to calculate the positions, we have:

$s(t_1)=2*(2)^{3}-21*(2)^{2}+60*2+3=16-84+120+3=55ft\\\\s(t_2)=2*(5)^{3}-21*(5)^{2}+60*5+3=250-525+300+3=28ft$

Have a nice day!

3 0

2 years ago

Jenny uses a roller to paint a wall. The roller has a radius of 1.75 inches and a height of 10 inches. In two rolls, what is the

Katen [24]

Answer:

The area of the wall that she will paint in two rolls is <u>219.8 inches²</u>.

Step-by-step explanation:

Given:

Jenny uses a roller to paint a wall. The roller has a radius of 1.75 inches and a height of 10 inches.

Now, to find the area of the wall that she will paint in two rolls.

So, we find the lateral surface area of roller.

Radius (r) = 1.75 inches.

Height (h) = 10 inches.

So, to get the lateral surface area we put formula:

$A_{L} =2\pi rh$

$A_L=2\times 3.14\times 1.75\times 10$

$A_L=109.9\ inches^2.$

Thus, the lateral surface area of the roller = 109.9 inches².

Now, to get the area of wall that she will paint in two rolls we multiply 2 by the lateral surface area of the roller:

$109.9\times 2\\\\=219.8\ inches^2.$

Therefore, the area of the wall that she will paint in two rolls is 219.8 inches².

3 0

2 years ago

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Does anyone know the answer???

Fiesta28 [93]

Answer:

-2

Step-by-step explanation:

41 - 8m - 57 = 0

41- 8m = 57

-8m = 57-41

-8m = 16

m = -16/8

m = -2

8 0

2 years ago

Read 2 more answers