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sammy [17]
2 years ago
15

Solve. 5) Solve 2y + 12 < 42. Show your work.

Mathematics
2 answers:
OLEGan [10]2 years ago
4 0

The answer is y < 15

Assoli18 [71]2 years ago
4 0

Answer:

y<15

Step-by-step explanation:

Subtract 12 from both sides.

2y+12−12<42−12

2y<30

Step 2: Divide both sides by 2.

2y

2

<

30

2

y<15

Answer:

y<15

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bagirrra123 [75]
This equation cannot be solved to get a single number. Instead just use normal algebra skills to get the w by itself by subtracting 2l from both sides and then dividing by 2.
5 0
3 years ago
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A cylinder has a diameter of 10 decimeters. What is the length of the curved surface?
shepuryov [24]
The length of the curved surface is the circumfrence of a circle with radius r = 10/2 = 5 decimeters.

Length of curved surface = 2π x 5 = 10π ≈ 31.4 decimeters.
6 0
2 years ago
PLEASE HELP ME <br>Multiply the monomials: <br><br>a2x5b, −0.6axb2 and 0.6a2b3
Tanzania [10]

Answer:

-0.36 a^{5}b^{6} x^{6}

Step-by-step explanation:

Given:

a^2x^5b,,\ -0.6axb^2,\ and\ 0.6a^2b^3

Required

Multiply

The above can be written as

a^2x^5b * -0.6axb^2\ * \ 0.6a^2b^3

Split the above expression

a^2*x^5*b * -0.6*a*x*b^2\ * \ 0.6*a^2*b^3

Collect like terms

-0.6 * 0.6 * a^2*a*a^2*b *b^2 *b^3*x^5*x

Apply first law on indices

-0.36  * a^{2+1+2} *b^{1+2+3} *x^{5+1}

-0.36  * a^{5} *b^{6} *x^{6}

-0.36 a^{5}b^{6} x^{6}

Hence, a^2x^5b * -0.6axb^2\ * \ 0.6a^2b^3 is equivalent to -0.36 a^{5}b^{6} x^{6}

8 0
3 years ago
A spring has a natural length of 7 m. If a 4-N force is required to keep it stretched to a length of 11 m, how much work W is re
bezimeni [28]

Answer:

18 J is the work required to stretch a spring from 7 m to 13 m.

Step-by-step explanation:

The work done is defined to be the product of the force F and the distance d  that the object moves:

W=Fd

If F is measured in newtons and d<em> </em>in meters, then the unit for is a newton-meter, which is called a joule (J).

This definition work as long as the force is constant, but if the force is variable like in this case, we have that the work done is given by

W=\int\limits^b_a {f(x)} \, dx

Hooke’s Law states that the force required to maintain a spring stretched x    units beyond its natural length is proportional to

f(x)=kx

where k is a positive constant (called the spring constant).

To find how much work W is required to stretch it from 7 m to 13 m you must:

Step 1: Find the spring constant

We know that the spring has a natural length of 7 m and a 4 N force is required to keep it stretched to a length of 11 m. So, applying Hooke’s Law

4=k(11-7)\\\\\frac{k\left(11-7\right)}{4}=\frac{4}{4}\\\\k=1

Thus F=x

Step 2: Find the the work done in stretching the spring from 7 m to 13 m.

Recall that the natural length is 7 m, so when we stretch the spring from 7 m to 13 m, we are stretching it by 6 m beyond its natural length.

Work needed to stretch it by 6 m beyond its natural length

W=\int\limits^6_0 {x} \, dx \\\\\mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1\\\\\left[\frac{x^{1+1}}{1+1}\right]^6_0\\\\\left[\frac{x^2}{2}\right]^6_0=18

18 J is the work required to stretch a spring from 7 m to 13 m.

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3 years ago
Seven more than the cube of a number is what
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7+x^3. It cannot be simplified further without knowing the value of x
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