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3 years ago

1. Find the VALUE of 7x + 3y, when x = 12 and y = -6.

Mathematics

2 answers:

denis23 [38]3 years ago

8 0

Answer:

1.7(12) + 3(-6) = 84 - 18 = 66

slavikrds [6]3 years ago

3 0

Answer:
1. 66
2. 3.87 x 10^-5
3. 81.67
4. 7y - 19

Explanation:

7(12) + 3(-6)
= 84 - 18
= 66

2. 0.0000387 = 3.87 x 10^-5

3.
Area = 3.14 x (5.1)^2
= 81.67cm^2

4.
6(2y - 3) - 5(y + 1)
= 12y - 18 - 5y - 1
= 7y - 19

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Mamont248 [21]

Given:

Consider the expression are

1) $\sqrt{36}$

2) $\sqrt[3]{-8}$

3) $-\sqrt{100}$

4) $\sqrt[3]{27}$

To find:

The simplified form of each expression.

Solution:

1. We have,

$\sqrt{36}=\sqrt{6^2}$

$\sqrt{36}=6$

Therefore, the value of this expression is 6.

2. We have,

$\sqrt[3]{-8}=(-8)^{\frac{1}{3}}$

$\sqrt[3]{-8}=((-2)^3)^{\frac{1}{3}}$

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Therefore, the value of this expression is -2.

3. We have,

$-\sqrt{100}=-\sqrt{10^2}$

$-\sqrt{100}=-10$

Therefore, the value of this expression is -10.

4. We have,

$\sqrt[3]{27}=(27)^{\frac{1}{3}}$

$\sqrt[3]{27}=(3^3)^{\frac{1}{3}}$

$\sqrt[3]{27}=(3)^{\frac{3}{3}}$

$\sqrt[3]{27}=3$

Therefore, the value of this expression is 3.

3 0

3 years ago

Two physics students are doing a side competition during a game of bowling, seeing who can toss a ball with the larger momentum.

dem82 [27]

Answer:

she should beat a speed of v₂ = 6.04 m/s in order to win the competition

Step-by-step explanation:

Since the momentum p is defined as

p = m*v

where

m= mass of the ball

v= velocity of the ball

denoting 1 and 2 as the first and second bowler , then to reach the momentum of the first bowler

p₂=p₁

therefore

m₁*v₁ = m₂*v₂

v₂ = v₁ * (m₁/m₂)

replacing values

v₂ = v₁ * (m₁/m₂) =8.6 m/s * (4.5 kg/6.4 kg ball) = 6.04 m/s

then since the momentum p = m*v increases with increasing v (at constant m) , she should beat a speed of v₂ = 6.04 m/s in order to win the competition

6 0

4 years ago

At total of 2 feet of snow falls over three days on the first day 1/6 of the snow falls on the second day 7/12 of the snow falls

ioda

Answer:

$Day\ 3 = 1\frac{3}{12}$