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4 years ago

1. Six plus "x" is greater than eleven.

Mathematics

1 answer:

kow [346]4 years ago

5 0

First of all you need to analyze each equation and then write it in an algebraic form , the just treat it as a regular equation

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Solve 9c + 4 = -23 -3 3 -2,1 4

irga5000 [103]

9c + 4 = -23
9c = -27
c = -3

6 0

3 years ago

Read 2 more answers

How do you get the same base for the power

zhannawk [14.2K]

64 = 2 ^6
16 = 2^4 =

2^ 6*(2x+4) = 2^4 * 5x

6(2x+4) = 20x

10/3 x = 2x +4

4/3 x = 4

x = 3

you can also use logarithms like so

(2x+4)ln64 = 5x ln16

ln64/ln16 = 3/2

3/2 * 2x + 3/2 * 4 = 5x

3x + 6 = 5x

2x = 6

x = 3

7 0

3 years ago

Select the correct figures.

Contact [7]

Answer:

3rd one and last one

Step-by-step explanation:

First one:

4 × 3 × 4 = 48

Second one:

1 × 2 × 4 = 8

Third one:

2 × 3 × 5 = 30

Fourth one:

2 × 5 × 4 = 40

Fifth one:

2 × 2 × 4 = 16

Process of elimination:

The first one is itself greater than 47 units³, so this is definitely not one of the two;

A combination of the fourth prism and the smallest one as the second will give a volume of 48 units³, which is greater than 47 so this prism is also not one of the two we are looking for;

This leaves the second, third and fifth;

The second and third sum to 38, which is less than 42 so it is not the correct combination;

Similarly, the second and last gives only 24 so it is not the correct combination;

This leaves the only remaining option of the third and fifth;

30 + 16 = 46, which confirms this to be the correct combination.

5 0

3 years ago

For what values of x would the following be true? -2(x-3)<4

emmainna [20.7K]

Answer:

x > 1

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Step-by-step explanation:

Step 1: Define

-2(x - 3) < 4

Step 2: Solve for x

Divide -2 on both sides: x - 3 > -2
Add 3 to both sides: x > 1

Here we see that any value x greater than 1 would work as a solution to the equation.

7 0

4 years ago

Suppose you have a water-balloon launcher. The balloon is 3 ft high when it leaves the launcher. Use the equation 0 = ?16t2 + 38

mario62 [17]

Answer:

t = 2.5 s

Step-by-step explanation:

Suppose you have a water-balloon launcher. The balloon is 3 ft high when it leaves the launcher. Its equation is :

$16t^2 + 38.8t + 3 =0$

The above equation is a quadratic equation. The general equation is :

$ax^2+bx+c=0$

Here, a = 16, b = 38.8 and c = 3

The solution of quadratic equation is given by :

$x=\dfrac{-b\pm \sqrt{b^2-4ac} }{2a}\\\\t=\dfrac{-38.8\pm \sqrt{(38.8)^2-4\times (-16)\times 3} }{2\times (-16)}\\\\t=\dfrac{-38.8+\sqrt{(38.8)^{2}-4\times-16\times3}}{-2\times16}, \dfrac{-38.8-\sqrt{(38.8)^{2}-4\times-16\times3}}{-2\times16}\\\\t=-0.075\ s,2.5\ s$

So, at t = 2.5 s the balloon is in air.

8 0

3 years ago