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3 years ago

Could somebody or someone help me with this please????

Mathematics

1 answer:

Murrr4er [49]3 years ago

6 0

No one knows the answer

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Are the equations 4-5x=24 and 5x=20 equivalent?

mina [271]

Answer:

not equivalent

Step-by-step explanation:

solve for x

4-5x=24

-5x=20

x = - 4

5x=20

x = 4

no they are not equivalent

8 0

2 years ago

3+7(0.7+1.3) What is the answer and how to solve it?

quester [9]

When you see this kind of question use « BODMAS »
1.Add the 2 digits which is 3+7=10
2.Add the digits inside which is 0.7+1.3=2
3.Add the inside and outside
3+7(0.7+1.3)
10(2)=20

5 0

2 years ago

What value of x makes the equation true? -1=1 – 4(-5+4x) – 4x

cluponka [151]

Answer:

x = 11/10

Step-by-step explanation:

we have from statement:

-1=1 - 4(-5+4x) – 4x

we apply distributive property:

-1 = 1 -4*(-5) -4*(4x) -4x

so we have:

-1= 1+20 -16x -4x

-1 = 21 -20x

-22=-20x

so we have:

x=22/20

x=11/10

we can substitute the x value found:

-1 = 1 - 4(-5+4*(11/10)) – 4*(11/10)

-1 = 1+20 -17.6 -4.4

-1 =21 -22

so we have:

-1 = -1

4 0

3 years ago

Use the figure below and 2-3 sentences to explain the following:

nydimaria [60]

Answer:

See below

Step-by-step explanation:

The side opposite angle E is side e (DF)

The side adjacent to angle E is side d (EF)

The hypotenuse is side f (ED)

5 0

3 years ago

Find a positive number for which the sum of it and its reciprocal is the smallest (least) possible. Let x be the number and let

Allisa [31]

Answer:

$S(x) = x + \frac{1}{x}$ --- Objective function

Interval = $\{x:x=1\}$

Step-by-step explanation:

Given

Represent the number with x

The required sum can be represented as:

$x + \frac{1}{x}$

Hence, the objective function is:

$S(x) = x + \frac{1}{x}$

To get the the interval, we start by differentiating w.r.t x

Using first principle, this gives:

$S'(x) = 1 - \frac{1}{x^2}$

Equate S'(x) to 0 in order to solve for x

$0 = 1 - \frac{1}{x^2}$

Subtract 1 from both sides

$0 -1 = 1 -1 - \frac{1}{x^2}$

$-1 = - \frac{1}{x^2}$

Multiply both sides by -1

$1 = \frac{1}{x^2}$

Cross Multiply

$x^2 * 1 = 1$

$x^2 = 1$

Take positive square root of both sides because x is positive

$\sqrt{x^2} = \sqrt{1$

$x = 1$

Representing x using interval notation, we have

Interval = $\{x:x=1\}$

To get the smallest sum, we substitute 1 for x in $S(x) = x + \frac{1}{x}$

$S(1) = 1 + \frac{1}{1}$

$S(1) = 1 + 1$

$S(1) = 2$

Hence, the smallest sum is 2

3 0

3 years ago