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2 years ago

How to solve this (complex fraction)

Mathematics

2 answers:

Leviafan [203]2 years ago

7 0

1/4+5/4=1.5
1.5 divided by 4 = 0.5625

skad [1K]2 years ago

4 0

$\frac{\frac{1}{4}+\frac{5}{4}}{4}=\frac{\frac{6}{4}}{4}=\frac{\not6^3}{4}*\frac{1}{\not4^2}=\frac{3}{8}=0.375$

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What is the value of x in theequation 2.5 (6x - 4) = 10 + 4 (1.5 +0.5 x)?

Zarrin [17]

Answer:

x = 2

Step-by-step explanation:

Let's solve your equation step-by-step.

2.5(6x−4)=10+4(1.5+0.5x)

Step 1: Simplify both sides of the equation.

2.5(6x−4)=10+4(1.5+0.5x)

(2.5)(6x)+(2.5)(−4)=10+(4)(1.5)+(4)(0.5x)(Distribute)

15x+−10=10+6+2x

15x−10=(2x)+(10+6)(Combine Like Terms)

15x−10=2x+16

Step 2: Subtract 2x from both sides.

15x−10−2x=2x+16−2x

13x−10=16

Step 3: Add 10 to both sides.

13x−10+10=16+10

13x=26

Step 4: Divide both sides by 13.

x=2

5 0

2 years ago

PLEASE HELP WITH ALGEBRA QUESTION NEED ASAP!!

Delvig [45]

Answer:

The function f(x) = ln(x - 4) is graphed the question options

Step-by-step explanation:

* Lets study the the information of the problem

- The graph of a logarithmic has a vertical asymptote at x=4

* That means the curve gets closer and closer to the vertical line x = 4

but does not cross it

- It contains the point (e+4, 1)

* That means if we substitute x = e + 4 in the equation the value

of y will be equal to 1

- It has an x-intercept of 5

* That means if we substitute y = 0 in the equation the value of x

will be equal to 5

* Lets find the right answer

∵ f(x) = ln(x - 4)

* To find the equation of the asymptote let x - 4 = 0

∵ x - 4 = 0

∴ x = 4

∴ f(x) has a vertical asymptote at x = 4

* Lets check the point (e + 4 , 1) lies on the graph of the f(x)

∵ x = e + 4

∴ f(e+4) = ln(e + 4 - 4) = ln(e)

∵ ln(e) = 1

∴ The point (e+4 , 1) lies on the graph of the function f(x)

* To find the x-intercept put y = 0

∵ f(x) = 0

∴ ln(x - 4) = 0

* Change the logarithmic function to the exponential function

- The base of the ln is e

∴ e^0 = x - 4

∵ e^0 = 1

∴ x - 4 = 1 ⇒ add 4 to the both sides

∴ x = 5

* The function f(x) = ln(x - 4) is graphed the question options

3 0

2 years ago

Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y

irina [24]

Answer:

Therefore the value of y(1)= 0.9152.

Step-by-step explanation:

According to the Euler's method

y(x+h)≈ y(x) + hy'(x) ....(1)

Given that y(0) =3 and step size (h) = 0.2.

$y'(x)= x^2y(x)-\frac12y^2(x)$

Putting the value of y'(x) in equation (1)

$y(x+h)\approx y(x) +h(x^2y(x)-\frac12y^2(x))$

Substituting x =0 and h= 0.2

$y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]$

$\Rightarrow y(0.2)\approx 3+0.2[-\frac12 \times3]$ [∵ y(0) =3 ]

$\Rightarrow y(0.2)\approx 2.7$

Substituting x =0.2 and h= 0.2

$y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2]$

$\Rightarrow y(0.4)\approx 2.7+0.2[(0.2)^2\times 2.7- \frac12(2.7)^2]$

$\Rightarrow y(0.4)\approx 1.9926$

Substituting x =0.4 and h= 0.2

$y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2]$

$\Rightarrow y(0.6)\approx 1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2]$

$\Rightarrow y(0.6)\approx 1.6593$

Substituting x =0.6 and h= 0.2

$y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]$

$\Rightarrow y(0.8)\approx 1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]$

$\Rightarrow y(0.6)\approx 0.8800$

Substituting x =0.8 and h= 0.2

$y(0.8+0.2)\approx y(0.8)+0.2[(0.8)^2\times y(0.8)-\frac12 (y(0.8))^2]$

$\Rightarrow y(1.0)\approx 0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]$

$\Rightarrow y(1.0)\approx 0.9152$

Therefore the value of y(1)= 0.9152.

4 0

3 years ago

Does anyone know what the √2.25 is when rewritten as a fraction?

Zinaida [17]

Answer: $\sqrt{\frac{9}{4} }$

Step-by-step explanation:

This is what it is as a mixed number

$\sqrt{2.25}$

= $\sqrt{2\frac{1}{4} }$

= $\sqrt{\frac{9}{4} }$

If you want it written in exponential form it would look like this:

$(\frac{9}{4}) ^{\frac{1}{2} }$

5 0

1 year ago

Read 2 more answers

In the science lab, Will is testing the freezing point of a substance, which should be −24 ∘C. He is changing the temperature at

mars1129 [50]

Answer:

The answer is -3t>-24 which equals t<8

Step-by-step explanation:

4 0

3 years ago