1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
postnew [5]
3 years ago
9

What is 2/3 x 9/8 as a fraction in simplest form

Mathematics
1 answer:
tresset_1 [31]3 years ago
7 0

Answer: \frac{3}{4}

Have attached the picture for explanation...

You might be interested in
Whats the intrgral of <img src="https://tex.z-dn.net/?f=%20%5Cint%20%20%5Cfrac%7Bx%5E2%2Bx-3%7D%7B%28x%5E3%2Bx%5E2-4x-4%29%5E2%7
rosijanka [135]
\displaystyle\int\frac{x^2+x-3}{(x^3+x^2-4x-4)^2}\,\mathrm dx

Notice that x^3+x^2-4x-4=x^2(x+1)-4(x+1)=(x-2)(x+2)(x+1). Decompose the integrand into partial fractions:

\dfrac{x^2+x-3}{(x-2)^2(x+2)^2(x+1)^2}
=\dfrac1{3(x+1)}-\dfrac{11}{32(x+2)}-\dfrac1{3(x+1)^2}-\dfrac1{16(x+2)^2}+\dfrac1{96(x-2)}+\dfrac1{48(x-2)^2}

Integrating term-by-term, you get

\displaystyle\int\frac{x^2+x-3}{(x^3+x^2-4x-4)^2}\,\mathrm dx
=-\dfrac1{48(x-2)}+\dfrac1{3(x+1)}+\dfrac1{16(x+2)}+\dfrac1{96}\ln|x-2|+\dfrac13\ln|x+1|-\dfrac{11}{32}\ln|x+2|+C
5 0
3 years ago
PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!
Nikitich [7]

Answer:   B. (-4, 0)

<u>Step-by-step explanation:</u>

The question is: when is the <u>function</u> constant?

function is also called f(x), which is also called "y".

So the question is asking: when is the y-value constant?

the y-value is constant (not increasing or decreasing) between x = -4 and x = 0. So the interval is (-4, 0)

7 0
3 years ago
Find the value of this expression if x = -2.<br>x^2 – 8/x + 6​
xz_007 [3.2K]

Answer:

-1

Step-by-step explanation:

x^2 – 8

-----------

x + 6​

Let x = -2

(-2)^2 – 8

-----------

-2 + 6​

Exponents first

4 – 8

-----------

-2 + 6​

Then complete the numerator and denominator

-4

-----

4

Then divide

-1

5 0
3 years ago
Read 2 more answers
Pls help if you actually can
Setler79 [48]

The triangles that are similar would be ΔGCB and ΔPEB  due to Angle, Angle, Angle similarity theorem.

<h3>How to identify similar triangles?</h3>

From the image attached, we see that we are given the Parallelogram GRPC. Thus;

A.  The triangles that are similar would be ΔGCB and ΔPEB  due to Angle, Angle, Angle similarity theorem.

B. The proof of the fact that ΔGCB and ΔPEB are similar pairs of triangles is as follow;

∠CGB ≅ ∠PEB  (Alternate Interior Angles)

∠BPE ≅ ∠BCG  (Alternate Interior Angles)

∠GBC ≅ ∠EBP  (Vertical Angles)

C.  To find the distance from B to E and from P to E, we will first find PE and then BE by proportion;

225/325 = PE/375  

PE = 260 ft

BE/425 = 225/325  

BE = 294 ft

Read more about Similar Triangles at; brainly.com/question/14285697

#SPJ1

5 0
2 years ago
Answer very Quick i need it like right away
Julli [10]

Answer:

c + (a + 1.70) = 6.70

Step-by-step explanation:

you start the equation by Subtracting how much more alan has to the total

(this will only be set as an equation do not actually solve it)

then add 1.70 to a

"

6.70 - 1.70 = a + c

(1.70 + a)

c + (a + 1.70)  = 6.70 "

8 0
3 years ago
Other questions:
  • Using the greatest common factor for the terms, how can you write 80 + 32 as a product?
    9·1 answer
  • Please help I need answer to this problem.
    10·2 answers
  • Find the midpoint between (4,-1) and (3,2)
    7·1 answer
  • Solve this equation for A: A ÷ 2 = 4
    10·2 answers
  • Can anyone answer this question for me ?
    8·2 answers
  • I need help asap ple​
    9·1 answer
  • Gabriella spent half of her weekly allowance buying pizza. To earn more money her parents let her clean the windows in the house
    11·1 answer
  • 10= 4+2s pls give me the answer tyy
    5·2 answers
  • EASY QUESTION! MARK BRAINLIEST!<br><br> if i pay $3.50 a month how much do i pay a year? :)))
    5·1 answer
  • 25 is 55% of what number
    9·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!