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3 years ago

Bonita said that the product of 5/6* 1 2/3 is 7/3 How can u tell that her answer is wrong

Mathematics

2 answers:

jarptica [38.1K]3 years ago

8 0

Answer:

Step-by-step explanation:

what was the answer?

11111nata11111 [884]3 years ago

3 0

Because if u plug it in to a calculator, hwr andwet is wrong so she’s stupif

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jonny [76]

He switched some of the numbers around

he can fix it by switching them back to 5,650,000

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3 years ago

Simplify.<br><br> −5x4(−3x2+4x−2)<br> *both x's are squared/raised to the power of

docker41 [41]

-5x^4 (-3x^2 + 4x - 2) =

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3 years ago

Graph a line with a y-intercept of -3 and a slope of 4/3

ollegr [7]

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Step-by-step explanation:

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3 years ago

A cyclist rode for 3.5 hours and completed a distance of 60.9 miles. If she kept the same average speed for each hour, how far d

WINSTONCH [101]

Answer:

She rode 17.4 miles in 1 hour

Step-by-step explanation:

A cyclist rode for 3.5 hours and completed a distance of 60.9 miles. If she kept the same average speed for each hour, how far did she ride in 1 hour?

Part A Estimate the quotient. Include an equation to show your work. Explain your thinking.?

The quotient is the average speed and this is calculated as:

Speed = Distance(miles)/Time(hours)

Where speed = quotient

Distance = Dividend

Time = Divisor

Speed = 60.9 miles/3.5 hours

= 17.4 miles/hour

Part B Find the exact distance she rode in 1 hour by completing the division. PLEASE HELP!

If :

3.5 hours = 60.9 miles

1 hour = x miles

Cross Multiply =

3.5x = 60.9 miles

x = 60.9/3.5

x = 17.4 miles

Hence, she rode 17.4 miles in 1 hour

6 0

3 years ago

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

$\displaystyle\int\frac{x^2+x-3}{(x^3+x^2-4x-4)^2}\,\mathrm dx$
$=-\dfrac1{48(x-2)}+\dfrac1{3(x+1)}+\dfrac1{16(x+2)}+\dfrac1{96}\ln|x-2|+\dfrac13\ln|x+1|-\dfrac{11}{32}\ln|x+2|+C$

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4 years ago