All categories

3 years ago

Solve the quadratic equation y^2-5y=-4 by factoring.

Mathematics

1 answer:

yan [13]3 years ago

8 0

$y^2-5y=-4\\
y^2-5y+4=0\\
y^2-y-4y+4=0\\
y(y-1)+4(y-1)=0\\
(y+4)(y-1)=0\\
y=-4 \vee y=1
$

You might be interested in

Which is the best estimate for the expression?<br> 49% of 15

irakobra [83]

49%=0.49

0.49x15=7.35

49% of 15 is 7.35

3 0

3 years ago

Read 2 more answers

the sum of two polynomials is –yz2 – 3z2 – 4y 4. if one of the polynomials is y – 4yz2 – 3, what is the other polynomial?

nikklg [1K]

Is the last digit plus 4 or minus 4

4 0

3 years ago

What's the least common denominator of 3/4 4/5 and 2/3

jok3333 [9.3K]

LCD(34, 45, 23) = LCM(4, 5, 3) = 22×3×5 = 60

34 = 4560

45 = 4860

23 = 4060

7 0

3 years ago

Read 2 more answers

Let S be the solid beneath z = 12xy^2 and above z = 0, over the rectangle [0, 1] × [0, 1]. Find the value of m > 1 so that th

jonny [76]

Answer:

The answer is $\sqrt{\frac{6}{5}}$

Step-by-step explanation:

To calculate the volumen of the solid we solve the next double integral:

$\int\limits^1_0\int\limits^1_0 {12xy^{2} } \, dxdy$

Solving:

$\int\limits^1_0 {12x} \, dx \int\limits^1_0 {y^{2} } \, dy$

$[6x^{2} ]{{1} \atop {0}} \right. * [\frac{y^{3}}{3}]{{1} \atop {0}} \right.$

Replacing the limits:

$6*\frac{1}{3} =2$

The plane y=mx divides this volume in two equal parts. So volume of one part is 1.

Since m > 1, hence mx ≤ y ≤ 1, 0 ≤ x ≤ $\frac{1}{m}$

Solving the double integral with these new limits we have:

$\int\limits^\frac{1}{m} _0\int\limits^{1}_{mx} {12xy^{2} } \, dxdy$

This part is a little bit tricky so let's solve the integral first for dy:

$\int\limits^\frac{1}{m}_0 [{12x \frac{y^{3}}{3}}]{{1} \atop {mx}} \right.\, dx =\int\limits^\frac{1}{m}_0 [{4x y^{3 }]{{1} \atop {mx}} \right.\, dx$

Replacing the limits:

$\int\limits^\frac{1}{m}_0 {4x(1-(mx)^{3} )\, dx =\int\limits^\frac{1}{m}_0 {4x-4x(m^{3} x^{3} )\, dx =\int\limits^\frac{1}{m}_0 ({4x-4m^{3} x^{4}) \, dx$

Solving now for dx:

$[{\frac{4x^{2}}{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right. = [{2x^{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right.$

Replacing the limits:

$\frac{2}{m^{2} }-\frac{4m^{3}\frac{1}{m^{5}}}{5} =\frac{2}{m^{2} }-\frac{4\frac{1}{m^{2}}}{5} \\ \frac{2}{m^{2} }-\frac{4}{5m^{2} }=\frac{10m^{2}-4m^{2} }{5m^{4}} \\ \frac{6m^{2} }{5m^{4}} =\frac{6}{5m^{2}}$