Let the numbers be x and y
According to the data : xy=-15 and x+y=-7
xy=-15
x+y=-7

y=-7-x
x(-x-7)=-15

y=-7-x
-x^2-7x=-15

y=-7-x
x^2+7x-15=0

y=-7-x
x=(-7±sqrt(7*7+4*15))/2

y=-7-x
x=(-7±sqrt(109))/2

Goes to two systems.
1)
y=-7-x
x=(-7+sqrt(109))/2


y=-7+3.5-sqrt(109))/2
x=(-7+sqrt(109))/2

y=-3.5-sqrt(109))/2
x=(-7+sqrt(109))/2

2)
y=-7-x
x=(-7-sqrt(109))/2

y=-7+3.5+sqrt(109))/2
x=(-7-sqrt(109))/2

y=-3.5+sqrt(109))/2
x=(-7-sqrt(109))/2

8 0

4 years ago

For each ordered pair determine whether it’s a solution to Y=6x-4

prisoha [69]

Answer:

The answer is in the image

Step-by-step explanation:

8 0

2 years ago

Evaluate triple integral

kaheart [24]

Answer:

$\\ \frac{1}{8} e^{4a}-\frac{3}{4}e^{2a}+e^{a} -\frac{3}{8} \\\\or\\\\ \frac{e^{4a}-6e^{2a}+8e^{a}-3}{8}$

Step-by-step explanation:

$\\ \int\limits^{a}_{0} \int\limits^{x}_{0} \int\limits^{x+y}_{0} {e^{x+y+z}} \, dzdydx \\\\=\int\limits^{a}_{0} \int\limits^{x}_{0} [\int\limits^{x+y}_{0} {e^{x+y}e^z} \, dz]dydx \\\\\\=\int\limits^{a}_{0} \int\limits^{x}_{0} [e^{x+y}\int\limits^{x+y}_{0} {e^z} \, dz]dydx\\\\=\int\limits^{a}_{0} \int\limits^{x}_{0} [e^{x+y}e^z\Big|_0^{x+y}]dydx \\\\\\=\int\limits^{a}_{0} \int\limits^{x}_{0} [e^{x+y}e^{x+y}-e^{x+y}]dydx \\\\\\=\int\limits^{a}_{0} \int\limits^{x}_{0} e^{2x+2y}-e^{x+y}dydx \\\\\\$

$\\=\int\limits^{a}_{0} [\int\limits^{x}_{0} e^{2x}e^{2y}-e^{x+y}dy]dx \\\\\\=\int\limits^{a}_{0} [\int\limits^{x}_{0} e^{2x}e^{2y}dy- \int\limits^{x}_{0}e^{x}e^{y}dy]dx \\\\\\u=2y\\du=2dy\\dy=\frac{1}{2}du\\\\\\=\int\limits^{a}_{0} [\frac{e^{2x}}{2}\int e^{u}du- e^x\int\limits^{x}_{0}e^{y}dy]dx \\\\\\=\int\limits^{a}_{0} [\frac{e^{2x}}{2}\cdot e^{2y}\Big|_0^x- e^xe^{y}\Big|_0^x]dx \\\\\\=\int\limits^{a}_{0} [\frac{e^{2x+2y}}{2} - e^{x+y}\Big|_0^x]dx \\\\$

$\\=\int\limits^{a}_{0} [\frac{e^{4x}}{2} - e^{2x}-\frac{e^{2x}}{2} + e^{x}]dx \\\\\\=\int\limits^{a}_{0} \frac{e^{4x}}{2} -\frac{3e^{2x}}{2} + e^{x}dx \\\\\\=\int\limits^{a}_{0} \frac{e^{4x}}{2}dx -\int\limits^{a}_{0}\frac{3e^{2x}}{2}dx + \int\limits^{a}_{0}e^{x}dx \\\\\\u_1=4x\\du_1=4dx\\dx=\frac{1}{4}du_1\\\\\u_2=2x\\du_2=2dx\\dx=\frac{1}{2}du_2\\\\\\=\frac{1}{8}\int e^{u_1}du_1 -\frac{3}{4}\int e^{u_2}du_2 + \int\limits^{a}_{0}e^{x}dx \\\\\\$

$\\=\frac{1}{8}e^{u_1}\Big| -\frac{3}{4}e^{u_2}\Big| + e^{x}\Big|_0^a \\\\\\=\frac{1}{8}e^{4x}\Big|_{0}^a -\frac{3}{4}e^{2x}\Big|_{0}^a + e^{x}\Big|_0^a \\\\\\=\frac{1}{8}e^{4x} -\frac{3}{4}e^{2x} + e^{x}\Big|_0^a \\\\\\=\frac{1}{8}e^{4a} -\frac{3}{4}e^{2a} + e^{a}-\frac{1}{8} +\frac{3}{4} -1\\\\\\=\frac{1}{8}e^{4a} -\frac{3}{4}e^{2a} + e^{a}-\frac{3}{8}\\\\\\$

Sorry if that took a while to finish. I am in AP Calculus BC and that was my first time evaluating a triple integral. You will see some integrals and evaluation signs with blank upper and lower boundaries. I just had my equation in terms of u and didn't want to get any variables confused. Hope this helps you. If you have any questions let me know. Have a nice night.

6 0

3 years ago

Identify a ray in the following figure. WILL MAKE BRAINLIEST

musickatia [10]

Answer:

The answer is A

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers