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3 years ago

Solve for v. -8 = v + 4

Mathematics

2 answers:

Mkey [24]3 years ago

7 0

Answer:

v=-12

Step-by-step explanation:

-8=v+4

-4 -4

-12=v

vampirchik [111]3 years ago

7 0

Answer:

v = -12

Step-by-step explanation:

We have the equation v + 4 = -8

Subtract 4 from both sides.

v + 4 - 4 = -8 - 4

+4 and -4 cancel one another out.

We get v = -12.

I would appreciate brainliest but if not that's ok!

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18. MP Reason Inductively The product of

timama [110]

Answer:

- 2 and 12

Step-by-step explanation:

let the 2 integers be x and y , x > y , then

x - y = 14 → (1)

x + y = 10 → (2)

add the 2 equations term by term to eliminate y

2x = 24 ( divide both sides by 2 )

x = 12

Substitute x = 12 into (2)

12 + y = 10 ( subtract 12 from both sides )

y = - 2

As a check

x + y = 12 + (- 2) = 12 - 2 = 10

x - y = 12 - (- 2) = 12 + 2 = 14

xy = 12 × - 2 = - 24

6 0

2 years ago

How do i check if 12=x-4 is 16

My name is Ann [436]

Answer:

x=16

Step-by-step explanation:

12 = x - 4

+4 +4

By adding 4 to both sides it cancels out the -4 leaving the x alone and makes x=16

8 0

3 years ago

Read 2 more answers

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

PLEASE HELP!!!!!!!!!!!!!!!!!!!!!!!!

Kamila [148]

Answer:

ok imma be honest i dont have a clue jk

Step-by-step explanation:

use a calculator

7 0

3 years ago

Solve for x. x/8+x/10=9/8 Simplify your answer as much as possible.

BabaBlast [244]

Answer:

I hope it's helps you

have a nice day

6 0

3 years ago