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2 years ago

Identify the solution(s) of v2+11v−26=0.

Mathematics

1 answer:

miss Akunina [59]2 years ago

4 0

Answer:

i solved for v and got

v = 2, -13

i hope this helps :]

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What is the percentage equivalent of 1.48?

Vinvika [58]

Answer:

A. 148%

To find percent you just have to multiply your decimal by one hundred, which in this case moves the decimal to the very back and makes 1.48 = 148%

5 0

2 years ago

A human resource manager is researching the entry-level salary for a technician position. She determined that other companies in

Sever21 [200]

$15.6375, but if rounded up, $15.64, and $15.63 if rounded down

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2 years ago

Correct answer only need urgent help

Andreas93 [3]

Answer:

Step-by-step explanation:

Use the formule V=4/3 *Pi*R^3

where R =10

so V=4/3*Pi*10^3

v= 4000Pi/3 =4188.7902

8 0

1 year ago

A quantity P is an exponential function of time I, such that P = 160 when t = 6 and P = 150 when I = 4. Use the given informatio

Klio2033 [76]

Answer:

k = 0.032
P0 = 131.836

Step-by-step explanation:

Perhaps you want to use the points (t, P) = (4, 150) and (6, 160) to find the parameters P0 and k in the equation ...

$P(t)=P_0\cdot e^{kt}$

We know from the given points that we can write the equation as ...

$P(t)=150\left(\dfrac{160}{150}\right)^{(t-4)/(6-4)}=150\left(\dfrac{16}{15}\right)^{\frac{t}{2}-2}\\\\=150\left(\dfrac{16}{15}\right)^{-2}\times\left(\left(\dfrac{16}{15}\right)^{\frac{1}{2}}\right)^t$

Comparing this to the desired form, we see that ...

$P_0=150\left(\dfrac{16}{15}\right)^{-2}\approx 131.836\\\\e^{k}=\left(\dfrac{16}{15}\right)^{1/2}\rightarrow k=\dfrac{1}{2}(\ln{16}-\ln{15})\approx 0.0322693$

So, the approximate equation for P is ...

$P(t)=131.836\cdote^{0.032t}$

And the parameters of interest are ...

k = 0.032
P0 = 131.836

4 0

2 years ago

Tan 48 degrees 19'23"

asambeis [7]

Answer:

Step-by-step explanation:

y = tan(48 + 19/60 + 23/3600)

y = tan(48.323055555...)

y = 1.1232851096924531815531574029372...

3 0

2 years ago