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Maru [420]
3 years ago
6

Convert -125 degrees to radians

Mathematics
1 answer:
miv72 [106K]3 years ago
7 0

Answer:

-125° × π / 180° = -2.181661565 rad

Step-by-step explanation:


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How do I decide if it’s a real solutions or extraneous solutions?
olga55 [171]

Step-by-step explanation:

So you can check for extraneous solutions by plugging in the 5 and -2 as x.

Plug 5 in as x

5=\sqrt{3(5)+10}\\5=\sqrt{15+10}\\5=\sqrt{25}\\5=5

I'm not sure if this is exactly considered an extraneous solution, but I'm assuming it is, since technically the radical doesn't have a plus/minus in front of it, so it's the principle square root which means it only outputs the positive values. This would mean that the -2 wouldn't be a solution, since the principle square root doesn't output negative values, so it's a function, since if it outputted 2 values, then it wouldn't be a function by definition.

3 0
2 years ago
A privately owned lake contains two types of game fish, bass and trout. The owner provides two types of food, A and B, for these
Vinvika [58]

Answer:

133 fishes

Step-by-step explanation:

Units of food A = 400 units

Units of food B = 400 units

Fish Bass required 2 units of A and 4 units of B.

Fish Trout requires 5 units of A and 2 units of B.

i. For food A,

total units of food A required = 2 + 5

                                                 = 7 units

number of bass and trout that would consume food A = 2 x \frac{400}{7}

                                               = 114.3

number of bass and trout that would consume food A = 114

ii. For food B,

total units of food B required = 4 + 2

                                                = 6 units

number of bass and trout that would consume food B = 2 x \frac{400}{6}

                                                 = 133.3

number of bass and trout that would consume food B = 133

Thus, the maximum number of fish that the lake can support is 133.

4 0
3 years ago
Find the area of the figures by counting the squares
statuscvo [17]

Answer:

Six

six

nine

mark me as BRAINLIEST

6 0
3 years ago
I Need help plzzzzzz
Marianna [84]

Answer:

With what?

Step-by-step explanation:

5 0
4 years ago
Read 2 more answers
Let h(x)=e−x+kx, where k is any constant. For what value(s) of k does h have (a) No critical points? (b) One critical point? (c)
mrs_skeptik [129]

Answer:

a) for k≤0 , h has no critical point

b) for k>0 , h has a critical point

c) for k=0 , has a horizontal asymptote

Step-by-step explanation:

for the function

h(x)=e^(−x)+k

h has a critical point when the first derivative is =0 or is undefined. Since e^(−x) and k*x are continuos functions for all x then the second case is discarded. Then

dh/dx = -e^(−x)+k = 0

k = e^(−x)

x = ln (1/k)

since ln (1/k) should be possitive then k should be >0 . Thus h(x) has a critical point when k>0 and do not have any when  k≤0

h has a horizontal asymptote when

lim h(x)=a when x→∞ (or -∞)

then

when x→∞, lim h(x)= lim e^(−x)+k*x = lim e^(−x) + k* lim x = 0 + k*∞ = ∞

on the other hand , when k=0 , lim h(x)= lim e^(−x)= 0 , then h has a horizontal  asymptote for k=0

for x→(-∞) , e^(-x) rises exponentially , thus there is no k such that h has an horizontal asymptote when x→(-∞)

6 0
4 years ago
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