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Leni [432]
3 years ago
14

Which of the following is equal to the expression below?

Mathematics
1 answer:
erik [133]3 years ago
8 0
(4^-5)^3 =
4^(-5*3) =
4^-15 =
1/(4^15).....answer C
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7/12 × 6\14 please help
Mama L [17]
7/12*6/14
  7*6=42
  12*14=168
=42/168
=0.25 =1/4
6 0
3 years ago
A dog owner is told by the veterinarian that his dog, Max, is overweight and needs to lose 4 to 5 pounds in order to live a heal
hodyreva [135]
This is a linear function, of the kind y = mx + b

where x is the number of visit,  b is the weight when x = 0, and m is the predicted change of weight for every visit.

m = - 4 ounces / visit, which must be converted to pounds (the negative sign indicates that the change is a decrease)

1 lb = 16 ounces = 4 ounces = 0.25 lb

Then m = - 0.25 lb / visit.

Now, for x = 1, y = 126 => 126 = - 0.25(1) + b => b = 126 + 0.25 = 126.25

Then the function is y = 126.25 - 0.25x

Now round to the nearest tenth:

y = 126.3 - 0.3x

Answer: y = 126.3 - 0.3x

 

6 0
3 years ago
Aliyah bought a car for $32,000. It is now worth $5000 less than half of what she purchased it for. To the nearest tenth of a pe
N76 [4]

Answer:

65.6% is the answer.........

4 0
3 years ago
Sinx = 1/2, cosy = sqrt2/2, and angle x and angle y are both in the first quadrant.
Leviafan [203]

Answer:

Option D. 3.73​

Step-by-step explanation:

we know that

tan(x+y)=\frac{tan(x)+tan(y)}{1-tan(x)tan(y)}

and

sin^{2}(\alpha)+cos^{2}(\alpha)=1

step 1

Find cos(X)

we have

sin(x)=\frac{1}{2}

we know that

sin^{2}(x)+cos^{2}(x)=1

substitute

(\frac{1}{2})^{2}+cos^{2}(x)=1

cos^{2}(x)=1-\frac{1}{4}

cos^{2}(x)=\frac{3}{4}

cos(x)=\frac{\sqrt{3}}{2}

step 2

Find tan(x)

tan(x)=sin(x)/cos(x)

substitute

tan(x)=1/\sqrt{3}

step 3

Find sin(y)

we have

cos(y)=\frac{\sqrt{2}}{2}

we know that

sin^{2}(y)+cos^{2}(y)=1

substitute

sin^{2}(y)+(\frac{\sqrt{2}}{2})^{2}=1

sin^{2}(y)=1-\frac{2}{4}

sin^{2}(y)=\frac{2}{4}

sin(y)=\frac{\sqrt{2}}{2}

step 4

Find tan(y)

tan(y)=sin(y)/cos(y)

substitute

tan(y)=1

step 5      

Find tan(x+y)

tan(x+y)=\frac{tan(x)+tan(y)}{1-tan(x)tan(y)}

substitute

tan(x+y)=[1/\sqrt{3}+1}]/[{1-1/\sqrt{3}}]=3.73

7 0
3 years ago
Question 11
mamaluj [8]
The first one is the correct answer

Y-7=-3(x+2)
6 0
3 years ago
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