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3 years ago

6

Use a calculator to evaluate sec 85 degrees

1 answer:

faltersainse [42]3 years ago

3 0

OK. I used my calculator to evaluate sec(85 degrees).

My calculator doesn't have a "sec" button on it.
But I remembered that

sec of an angle = 1 / (cosine of the same angle) .

So I used my calculator to find cos(85), and then I hit the
" 1/x " key, and got 11.474, which I knew to be sec(85).

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Question #7multiplechoice score: in a certain candy store, 3 pounds of candy and 2 pounds of mints cost $10.80, and 1 pound of c

irga5000 [103]

Let the cost of one pound of candy be x and one pound of mint be y.

3x+2y=10.8 a
x+3y = 5.35 b
B*3 = 3x+9y = 16.05 b

b-a = 7y = 16.05-10.8= 5.25
y = 5.25/7 = 0.75

The cost per pound of mints is $0.75

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3 years ago

Which shape is similar to the original shape?<br> A) A <br> B) B <br> C) C <br> D) D

Anna35 [415]

Answer:

I think the answer is A.

Step-by-step explanation:

Original shape: 7 x 5 x 225 x 6 x 12 x 45 = 25,515,000

A: 9 x 5 x 225 x 6 x 35 x 12 = 25,515,000

B: 14 x 10 x 255 x 12 x 24 x 45 = 462,672,000

C: 9 x 7 x 225 x 8 x 15 x 45 = 76,545,000

D: 8 x 4 x 225 x 6 x 12 x 35 = 18,144,000

This is probably not the way you would solve it but this is how I did it. Basically I multiplied all the numbers in each shape together.

<em>(also i'm not quite sure if this is right >.< sorry!)</em>

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3 years ago

John borrowed a certain amount of money from Tebogo at a simple interest rate of 8,7% per year.

Anestetic [448]

Answer:

Step-by-step explanation:

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3 years ago

Convert and round to the proper Precision or accuracy. 21.4 grams= mg? * O .0214 mg О 214 mg О 214.0 mg O 21,400 mg 00214 mg

KengaRu [80]

Answer: 21,400 mg

Step-by-step explanation: Multiply the mass value by 1000 .

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3 years ago

This extreme value problem has a solution with both a maximum value and a minimum value. use lagrange multipliers to find the ex

noname [10]

$L(x,y,z,\lambda)=10x+10y+2z+\lambda(5x^2+5y^2+2z^2-42)$

$L_x=10+10\lambda x=0\implies1+\lambda x=0$

$L_y=10+10\lambda y=0\implies1+\lambda y=0$

$L_z=2+4\lambda z=0\implies1+2\lambda z=0$

$L_\lambda=5x^2+5y^2+2z^2-42=0$

$\begin{cases}L_x=0\\L_y=0\\L_z=0\end{cases}\implies1+\lambda x=1+\lambda y=1+2\lambda z=0\implies x=y=2z$

$5x^2+5y^2+2z^2=5(2z)^2+5(2z)^2+2z^2=42z^2=42\implies z^2=1$

$z^2=1\implies z=\pm1\implies x=y=\pm2$

There are two critical points, at which we have

$f\left(2,2,1\right)=42\text{ (a maximum value)}$

$f\left(-2,-2,-1\right)=-42\text{ (a minimum value)}$

3 0

3 years ago