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Vadim26 [7]
4 years ago
10

Plz some one look at this

Mathematics
1 answer:
aleksklad [387]4 years ago
4 0
5 : - 1/6 = 0.167     out of 10 = 1.67  out of 100 = 16.7

even number  0.33 out of 10:-  3.3 out of 100:-  33

a prime number  0.5 out of 10: 5 and out of 100:- 50

multipl of 3:- 0/33  out of 10:- 3.3 and out of 100: 33
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Identify an equation in point-slope form for the line perpendicular to y= 3x+ 5.
Vedmedyk [2.9K]

Answer:

y+1=-1/3(x-4)

Step-by-step explanation:

5 0
3 years ago
Please help me solve this
dalvyx [7]

Answer:

each \: ribeye \: costs \:  \boxed{(r) \to \: \$23.3664} ......................... .. \\  each \: grilled \:  salmon \:  dinners \: costs \:  \boxed{(r) \to \:  \$18.29912}

Step-by-step explanation:

let \: ribeye \: be \to \: (r).......................... \\ let \: grilled  \: salmon \:  dinners \: be \to \: (g) \\  \tt{hence}  \to\\ (5) \times  17r + 10g =  \$580.22 \\(10)  \times 21r + 5g =  \$582.19 \\  \to \\ (85r + 50g = 2,901.1) \\  - (minus)\\  (210r + 50g =5,821.9 ) \\  - 125r =  - 2,920.8 \\ r =  \frac{ -2,920.8 }{ - 125}  \\  \boxed{r = \$23.3664} \\  \tt{hence}  \to\\17r + 10g =  \$580.22 \\ 17(23.3664) + 10g =  \$580.22 \\1 0g = \$580.22 - \$397.2288 \\ 10g = \$182.9912 \\ g =  \frac{\$182.9912}{10}  \\  \boxed{g = \$18.29912}

3 0
3 years ago
Plz solve anyone question​
USPshnik [31]
Here you go. Should be solved

3 0
3 years ago
Given f(x)=-1/7√16-x^2 find f^-1(x)
Vera_Pavlovna [14]

Answer:

f^{-1}(x)=\pm \sqrt{49x^2-16}

Not a function.

Step-by-step explanation:

The given function is;

f(x)=-\frac{1}{7}\sqrt{16-x^2}

Let y=-\frac{1}{7}\sqrt{16-x^2}

Interchange x and y;

x=-\frac{1}{7}\sqrt{16-y^2}

Solve for y;

-7x=\sqrt{16-y^2}

Square both sides

(-7x)^2=(\sqrt{16-y^2})^2

49x^2^2=16-y^2

49x^2^2-16=y^2

y=\pm \sqrt{49x^2-16}

The inverse is

f^{-1}(x)=\pm \sqrt{49x^2-16}

f^{-1}(x)=\pm \sqrt{49x^2-16} is not a function because one x-value maps onto to different y-values.

6 0
4 years ago
Solve the system of equation by graphing
Zigmanuir [339]
It is A because I saw your other post
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