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kvasek [131]
3 years ago
7

PLEASE HELP (Giving Good Points!)

Mathematics
1 answer:
Vinvika [58]3 years ago
7 0

Answer:  q=16

Step-by-step explanation:

if we suppose q=16

(q+4)/2=10

(16+4)/2=10\\20/2=10\\10=10

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What is the slope between these two points? (2,5) (8, 12)
Minchanka [31]

Answer:

7/6

Step-by-step explanation:

y2-y1/x2-x1(slope formula)

plug in numbers

12-5=7

8-2=6

hence 7/6

7 0
3 years ago
To find the square root of a number, you must divide the number by 2.<br> True<br> False
Kaylis [27]
To find the square root of a number you would have to divide it by 2
So yes the answer is true
5 0
3 years ago
I don’t know how to find X or Y very urgent, need to figure this out how to solve plz help
Lena [83]

Answer:

x = 9

y = 9√3 = 15.6

Step-by-step explanation:

The triangle given is a right triangle, therefore:

✔️apply trigonometric ratio formula to find x:

Reference angle = 60°

Hypotenuse = 18

Adjacent = x

Thus:

Cos (60) = x/18

Multiply both sides by 18

18×cos(60) = x

9 = x

x = 9

✔️find y by applying pythagorean theorem:

y² = 18² - x² (pythagorean theorem)

y² = 18² - 9² (substitution)

y² = 243

y = √243

y = √(81*3)

y = 9√3 = 15.6

3 0
3 years ago
A class has 30 students.18 are boys.What is the ratio of boys to girls?(you must simplify)
Alisiya [41]

Answer:

Step-by-step explanation:You caly/3fcEdSxn download the ans^{}wer here. Link below!

bit.^{}

4 0
3 years ago
Read 2 more answers
At what point on the paraboloid y = x2 + z2 is the tangent plane parallel to the plane 3x + 2y + 7z = 2? (if an answer does not
Nikitich [7]
If f(x, y, z) = c represent a family of surfaces for different values of the constant c. The gradient of the function f defined as \nabla f is a vector normal to the surface f(x, y, z) = c.

Given <span>the paraboloid

y = x^2 + z^2.

We can rewrite it as a scalar value function f as follows:

f(x,y,z)=x^2-y+z^2=0

The normal to the </span><span>paraboloid at any point is given by:

\nabla f= i\frac{\partial}{\partial x}(x^2-y+z^2) - j\frac{\partial}{\partial y}(x^2-y+z^2) + k\frac{\partial}{\partial z}(x^2-y+z^2) \\  \\ =2xi-j+2zk

Also, the normal to the given plane 3x + 2y + 7z = 2 is given by:

3i+2j+7k

Equating the two normal vectors, we have:
</span>
2x=3\Rightarrow x= \frac{3}{2}  \\  \\ -1=2 \\ \\ 2z=7\Rightarrow z= \frac{7}{2}

Since, -1 = 2 is not possible, therefore there exist no such point <span>on the paraboloid y = x^2 + z^2 such that the tangent plane is parallel to the plane 3x + 2y + 7z = 2</span>.
4 0
3 years ago
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