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2 years ago

Find the diameter of circle N if PQ = 14 and PR is congruent to RQ. NR is 5. Round the answer to the tenths place.

Mathematics

2 answers:

Arlecino [84]2 years ago

8 0

Radius = sqrt(7^2 - 5^2) = sqrt(49 - 25) = sqrt(24) = 4.90
diameter = 2 * radius = 2(4.90) = 9.8

ICE Princess25 [194]2 years ago

5 0

I hope this helps you

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Given PQR is congruent to STU, find the value of x

lukranit [14]

Answer:X=15 Y=4

Step-by-step explanation:

1. PR=US or 3y=12 divide 3y by 3 and cancel 3 out then divide 12 and get Y=4.

2. QR=TU or 2x=30, divide 2x by 2 and cancel out 2, then divide 30 by 2 and get X=15.

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2 years ago

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3 years ago

4 - (-8) = What is the answer

lianna [129]

Answer:

Step-by-step explanation:

When you subtract a negative number, the two negatives make a plus sign.

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3 years ago

Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software,

vichka [17]

Answer:

A(-1,0) is a local maximum point.

B(-1,0) is a saddle point

C(3,0) is a saddle point

D(3,2) is a local minimum point.

Step-by-step explanation:

The given function is

$f(x,y)=x^3+y^3-3x^2-3y^2-9x$

The first partial derivative with respect to x is

$f_x=3x^2-6x-9$

The first partial derivative with respect to y is

$f_y=3y^2-6y$

We now set each equation to zero to obtain the system of equations;

$3x^2-6x-9=0$

$3y^2-6y=0$

Solving the two equations simultaneously, gives;

$x=-1,x=3$ and $y=0,y=2$

The critical points are

A(-1,0), B(-1,2),C(3,0),and D(3,2).

Now, we need to calculate the discriminant,

$D=f_{xx}(x,y)f_{yy}(x,y)-(f_{xy}(x,y))^2$

But, we would have to calculate the second partial derivatives first.

$f_{xx}=6x-6$

$f_{yy}=6y-6$

$f_{xy}=0$

$\Rightarrow D=(6x-6)(6y-6)-0^2$

$\Rightarrow D=(6x-6)(6y-6)$

At A(-1,0),

$D=(6(-1)-6)(6(0)-6)=72\:>\:0$ and $f_{xx}=6(-1)-6=-18\:$

Hence A(-1,0) is a local maximum point.

See graph

At B(-1,2);

$D=(6(-1)-6)(6(2)-6)=-72\:$

Hence, B(-1,0) is neither a local maximum or a local minimum point.

This is a saddle point.

At C(3,0)

$D=(6(3)-6)(6(0)-6)=-72\:$

Hence, C(3,0) is neither a local minimum or maximum point. It is a saddle point.

At D(3,2),

$D=(6(3)-6)(6(2)-6)=72\:>\:0$ and $f_{xx}=6(3)-6=12\:>\:0$

Hence D(3,2) is a local minimum point.

See graph in attachment.

3 0

3 years ago

Find the solution set for the following equation. <img src="https://tex.z-dn.net/?f=-%5Csqrt%28x%29%3D5" id="TexFormula1" title=

andrey2020 [161]

Answer:

Solution set = {25}

Step-by-step explanation:

=> $-\sqrt{x} =5$

Dividing both sides by -1

=> $\sqrt{x} = -5$

Taking square on both sides

=> x = 25

Solution set = {25}

3 0

2 years ago