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3 years ago

The three lengths of four triangles are given. Which triangle is a right triangle?

1 answer:

3 0

A 3-4-5 right triangle is a triangle whose side lengths are in the ratio of 3:4:5. In other words, a 3-4-5 triangle has the ratio of the sides in whole numbers called Pythagorean Triples.

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See figure in triangle PQR,ANGLE PQR=90°,segQS seg PR find x,y and z

e-lub [12.9K]

Answer:

x=4√5

y=12

z=6√5

is your answer look it once

3 0

3 years ago

Find the curl of ~V ~V = sin(x) cos(y) tan(z) i + x^2y^2z^2 j + x^4y^4z^4 k

ch4aika [34]

Given

$\vec v = f(x,y,z)\,\vec\imath+g(x,y,z)\,\vec\jmath+h(x,y,z)\,\vec k \\\\ \vec v = \sin(x)\cos(y)\tan(z)\,\vec\imath + x^2y^2z^2\,\vec\jmath+x^4y^4z^4\,\vec k$

the curl of $\vec v$ is

$\displaystyle \nabla\times\vec v = \left(\frac{\partial h}{\partial y}-\frac{\partial g}{\partial z}\right)\,\vec\imath - \left(\frac{\partial h}{\partial x}-\frac{\partial f}{\partial z}\right)\,\vec\jmath + \left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)\,\vec k$

$\nabla\times\vec v = \left(4x^4y^3z^4-2x^2y^2z\right)\,\vec\imath \\\\ - \left(4x^3y^4z^4-\sin(x)\cos(y)\sec^2(z)\right)\,\vec\jmath \\\\ + \left(2xy^2z^2+\sin(x)\sin(y)\tan(z)\right)\,\vec k$

$\nabla\times\vec v = \left(4x^4y^3z^4-2x^2y^2z\right)\,\vec\imath \\\\ + \left(\sin(x)\cos(y)\sec^2(z)-4x^3y^4z^4\right)\,\vec\jmath \\\\ + \left(2xy^2z^2+\sin(x)\sin(y)\tan(z)\right)\,\vec k$

7 0

3 years ago

What is the best next step in the construction of the perpendicular bisector AB?

qaws [65]

“A” is the answer i believe !

hope this helps

3 0

3 years ago

Help! im awful at these problems!! solve the formula: t=v+k/g for g.

Sedbober [7]

Answer:

g = $\frac{k}{t-v}$

Step-by-step explanation:

Given

t = v + $\frac{k}{g}$

Multiply through by g to clear the fraction

tg = vg + k ( subtract vg from both sides )

tg - vg = k ← factor out g from each term on the left side

g(t - v) = k ← divide both sides by (t - v)

g = $\frac{k}{t-v}$

4 0

3 years ago

Nina purchased apples and strawberries. She purchased a total of 9 pounds of fruit and spent a total of $16.35. Strawberries cos

irina [24]

Answer:

4 pounds of strawberries and 5 pounds of apples are bought.

Step-by-step explanation:

Given:

Total number of pounds of fruit = 9 pounds

Total money spent = $16.35

Cost of 1 pound of strawberry = $1.60

Cost of 1 pound of apple = $1.99

Let 'x' pounds of strawberries and 'y' pounds of apples are bought.

So, as per question:

The sum of the pounds is 9. So,

$x+y=9\\\\y=9-x----1$

Now, total sum of the fruits is equal to the sum of 'x' pounds of strawberries and 'y' pounds of apples. So,

$1.60x+1.99y=16.35----2$

Now, plug in the 'y' value from equation (1) in to equation (2). This gives,

$1.60x+1.99(9-x)=16.35\\\\1.60x+17.91-1.99x=16.35\\\\Combining\ like\ terms, we get:\\\\1.60x-1.99x=16.35-17.91\\\\-0.39x=-1.56\\\\x=\frac{-1.56}{-0.39}=4\ pounds$

Now, from equation 1, we have:

$y=9-4=5\ pounds$

Therefore, 4 pounds of strawberries and 5 pounds of apples are bought.

4 0

3 years ago