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3 years ago

Find the hypotenuse of a right triangle given the sides 8cm and 6cm

Mathematics

2 answers:

marusya05 [52]3 years ago

4 0

To solve this you must use Pythagorean theorem:

$a^{2} +b^{2} =c^{2}$

a and b are the legs (the sides that form a perpendicular/right angle)

c is the hypotenuse (the side opposite the right angle)

In this case...

a = 8 cm

b = 6 cm

c = x

^^^Plug these numbers into the theorem

$8^{2} +6^{2} =x^{2}$

simplify

64 + 36 = $x^{2}$

100 = $x^{2}$

To remove the square from x take the square root of both sides to get you...

√100 = x

Which is when simplified:

(choice C)

10 cm

Hope this helped!

~Just a girl in love with Shawn Mendes

ivann1987 [24]3 years ago

4 0

Answer: C. 10 cm

Step-by-step explanation:

the hypotenuse of a right triangle will always be longer than the other 2 sides. So 10 cm

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elena55 [62]

You would have to determine if the following lengths make an acute, right or obtuse triangle. Plug in each set of lengths into the Pythagorean Theorem.

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3 years ago

Ashley can bake 64 cakes in 12 hours how many can she bake in 36 hours

mrs_skeptik [129]

Answer:

192 cakes

Step-by-step explanation:

36/12= 3

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192 cakes in 36 hours

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3 years ago

Find a and b. А E b B 409 с D

babymother [125]

Answer:

a = 70°, b = 140°

Step-by-step explanation:

use alternate interior angles

The interior angles of a straight line are 180°

a = 180°-110° = 70°

b = 180°-40° = 140°

//I'm sorry, but I'm not very fluent in English.

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2 years ago

What does the product of the two circled numbers represent?

Stella [2.4K]

Answer:

Step-by-step explanation:

Product is another word for multiplication, and 2 * 9 = 18.

Hope this helps! :D

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2 years ago

Read 2 more answers

Find the absolute maximum and absolute minimum values of the function f(x, y) = x 2 + y 2 − x 2 y + 7 on the set d = {(x, y) : |

dsp73

Looks like $f(x,y)=x^2+y^2-x^2y+7$ .

$f_x=2x-2xy=0\implies2x(1-y)=0\implies x=0\text{ or }y=1$

$f_y=2y-x^2=0\implies2y=x^2$

If $x=0$ , then $y=0$ - critical point at (0, 0).
If $y=1$ , then $x=\pm\sqrt2$ - two critical points at $(-\sqrt2,1)$ and $(\sqrt2,1)$

The latter two critical points occur outside of $D$ since $|\pm\sqrt2|>1$ so we ignore those points.

The Hessian matrix for this function is

$H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}2-2y&-2x\\-2x&2\end{bmatrix}$

The value of its determinant at (0, 0) is $\det H(0,0)=4>0$ , which means a minimum occurs at the point, and we have $f(0,0)=7$ .

Now consider each boundary:

If $x=1$ , then

$f(1,y)=8-y+y^2=\left(y-\dfrac12\right)^2+\dfrac{31}4$

which has 3 extreme values over the interval $-1\le y\le1$ of 31/4 = 7.75 at the point (1, 1/2); 8 at (1, 1); and 10 at (1, -1).

If $x=-1$ , then

$f(-1,y)=8-y+y^2$

and we get the same extrema as in the previous case: 8 at (-1, 1), and 10 at (-1, -1).

If $y=1$ , then

$f(x,1)=8$

which doesn't tell us about anything we don't already know (namely that 8 is an extreme value).

If $y=-1$ , then

$f(x,-1)=2x^2+8$

which has 3 extreme values, but the previous cases already include them.

Hence $f(x,y)$ has absolute maxima of 10 at the points (1, -1) and (-1, -1) and an absolute minimum of 0 at (0, 0).

3 0

3 years ago