All categories

3 years ago

HELP ME PLZZZZ!!!!! It’s just 21 and 22

Mathematics

1 answer:

galina1969 [7]3 years ago

7 0

Answer:

21) 12 cm

22) 5.9 cm

Step-by-step explanation:

21) The base is 10, and if the other two sides are congruent and the perimeter is 36, we can figure out with simple algebra that the sides are 13 cm long.

Half of 10 is 5, so we can use the pythagorean theorem.

5^2+x^2=13^2

Rearranging the variables we have 169-25=x^2

144=x^2

x can be plus or minus 12, but since negative length is impossible we find that x is positive 12 cm.

22) We want to use sine, because we have opposite and hypotenuse. A simple and easy way to memorize this is the SohCahToa method. If we have opposite (O) and hypotenuse (H) we have OH. Soh has the letters O and H, and the S means we should use sine.

sine 36=a/10

Plug this into a calculator or desmos scientific calculator to get a=5.9

You might be interested in

Translate the following into an algebraic expression:

Nata [24]

For this case, the first thing we must do is define variables.

We have then:

x: number of minutes

y: total cost

We write the algebraic expression that models the problem.

We have then:

$y = 10 * (\frac{20}{60}) x$

Simplifying we have:

$y = (\frac{10}{3}) x$

Then, by the time the cost is equal to $ 300 we have:

$300 = (\frac{10}{3}) x$

From here, we clear the value of x.

Answer:

an algebraic expression for the problem is:

$300 = (\frac{10}{3}) x$

3 0

3 years ago

Let C = C1 + C2 where C1 is the quarter circle x^2+y^2=4, z=0,from (0,2,0) to (2,0,0), and where C2 is the line segment from (2,

trapecia [35]

Not much can be done without knowing what $\mathbf F(x,y,z)$ is, but at the least we can set up the integral.

First parameterize the pieces of the contour:

$C_1:\mathbf r_1(t_1)=(2\sin t_1,2\cos t_1,0)$
$C_2:\mathbf r_2(t_2)=(1-t_2)(2,0,0)+t_2(3,3,3)=(2+t_2, 3t_2, 3t_2)$

where $0\le t_1\le\dfrac\pi2$ and $0\le t_2\le1$ . You have

$\mathrm d\mathbf r_1=(2\cos t_1,-2\sin t_1,0)\,\mathrm dt_1$
$\mathrm d\mathbf r_2=(1,3,3)\,\mathrm dt_2$

and so the work is given by the integral

$\displaystyle\int_C\mathbf F(x,y,z)\cdot\mathrm d\mathbf r$
$=\displaystyle\int_0^{\pi/2}\mathbf F(2\sin t_1,2\cos t_1,0)\cdot(2\cos t_1,-2\sin t_1,0)\,\mathrm dt_1$
${}\displaystyle\,\,\,\,\,\,\,\,+\int_0^1\mathbf F(2+t_2,3t_2,3t_2)\cdot(1,3,3)\,\mathrm dt_2$