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2 years ago

Joey built a model sailboat that is similar to his real sailboat. The triangular sail on

Mathematics

1 answer:

lord [1]2 years ago

3 0

Answer:

31.25

Step-by-step explanation:

25/12=2.083333

15 x 2.0833333=31.24999999995

round up to get 31.25

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Of his 94 compact discs, Raul plans to donate 17 to the thrift store. To the nearest tenth of a percent, what percent is this?

Masteriza [31]

For this case we can solve the problem by means of the following rule of three:
94 --------> 100%
17 --------> x
Clearing x we have:
x = (17/94) * (100)
x = 18,08510638%
To the nearest tenth of a percent:
x = 18.1%
Answer:
This is 18.1% of his compact discs.

6 0

3 years ago

Will give brainliest answer

yaroslaw [1]

Answer

Step-by-step explanation:

To find the area of a shape multiply its height by its width. For a square you only need to find the length of one of the sides (as each side is the same length) and then multiply this by itself to find the area.

please mark me as brainliest.

7 0

3 years ago

Read 2 more answers

Find X (Its a circle question, PLEASE HELP)

mario62 [17]

Step-by-step explanation:

Since angles in the same segment are equal, Angle ABC (x) = Angle AOC.

Since Angles AOC and COD are supplementary, Angle AOC = 180° - 130° = 50°

Therefore x = 50°.

7 0

3 years ago

If the image of a triangle is congruent to the pre-image, what is the scale factor of the dilation

Evgen [1.6K]

Answer:

Step-by-step explanation:

Scale factor is 1 because the triangle and image are congruent ( same shape and size).

3 0

3 years ago

Use Lagrange multipliers to find the maximum and minimum values of (i) f(x,y)-81x^2+y^2 subject to the constraint 4x^2+y^2=9. (i

sp2606 [1]

i. The Lagrangian is

$L(x,y,\lambda)=81x^2+y^2+\lambda(4x^2+y^2-9)$

with critical points whenever

$L_x=162x+8\lambda x=0\implies2x(81+4\lambda)=0\implies x=0\text{ or }\lambda=-\dfrac{81}4$

$L_y=2y+2\lambda y=0\implies2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_\lambda=4x^2+y^2-9=0$

If $x=0$ , then $L_\lambda=0\implies y=\pm3$ .
If $y=0$ , then $L_\lambda=0\implies x=\pm\dfrac32$ .
Either value of $\lambda$ found above requires that either $x=0$ or $y=0$ , so we get the same critical points as in the previous two cases.

We have $f(0,-3)=9$ , $f(0,3)=9$ , $f\left(-\dfrac32,0\right)=\dfrac{729}4=182.25$ , and $f\left(\dfrac32,0\right)=\dfrac{729}4$ , so $f$ has a minimum value of 9 and a maximum value of 182.25.

ii. The Lagrangian is

$L(x,y,z,\lambda)=y^2-10z+\lambda(x^2+y^2+z^2-36)$

with critical points whenever

$L_x=2\lambda x=0\implies x=0$ (because we assume $\lambda\neq0$ )

$L_y=2y+2\lambda y=0\implies 2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_z=-10+2\lambda z=0\implies z=\dfrac5\lambda$

$L_\lambda=x^2+y^2+z^2-36=0$

If $x=y=0$ , then $L_\lambda=0\implies z=\pm6$ .
If $\lambda=-1$ , then $z=-5$ , and with $x=0$ we have $L_\lambda=0\implies y=\pm\sqrt{11}$ .

We have $f(0,0,-6)=60$ , $f(0,0,6)=-60$ , $f(0,-\sqrt{11},-5)=61$ , and $f(0,\sqrt{11},-5)=61$ . So $f$ has a maximum value of 61 and a minimum value of -60.

5 0

3 years ago