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3 years ago

5

Consider triangle QRS. The legs each have a length of 10 units. Triangle Q R S is shown. Angle Q S R is 90 degrees. The length o

f Q X is 10 and the length of S R is 10. What is the length of the hypotenuse of the triangle?

1 answer:

Ipatiy [6.2K]3 years ago

5 0

Answer:

10 square root 2

Step-by-step explanation:

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Factor the polynomial expression 16y^4 - 625x^4

Helga [31]

Answer:

(25x2 + 4y2) (5x + 2y) (−5x + 2y)

Step-by-step explanation:

Factor 16y4−625x4

−625x4 + 16y4

=(25x2 + 4y2) (5x + 2y) (−5x + 2y)

6 0

3 years ago

The committee at Ann's apartment

xxMikexx [17]

Answer: D = 40 / C

Step-by-step explanation:

total amount of soil divided by soil in each bag. To find the total amount of soil multiply number of flower boxes (10) by amount of soil each box holds (4 kilograms). So, number of bags (D) equals 40 divided by kg of soil in each bag (C)

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3 years ago

There are as many girls as there are boys in a class. 8 girls went for a dance competition. now there are twice as many boys as

Rashid [163]

Let the number of girls be x and the number of boys be y

The number of boys is the same as the number of girls.
x = y
The number of boys is twice the number of girls when 8 girls left.
2(x - 8) = y

x = y ---------------- (1)
2(x - 8) = y -------- (2)

Sub (1) into (2)

2(y - 8) = y
2y - 16 = y
2y - y = 16
y = 16
There were 16 boys and 16 girls initially.

16 + 16 = 32
There were a total of 32 students in the class initially.

4 0

3 years ago

When the beam makes an angle of 40

egoroff_w [7]

A construction crew wants to hoist a heavy
beam so that it is standing up straight. They
tie a rope to the beam, secure the base, and
pull the rope through a pulley to raise one
end of the beam from the ground. When
the beam makes an angle of 40 degrees with the
ground, the top of the beam is 8 ft above
the ground.
Th e construction site has some telephone
wires crossing it. Th e workers are
concerned that the beam may hit the wires.
When the beam makes an angle of 60 degrees with
the ground, the wires are 2 ft above the top
of the beam. Will the beam clear the wires
on its way to standing up straight?

<span>Math - Steve Thursday, April 16, 2015 at 6:22pmwe see that the length of the beam is

8/sin40 = 12.45 ft

At 60 degrees, the top is

12.45sin60 = 10.78 ft high

So, the wire is 12.78 ft up.

Since the beam is only 12.45 ft long, it will not touch the wires.</span>

3 0

3 years ago

Help needed plez<br><br>Ans in (ii) is 1.5

Deffense [45]

OK first let's check the x=1.5.

$y = \frac x 6 (x^2 - 10)$

$y = 1-2x$

$1-2x = \frac x 6 (x^2 - 10)$

$6-12x = x^3 - 10 x$

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

Oh my, that's called a depressed cubic, no $x^2$ term. There's a formula for these very much like the quadratic formula but you're probably not quite old enough for that. Anyway, $x\approx 1.45616 \approx 1.5$ is a solution, but that's not what they're asking. They are asking us to compare

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

with

$x^3 + bx^2 + cx + d = 0$

and conclude $b=0, c= 2, d=-6$

It turns out we did need all the rest of it. Save those brain cells, there's lots more math coming.

~~~~~~~~~~~~~~

I love it when the student asks for more. Here's the formula for a depressed cubic. I won't derive it here (though I did earlier today, coincidentally, but I'm probably not allowed to link to my Quora answer "what led to the discovery of complex numbers" from here). We use the trick of putting coefficients on the coefficients to avoid fractions.

$x^3 + 3 p x = 2 q$

has solutions

$x = \sqrt[3] { q - \sqrt{p^3 + q^2} } + \sqrt[3] {q + \sqrt{p^3 + q^2} } 

$

That's pretty simple, though sometimes we end up having to take the cube roots of complex numbers, which isn't that helpful. Let's try it out on

$x^3 + 2x=6$

That's $p=2/3, q=3$ so

$x = \sqrt[3] { 3 - \sqrt{(2/3)^3+9} } + \sqrt[3] {3 + \sqrt{(2/3)^3+9} }$

$x = \sqrt[3] { 3 - \sqrt{753}/9 } +\sqrt[3]{3 + \sqrt{753}/9 }$

$x \approx 1.4561642461359084609748069666$

6 0

3 years ago