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2 years ago

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All of the students in Ms. Osbourne's class like pizza. Jeff is a student who likes pizza. Therefore, Jeff is a student in Ms. O

sbourne's class.

1 answer:

faust18 [17]2 years ago

8 0

Whats the question. I need the question to answer.

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Given triangle GHJ, the measure of angle G equals 110°, the measure of angle J equals 40°, and the measure of angle H equals 30°

tester [92]

Answer:

Since angle G is

✔ the largest

angle, the opposite side, JH, is

✔ the longest side

. The order of the side lengths from longest to shortest is

✔ HJ, GH, and GJ

.

Step-by-step explanation:

5 0

1 year ago

Read 2 more answers

The boiling temperature (in degrees Celsius) of platinum is 199 more than four times the boiling temperature $z$ (in degrees Cel

Zolol [24]

The correct answer is letter C

5 0

2 years ago

What are the solutions of the equation 9x4 – 2x2 – 7 = 0? Use u substitution to solve. tions of the equation 9

skelet666 [1.2K]

Answer:

Step-by-step explanation:

Let $u^2=x^4\\u = x^2$

Subbing in:

$9u^2-2u-7=0$

a = 9, b = -2, c = -7

The product of a and c is the aboslute value of -63, so a*c = 63. We need 2 factors of 63 that will add to give us -2. The factors of 63 are {1, 63}, (3, 21}, {7, 9}. It looks like the combination of -9 and +7 will work because -9 + 7 = -2. Plug in accordingly:

$9u^2-9u+7u-7=0$

Group together in groups of 2:

$(9u^2-9u)+(7u-7)=0$

Now factor out what's common within each set of parenthesis:

$9u(u-1)+7(u-1)=0$

We know this combination "works" because the terms inside the parenthesis are identical. We can now factor those out and what's left goes together in another set of parenthesis:

$(u-1)(9u+7)=0$

Remember that $u=x^2$

so we sub back in and continue to factor. This was originally a fourth degree polynomial; that means we have 4 solutions.

$(x^2-1)(9x^2+7)=0$

The first two solutions are found withing the first set of parenthesis and the second two are found in other set of parenthesis. Factoring $(x^2-1)$ gives us that x = 1 and -1. The other set is a bit more tricky. If

$9x^2+7=0$ then

$9x^2=-7$ and

$x^2=-\frac{7}{9}$

You cannot take the square root of a negative number without allowing for the imaginary component, i, so we do that:

$x=$ ± $\sqrt{-\frac{7}{9} }$

which will simplify down to

$x=$ ± $\frac{\sqrt{7} }{3}i$

Those are the 4 solutions to the quartic equation.

5 0

2 years ago

The table shows the blood pressures of 16 clinic patients. What is the interquartile range of the data?

Alex73 [517]

I’ve attached my work below....
I ran out of space so the step 5 the last step is subtract q1 and q3 which step 3 and 4
So 90.9-81.8=9.1

5 0

1 year ago

The variable complex number z is given by z=1+cos 2θ+isin2θ,where θ takes all values in the interval −1/2π<θ<1/2π

FinnZ [79.3K]

The given complex number is
z = 1 + cos(2θ) + i sin(2θ), for -1/2π < θ < 1/2π

Part (i)
Let V = the modulus of z.
Then
V² = [1 + cos(2θ)]² + sin²(2θ)
= 1 + 2 cos(2θ) + cos²2θ + sin²2θ

Because sin²x + cos²x = 1, therefore
V² = 2(1 + cos2θ)

Because cos(2x) = 2 cos²x - 1, therefore
V² = 2(1 + 2cos²θ - 1) = 4 cos²θ
Because -1/2π < θ < 1/2π,
V = 2 cosθ PROVEN

Part ii.
1/z = 1/[1 + cos2θ + i sin 2θ]
$\frac{1}{z} = \frac{(1+cos2\theta - i\, sin2\theta)}{(1 + cos 2\theta + i\, sin 2\theta)(1+cos2\theta - i \,sin2\theta)}\\ = \frac{1+cos2\theta - i \,sin 2\theta}{(1+cos2\theta)^{2} + sin^{2}2\theta}$

The denominator is
$(1+cos2\theta)^{2}+sin^{2}2\theta \\ = 1+2cos2\theta+cos^{2}2\theta+sin^{2}2\theta \\ =2cos2\theta+2 \\ = 2(1+cos2\theta)$

Therefore
$\frac{1}{z} = \frac{1}{2} -i \frac{sin2\theta}{2(1+cos2\theta)}$

The real part of 1/ = 1/ (constant).

6 0

2 years ago