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3 years ago

5. Write two inequalities to compare -17 and -22.

Mathematics

1 answer:

galina1969 [7]3 years ago

8 0

Answer:

-17>-22

-22<-17

Negative seventeen is closer to zero, therefore it is always larger than negative twenty-two.

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At what points does the helix r = left angle bracketsin(t), cos(t), tright angle bracket intersect the sphere x 2 + y 2 + z 2 =

r-ruslan [8.4K]

Answer:

the points are = (0.1045, 0.9945, 6) and ( -0.1045, 0.9945, -6)

Step-by-step explanation:

The detailed steps and appropriate substitution is as shown in the attached file.

4 0

3 years ago

What’s the value of x trigonometry question

Softa [21]

Answer:

x=9.2394

Step-by-step explanation:

It's tan we're going to use because we're looking for the opposite and we already have the angle

Therefore, x will be

Tan 57= x/6

1.5399= x/6

x= 6*1.5399

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You can approximate it to the nearest whole number which will be 9

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2 years ago

50 Points!

koban [17]

Answer:

The exact height of the actual building is 822.5 feet.

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3 years ago

If x-3 is a factor of P(x)=x^3- 7x²+15x – 9, which of the following represents the complete factorization for P(x)?

Ronch [10]

Answer:

$x \times x + x \times 3 + x \times x + x \times 1 - 3 \times x + 3 \times 3 + 3 \times x + 3 \times 1$

3 0

3 years ago

Hello I need help Finding the probability

Stolb23 [73]

Each petal of the region $R$ is the intersection of two circles, both of diameter 10. Each petal in turn is twice the area of a circular segment bounded by a chord of length $5\sqrt2$ , which implies the segment is subtended by an angle of $\dfrac\pi2$ . This means the area of the segment is

$\text{area}_{\text{segment}}=\text{area}_{\text{sector}}-\text{area}_{\text{triangle}}$
$\text{area}_{\text{segment}}=\dfrac{25\pi}4-\dfrac{25}2$

This means the area of one petal is $\dfrac{25\pi}2-25$ , and the area of $R$ is four times this, or $50\pi-100$ .

Meanwhile, the area of $G$ is simply the area of the square minus the area of $R$ , or $10^2-(50\pi-100)=200-50\pi$ .

So

$\mathbb P(X=R)=\dfrac{50\pi-100}{100}=\dfrac\pi2-1$
$\mathbb P(X=G)=\dfrac{200-50\pi}{100}=2-\dfrac\pi2$
$\mathbb P((X=R)\land(X=G))=0$ (provided these regions are indeed disjoint; it's hard to tell from the picture)
$\mathbb P((X=R)\lor(X=G))=\mathbb P(X=R)+\mathbb P(X=G)=1$

4 0

3 years ago