All categories

3 years ago

Find an equation of the sphere that passes through the origin and whose center is (1, -7, 9

1 answer:

6 0

The distance between any point on the sphere and its center is the radius. The origin is on the sphere, so the radius $r$ satisfies

$r^2=(1-0)^2+(-7-0)^2+(9-0)^2=131$

This means the equation of the circle is

$(x-1)^2+(y+7)^2+(z-9)^2=131$

You might be interested in

How to check division with multplication<br><img src="https://tex.z-dn.net/?f=3%20%5Ctimes%20292%20" id="TexFormula1" title="3 \

svetoff [14.1K]

Answer: By cross multiplication.

Step-by-step explanation: Given product is 3 × 292.

We know that after simple multiplication, we get 3 × 292 = 876.

Now, to check division with multiplication, either we need to divide 876 by 3 to get the answer 292,

we need to divide 876 by 292 to get the answer 3.

We will do that as follows -

$\dfrac{876}{3}=292~~~\textup{and}~~~\dfrac{876}{292}=3.$

Thus, doing cross-multiplication, we arrive at our conclusion.

6 0

3 years ago

Jack was 8 years older than pricillla. Together their ages totaled 166. what are their ages

mylen [45]

The answer is 58 I think 166-8= 158

4 0

3 years ago

Read 2 more answers

HELP PLEASE AND THANKS!!! EASY PROBLEM

sp2606 [1]

Answer:

52%

Step-by-step explanation:

13/25

5 0

3 years ago

Read 2 more answers

Find sin(a)&cos(B), tan(a)&cot(B), and sec(a)&csc(B).

Reil [10]

Answer:

Part A) $sin(\alpha)=\frac{4}{7},\ cos(\beta)=\frac{4}{7}$

Part B) $tan(\alpha)=\frac{4}{\sqrt{33}},\ tan(\beta)=\frac{4}{\sqrt{33}}$

Part C) $sec(\alpha)=\frac{7}{\sqrt{33}},\ csc(\beta)=\frac{7}{\sqrt{33}}$

Step-by-step explanation:

Part A) Find $sin(\alpha)\ and\ cos(\beta)$

we know that

If two angles are complementary, then the value of sine of one angle is equal to the cosine of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

$sin(\alpha)=cos(\beta)$

Find the value of $sin(\alpha)$ in the right triangle of the figure

$sin(\alpha)=\frac{8}{14}$ ---> opposite side divided by the hypotenuse

simplify

$sin(\alpha)=\frac{4}{7}$

therefore

$sin(\alpha)=\frac{4}{7}$

$cos(\beta)=\frac{4}{7}$

Part B) Find $tan(\alpha)\ and\ cot(\beta)$

we know that

If two angles are complementary, then the value of tangent of one angle is equal to the cotangent of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

$tan(\alpha)=cot(\beta)$

<em>Find the value of the length side adjacent to the angle alpha</em>

Applying the Pythagorean Theorem

Let

x ----> length side adjacent to angle alpha

$14^2=x^2+8^2\\x^2=14^2-8^2\\x^2=132$

$x=\sqrt{132}\ units$

simplify

$x=2\sqrt{33}\ units$

Find the value of $tan(\alpha)$ in the right triangle of the figure

$tan(\alpha)=\frac{8}{2\sqrt{33}}$ ---> opposite side divided by the adjacent side angle alpha

simplify

$tan(\alpha)=\frac{4}{\sqrt{33}}$

therefore

$tan(\alpha)=\frac{4}{\sqrt{33}}$

$tan(\beta)=\frac{4}{\sqrt{33}}$

Part C) Find $sec(\alpha)\ and\ csc(\beta)$

we know that

If two angles are complementary, then the value of secant of one angle is equal to the cosecant of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

$sec(\alpha)=csc(\beta)$

Find the value of $sec(\alpha)$ in the right triangle of the figure

$sec(\alpha)=\frac{1}{cos(\alpha)}$

Find the value of $cos(\alpha)$

$cos(\alpha)=\frac{2\sqrt{33}}{14}$ ---> adjacent side divided by the hypotenuse

simplify

$cos(\alpha)=\frac{\sqrt{33}}{7}$

therefore

$sec(\alpha)=\frac{7}{\sqrt{33}}$

$csc(\beta)=\frac{7}{\sqrt{33}}$

6 0

3 years ago

How do you divide fractions the way they taught it on 7th grade math edgenuidy * sorry I spelt it wrong it wouldn’t let me spell

Aleksandr [31]

Answer:

I'm old and the way I was taught about fraction division was:

INVERT AND MULTIPLY.

SO: (5/8) divided by (1/4) equals

(5/8) multiplied by (4/1) which equals (20/8) or 2.5

Step-by-step explanation:

7 0

3 years ago