All categories

2 years ago

What are the dimensions of the product? -3 2 4 -3 X-2 3 5

Mathematics

1 answer:

Brut [27]2 years ago

4 0

Answer:

Step-by-step explanation:

You might be interested in

So,some please help!

Leni [432]

4.6 is the correct answer. Hope I helped:)

5 0

2 years ago

Read 2 more answers

3. Let sin 21 = 0.358. What is the value of x for cos x=0.358?

r-ruslan [8.4K]

Answer:

$\cos(69)$

Step-by-step explanation:

We will use the co-function identity for sine, where:

$\sin(x)=\cos(90-x)$

Since sin(21)=0.358=cos(x). This means that:

$21=90-x$

Solve for x. Add x to both sides:

$x+21=90$

Subtract 21 from both sides:

$x=69\textdegree$

Hence:

$\sin(21)=\cos(69)\approx0.358$

6 0

3 years ago

Find an equation of the plane with x-intercept a, y-intercept b, and z-intercept c. What is the distance between the origin and

Elenna [48]

Answer:

The equation is:

(1/a)x + (1/b)y + (1/c)z = 1

Step-by-step explanation:

The direction vector between the points (a, 0, 0) and (0, b, 0) is given as:

<0 - a, b - 0, 0 - 0>

<-a, b, 0> .....................(1)

The direction vector between (0, 0, c) and (0, b, 0) is given as:

<0 - 0, b - 0, 0 - c>

= <0, b, -c> .....................(2)

To obtain the direction vector that is normal to the surface of the plane, we take the cross product of (1) and (2).

Doing this, we have:

<-a, b, 0> × <0, b, -c> = <-bc, -ac, -ab>

To find the scalar equation of the plane we can use any of the points that we know. Using (0, b, 0), we have:

(-bc)x + (-ac)y + (-ab)z = (-bc)0 + (-ac)b + (-ab)0

(bc)x + (ac)y + (ab)z = (ac)b

Dividing both sides by abc, we have:

(1/a)x + (1/b)y + (1/c)z = 1

3 0

3 years ago

PLEASE HELP, algebra 1

podryga [215]

By intuition, the function of y with respect to x will be:

y = 4x

You can prove this by substitute in each pair of values of x and y

So the constant of relationship is 4

4 0

3 years ago

A polynomial p(x) is divided by a binomial x-a the remainder is 0

schepotkina [342]

If this is the case, then you know by the polynomial remainder theorem that $x-a$ is a factor of $p(x)$ , so there is some polynomial $q(x)$ such that

$p(x)=(x-a)q(x)$

3 0

3 years ago