Find the product of 305.08 x 1.5 Help pls

Members of the cooking club would like to learn how to make peach ice cream. There are 14 people in the club. Each member will n

ohaa [14]

Answer:

Step-by-step explanation:

Total members = 14

Each member:

Peaches = 3

Pint of cream = 1

Cost

Peaches = x cent

Pint of cream = 46 cent

46 cent = $0.46

Cost per member = 3(x cent) + 0.46

= 3x cent + 0.46

Total cost for all members = 14(3x cent + 0.46)

= 42x cent + 6.44

8 0

3 years ago

Find the y-intercept and the x-intercept of the line 2x-4y=10

kompoz [17]

Finding y intercept and x intercept is easy:

X intercept will be of the form (x,0) and y intercept will be of the form (0,y)

● If you put x=0 in the equation, you will get y-intercept.

● If you put y=0 in the equation, you will get x-intercept.
______________________________

Given equation: 2x - 4y = 10

◆ Put x = 0
2×0 - 4y = 10
=> -4y = 10
=> y = 10/(-4)
=> y = -5/2

Thus y intercept is (0, -5/2)

◆Put y = 0
2x - 4×0 = 10
=> 2x = 10
=> x = 10/2
=> x = 5

Thus the x intercept is (5,0)

4 0

3 years ago

PLEASE ANSWER! WILL MARK BRAINLIEST!!

lord [1]

Answer:

Here you go.Hope this help!!

5 0

2 years ago

Read 2 more answers

How do I solve this?

Sonbull [250]

The question is somewhat poorly posed because the equation doesn't involve θ at all. I assume the author meant to use x.

sec(x) = csc(x)

By definition of secant and cosecant,

1/cos(x) = 1/sin(x)

Multiply both sides by sin(x) :

sin(x)/cos(x) = sin(x)/sin(x)

As long as sin(x) ≠ 0, this reduces to

sin(x)/cos(x) = 1

By definition of tangent,

tan(x) = 1

Solve for x :

x = arctan(1) + nπ

x = π/4 + nπ

(where n is any integer)

In the interval 0 ≤ x ≤ 2π, you get 2 solutions when n = 0 and n = 1 of

x = π/4 or x = 5π/4

5 0

2 years ago

in exercises 15-20 find the vector component of u along a and the vecomponent of u orthogonal to a u=(2,1,1,2) a=(4,-4,2,-2)

Nimfa-mama [501]

Answer:

The component of $\vec u$ orthogonal to $\vec a$ is $\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$ .

Step-by-step explanation:

Let $\vec u$ and $\vec a$ , from Linear Algebra we get that component of $\vec u$ parallel to $\vec a$ by using this formula:

$\vec u_{\parallel \,\vec a} = \frac{\vec u \bullet\vec a}{\|\vec a\|^{2}} \cdot \vec a$ (Eq. 1)

Where $\|\vec a\|$ is the norm of $\vec a$ , which is equal to $\|\vec a\| = \sqrt{\vec a\bullet \vec a}$ . (Eq. 2)

If we know that $\vec u =(2,1,1,2)$ and $\vec a=(4,-4,2,-2)$ , then we get that vector component of $\vec u$ parallel to $\vec a$ is:

$\vec u_{\parallel\,\vec a} = \left[\frac{(2)\cdot (4)+(1)\cdot (-4)+(1)\cdot (2)+(2)\cdot (-2)}{4^{2}+(-4)^{2}+2^{2}+(-2)^{2}} \right]\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\frac{1}{20}\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$

Lastly, we find the vector component of $\vec u$ orthogonal to $\vec a$ by applying this vector sum identity:

$\vec u_{\perp\,\vec a} = \vec u - \vec u_{\parallel\,\vec a}$ (Eq. 3)

If we get that $\vec u =(2,1,1,2)$ and $\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$ , the vector component of $\vec u$ is:

$\vec u_{\perp\,\vec a} = (2,1,1,2)-\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$

$\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$

The component of $\vec u$ orthogonal to $\vec a$ is $\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$ .

4 0

3 years ago