Determine the distance between point A(-3,7) and the line y = 1/3x-2.

What is your estimate for the total amount of time students in your school spent during this past semester studying for final ex

MA_775_DIABLO [31]

Answer:

https://prezi.com/

Step-by-step explanation:

search up the answer plus there is a video too

5 0

3 years ago

The density of mercury is 13.6 grams per cubic centimeter. Complete the steps for converting 13.6 g/cm^3 to kg/m^3

Mama L [17]

1 cubic meter = 100 cm * 100 cm * 100 cm = 1 x 10^6 cc

1 kilogram = 1,000 grams

13.6 g / cc

If we had a cubic meter of mercury, its mass would be (or it would "weigh") 13.6 * 1,000,000 = 13,600,000 grams or 13,600 kilograms.

And so its density would be 13,600 kg / cubic meter.

4 0

3 years ago

Write the slope intercept form of a line going through

anyanavicka [17]

Perpendicular = opposite sign and reciprocal slope
Slope 2 turns into -1/2
Y = -1/2x + b
Plug in the point
-5 = -1/2(2) + b, b = -4
Solution: y = -1/2x - 4

5 0

3 years ago

In ΔWXY, the measure of ∠Y=90°, XW = 53, YX = 28, and WY = 45. What is the value of the cosine of ∠X to the nearest hundredth?

kotegsom [21]

The value of ∠X = 58.11°, If ΔWXY, the measure of ∠Y=90°, XW = 53, YX = 28, and WY = 45.

Step-by-step explanation:

The given is,

In ΔWXY, ∠Y=90°

XW = 53

YX = 28

WY = 45

Step:1

Ref the attachment,

Given triangle XWY is right angled triangle.

Trigonometric ratio's,

$Cos$ ∅ $= \frac{Adj}{Hyp}$

For the given attachment, the trigonometric ratio becomes,

$Cos$ ∅ $= \frac{XY}{XW}$ .....................................(1)

Let, ∠X = ∅

Where, XY = 28

XW = 53

Equation (1) becomes,

$Cos$ ∅ $= \frac{28}{53}$

$Cos$ ∅ = 0.5283

∅ = $cos^{-1}$ (0.5283)

∅ = 58.109°

Result:

The value of ∠X = 58.11°, If ΔWXY, the measure of ∠Y=90°, XW = 53, YX = 28, and WY = 45.

4 0

3 years ago

Can someone help me with this, please? No links or false answers, please

VARVARA [1.3K]

The exact value is found by making use of order of operations. The

functions can be resolved using the characteristics of quadratic functions.

Correct responses:

$\displaystyle 1\frac{4}{7} \div \frac{2}{3} - 1\frac{5}{7} =\frac{9}{14}$

x = -4, y = 12

When P = 1, V = 6

$\displaystyle x = -3 \ or \ x = \frac{1}{2}$

$\displaystyle i. \hspace{0.1 cm} \underline{ f(x) = 2 \cdot \left(x - 1.25 \right)^2 + 4.875 }$

ii. The function has a minimum point

iii. The value of x at the minimum point, is 1.25

iv. The equation of the axis of symmetry is x = 1.25

<h3>Methods by which the above responses are found</h3>

First part:

The given expression, $\displaystyle \mathbf{ 1\frac{4}{7} \div \frac{2}{3} -1\frac{5}{7}}$ , can be simplified using the algorithm for arithmetic operations as follows;

$\displaystyle 1\frac{4}{7} \div \frac{2}{3} - 1\frac{5}{7} = \frac{11}{7} \div \frac{2}{3} - \frac{12}{7} = \frac{11}{7} \times \frac{3}{2} - \frac{12}{7} = \frac{33 - 24}{14} =\underline{\frac{9}{14}}$

Second part:

y = 8 - x

2·x² + x·y = -16

Therefore;

2·x² + x·(8 - x) = -16

2·x² + 8·x - x² + 16 = 0

x² + 8·x + 16 = 0

(x + 4)·(x + 4) = 0

x = -4

y = 8 - (-4) = 12

y = 12

Third part:

(i) P varies inversely as the square of V

Therefore;

$\displaystyle P \propto \mathbf{\frac{1}{V^2}}$

$\displaystyle P = \frac{K}{V^2}$

V = 3, when P = 4

Therefore;

$\displaystyle 4 = \frac{K}{3^2}$

K = 3² × 4 = 36

$\displaystyle V = \sqrt{\frac{K}{P}$

When P = 1, we have;

$\displaystyle V =\sqrt{ \frac{36}{1} } = 6$

When P = 1, V = 6

Fourth Part:

Required:

Solving for x in the equation; 2·x² + 5·x - 3 = 0

Solution:

The equation can be simplified by rewriting the equation as follows;

2·x² + 5·x - 3 = 2·x² + 6·x - x - 3 = 0

2·x·(x + 3) - (x + 3) = 0

(x + 3)·(2·x - 1) = 0

$\displaystyle \underline{x = -3 \ or\ x = \frac{1}{2}}$

Fifth part:

The given function is; f(x) = 2·x² - 5·x + 8

i. Required; To write the function in the form a·(x + b)² + c

The vertex form of a quadratic equation is f(x) = a·(x - h)² + k, which is similar to the required form

Where;

(h, k) = The coordinate of the vertex

Therefore, the coordinates of the vertex of the quadratic equation is (b, c)

The x-coordinate of the vertex of a quadratic equation f(x) = a·x² + b·x + c, is given as follows;

$\displaystyle h = \mathbf{ \frac{-b}{2 \cdot a}}$

Therefore, for the given equation, we have;

$\displaystyle h = \frac{-(-5)}{2 \times 2} = \mathbf{ \frac{5}{4}} = 1.25$

Therefore, at the vertex, we have;

$k = \displaystyle f\left(1.25\right) = 2 \times \left(1.25\right)^2 - 5 \times 1.25 + 8 = \frac{39}{8} = 4.875$

a = The leading coefficient = 2

b = -h

c = k

Which gives;

$\displaystyle f(x) \ in \ the \ form \ a \cdot (x + b)^2 + c \ is \ f(x) = 2 \cdot \left(x + \left(-1.25 \right) \right)^2 +4.875$

Therefore;

$\displaystyle \underline{ f(x) = 2 \cdot \left(x -1.25\right)^2 + 4.875}$

ii. The coefficient of the quadratic function is 2 which is positive, therefore;

The function has a minimum point.

iii. The value of x for which the minimum value occurs is -b = h which is therefore;

The x-coordinate of the vertex = h = -b = 1.25

iv. The axis of symmetry is the vertical line that passes through the vertex.

Therefore;

The axis of symmetry is the line x = 1.25.

Learn more about quadratic functions here:

brainly.com/question/11631534

6 0

2 years ago

Read 2 more answers