Ask question

Ask question

All categories

3 years ago

11

The equation of line WX is y = −2x − 5. Write an equation of a line perpendicular to line WX in slope-intercept form that contai

ns point (−1, −2). y = 1 over 2x + 3 over 2 y = negative 1 over 2x + 3 over 2 y = 1 over 2x − 3 over 2 y = − 1 over 2x − 3 over 2

1 answer:

Nezavi [6.7K]3 years ago

4 0

Good job im not in high school yet but yea

You might be interested in

How can I solve these problems

Anuta_ua [19.1K]

Here’s the first few and you can do the rest with the same principles. If you need more help with the last ones then let me know

4 0

3 years ago

Help me with these questions

Elena L [17]

Answer: a = 4, b = 163,422

<u>Step-by-step explanation:</u>

$P = P_{0} e^{kt}$ P is the new population, P₀ is the initial population, k is the growth rate, t is the time elapsed.

Part a) What do we know?

P = 7003

P₀ = unknown

k = .21 <em>(converted 21% into a decimal)</em>

t = 36 yrs <em>(2006 - 1970) </em>

$P = P_{0} e^{kt}$ <em>need to solve for P₀</em>

$7003 = P_{0} e^{(.21)(36)}$

$7003 = P_{0} e^{7.56}$

7003 = P₀ (1920)

$\frac{7003}{1920} = P_{0}$

3.6 = P₀

rounded to the nearest whole number = 4

Part b) What do we know?

P = unknown

P₀ = 7003

k = .21 <em>(converted 21% into a decimal)</em>

t = 15 yrs <em>(2021 - 2006) </em>

$P = P_{0} e^{kt}$ <em>need to solve for P₀</em>

$P = 7003 e^{(.21)(15)}$

$P = 7003 e^{3.15}$

P = 7003(23.33)

P = 163,422.46

rounded to the nearest whole number = 163,422

*************************************************************

Answer: 2, -1

<u>Step-by-step explanation:</u>

eˣ² = eˣ * e²

eˣ² = eˣ⁺²

x² = x + 2

x² - x - 2 = 0

(x - 2)(x + 1) = 0

x - 2 = 0 or x + 1 = 0

x = 2 or x = -1

Check both answer for validity:

eˣ² = eˣ⁺²

e⁴ = e⁴ or e¹ = e¹

TRUE TRUE

*****************************************

Answer: 0 = ln 1

<u>Step-by-step explanation:</u>

$e^{-10} = \frac{1}{e^{10}}$

$ln(e^{-10}) = ln(\frac{1}{e^{10}})$

-10 = ln 1 - 10 <em>log rules say division is subtraction</em>

0 = ln 1

5 0

3 years ago

Can atriangle be formed with side lengths of 5cm. 10cm and 15cm

andreyandreev [35.5K]

Theorem of cosine:
a²=b²+c²-2bc(cos α) ⇒cos α=-(a²-b²-c²) / 2bc

In this case:
a=15 cm
b=10 cm
c=5 cm

cos α=-(15²-10²-5²) / 2*10*5
cos α=-100 / 100
cos α=-1

A=arc cos -1=180º This is impossible, because:

A+B+C=180º; then B=C=0º This is impossible for make a triangle (B>0 and C>0 if we want to make a triangle).

Therefore: it is not possible can make a triangle with side lengths of 5 cm, 10 cm and 15 cm.

3 0

3 years ago

Read 2 more answers

What is standard form of y=2/5x - 3?

vlabodo [156]

Answer:

Bring 3x to the left: −3x + y = 2. Multiply all by −1: 3x − y = −2. Note: A = 3, B = −1, C = −2. This form: Ax + By + C = 0. is sometimes called "Standard Form", but is more properly called the "General Form".

Step-by-step explanation:

dat is only exsample so i thingk i help u

3 0

3 years ago

X2 + 4x - 21 = 0 quadratic formula

Elena-2011 [213]

Answer:

0

Step-by-step explanation:

x2 +4x-21=0. solve for x

$x = \frac{7}{2}$

$x = 3 \frac{1}{2} \ x = 3.5$

6 0

3 years ago