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2 years ago

3. Find the slope (it says its too short thats why im writing this)

Mathematics

1 answer:

ad-work [718]2 years ago

7 0

Answer:

$\displaystyle m=\frac{2}{3}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Slope Formula: $\displaystyle m=\frac{y_2-y_1}{x_2-x_1}$

Step-by-step explanation:

Step 1: Define

Find points from graph.

Point (0, 3)

Point (-3, 1)

Step 2: Find slope m

Simply plug in the 2 coordinates into the slope formula to find slope m

Substitute in points [SF]: $\displaystyle m=\frac{1-3}{-3-0}$
[Fraction] Subtract: $\displaystyle m=\frac{-2}{-3}$
[Fraction] Simplify: $\displaystyle m=\frac{2}{3}$

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Given right triangle ABC with altitude BD drawn to hypotenuse AC. If AC =16 and DC=5 what is the length of BC in the simplest ra

Nana76 [90]

The length of BC is $4 \sqrt{5}$ .

Solution:

Given ABC is a right triangle.

AC is the hypotenuse and BD is the altitude.

AB and BC are legs of the triangle ABC.

AC = 16 and DC = 5

Leg rule of geometric mean theorem:

$\frac{\text { hypotenuse }}{\text { leg }}=\frac{\text { leg }}{\text { part }}$

$\Rightarrow \frac{AC}{BC}=\frac{BC}{DC}$

$\Rightarrow \frac{16}{x}=\frac{x}{5}$

Do cross multiplication.

$\Rightarrow 16\times 5 = x\times x$

$\Rightarrow 80= x^2$

$\Rightarrow 16\times 5= x^2$

Taking square root on both sides.

$\Rightarrow \sqrt{16\times 5} = \sqrt{x^2}$

$\Rightarrow \sqrt{4^2\times 5} = \sqrt{x^2}$

square and square roots get canceled, we get

$\Rightarrow 4\sqrt{ 5} = x$

The length of BC is $4 \sqrt{5}$ .

7 0

3 years ago

Help. It’s due today plz I’m begging u :/ click the picture to make up bigger

Pani-rosa [81]

Answer:

im pretty sure B 7.5 units

Step-by-step explanation:

6 0

3 years ago

Solve the linear equation. 2x+3(x-4)=(2x+3)

I am Lyosha [343]

Answer:

X=5

Step-by-step explanation:

Cancel equal terms remove parentheses.

Move the constant to the right.

Add the numbers.

Divide by 3 on both side. and X=5

Hope this helps

3 0

3 years ago

Find the value of x.

konstantin123 [22]

Answer:

x= 37.5°

Step-by-step explanation:

∠CBD

= 180° -75° (adj. ∠s on a str. line)

= 105°

∠BCD= ∠BDC (base ∠s of isos. △BCD)

∠BCD= x

∠BCD +∠BDC +∠CBD= 180° (∠ sum of △BCD)

x +x +105°= 180°

2x= 180° -105°

2x= 75°

x= 37.5°

Alternative working:

∠BDA= (180° -75°) ÷2 (base ∠s of isos. △ABD)

∠BDA= 52.5°

∠BDA +∠BDC= 90°

52.5° +x= 90°

x= 90° -52.5°

x= 37.5°

6 0

3 years ago

Suppose there are two lakes located on a stream. Clean water flows into the first lake, then the water from the first lake flows

never [62]

Answer:

a. For the first lake;

c = (m·t - 0.005·m·t²)/100000

For the second tank, we have;

c = m·t/200 - m·t²/80000

b. t ≈ 1.00505 hours

c. 200 hours

Step-by-step explanation:

The flow rate of water in and out of the lakes = 500 liters/hour

The volume of water in the first lake = 100 thousand liters

The volume of water in the second lake = 200 thousand liters

The mass of toxic substances that entered into the first lake = 500 kg

The concentration of toxic substance in the first lake = m₁/(100000)

Therefore, we have;

The quantity of fresh water supplied at t hours = 500 × t

The change

The change in the mass of the toxic substance with time is given as follows

dm/dt = (m - m/100000 × 500 × t)/100000

c = (m·t - 0.005·m·t²)/100000

For the second tank, we have;

c = m/100000 × 500 × t - (m/100000 × 500 × t)/200000 × 500 × t

Using an online tool, we have;

c = m·t/200 - m·t²/80000

b. When c < 0.001 kg per liter, we have m < 0.001 × 100000, which gives m < 100

Substituting gives;

0.001 = (100·t - 0.005·100·t²)/100000, solving with an online tool, gives;

t ≈ 1.00505 hours

c. For maximum concentration, we have;

c = m·t/200 - m·t²/80000

m/200000 = m·t/200 - m·t²/80000

1/200000 = t/200 - t²/80000

dc/dt = d(t/200 - t²/80000)/dt = 0

Solving with an online tool gives t = 200 hours

6 0

2 years ago