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3 years ago

WILL MARK BRAINLIEST! WORTH 20 POINTS PLZ HELP ME OUT!

Mathematics

1 answer:

dybincka [34]3 years ago

3 0

Answer:

(0,12) and (5,0)

Step-by-step explanation:

Since the x is the small pot(5) it would come first when writing it out. And 12 is the large pot(y)

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A ramp with a one meter distance from edge to edge reaches a height of 0.643 meters. What is the value of the acute angle?

natita [175]

The value of the acute angle is 40°

Step-by-step explanation:

Let the angle be 'θ'

Length of the ramp = 1 m

Height = 0.643 m

To find :

The angle of elevation.

we can use sine to find the angle of elevation.

sin θ = 0.643/1

sin θ = 0.643

0.643 = sin 40°

sin θ = sin 40°

So, θ = 40°

The angle of elevation is 40°

4 0

3 years ago

Can some help with this question

marissa [1.9K]

Answer:

y + 1 = 3(x - 1)

Step-by-step explanation:

The given point is (1,-1) meaning that x1 = 1 and y1 = -1

The slope of this line is m = 3 because we go up 3 and over to the right 1 (eg: go from the point (1,-1) to (2,2) to see this in action)

Plug these three pieces of info into the point slope formula below

y - y1 = m(x - x1)

y - (-1) = 3(x - 1)

y + 1 = 3(x - 1)

6 0

3 years ago

Read 2 more answers

4. Describe the proportion method for solving a percentage problem:

kondor19780726 [428]

Step-by-step explanation:

Consider the provided information.

For the proportion method first set up the equation like this:

$\frac{Part}{Whole}=\frac{Percentage}{100}$

Perform the cross multiplication and then solve for the missing part.

For example:

Find 80 percentage of 10.

Step 1: Set up the equation.

$\frac{Part}{10}=\frac{80}{100}$

$\frac{Part}{10}=\frac{4}{5}$

Step 2: Perform the cross multiplication

$Part=\frac{4}{5}\times 10$

Step 3: Solve for the missing part.

$Part=4 \times 2$

$Part=8$

3 0

3 years ago

Read 2 more answers

Find the 2th term of the expansion of (a-b)^4.

vladimir1956 [14]

The second term of the expansion is $-4a^3b$ .

Solution:

Given expression:

$(a-b)^4$

To find the second term of the expansion.

$(a-b)^4$

Using Binomial theorem,

$(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}$

Here, a = a and b = –b

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

Substitute i = 0, we get

$$\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}=1 \cdot \frac{4 !}{0 !(4-0) !} a^{4}=a^4$

Substitute i = 1, we get

$$\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}=\frac{4 !}{3!} a^{3}(-b)=-4 a^{3} b$

Substitute i = 2, we get

$$\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}=\frac{12}{2 !} a^{2}(-b)^{2}=6 a^{2} b^{2}$

Substitute i = 3, we get

$$\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}=\frac{4}{1 !} a(-b)^{3}=-4 a b^{3}$

Substitute i = 4, we get

$$\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}=1 \cdot \frac{(-b)^{4}}{(4-4) !}=b^{4}$

Therefore,

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

$=\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}+\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}+\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}+\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}+\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}$ $=a^{4}-4 a^{3} b+6 a^{2} b^{2}-4 a b^{3}+b^{4}$

Hence the second term of the expansion is $-4a^3b$ .

3 0

3 years ago

6x-9=y ; y=-3x linear equations solution for x and y I need help can someone explain

snow_lady [41]

The answer is
x = 1 and y = -3
<span>[1] 6x-9=y [2] y=-3x</span>

7 0

3 years ago