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3 years ago

HELP PLS QUICK

Mathematics

1 answer:

Daniel [21]3 years ago

3 0

Answer:

$\displaystyle d \approx 7.07$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Reading a Cartesian plane
Coordinates (x, y)

Algebra II

Distance Formula: $\displaystyle d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Step-by-step explanation:

Step 1: Define

Point C (-2, 0)

Point D (3, 5)

Step 2: Find distance d

Simply plug in the 2 coordinates into the distance formula to find distance d

Substitute in points [Distance Formula]: $\displaystyle d = \sqrt{(3+2)^2+(5-0)^2}$
[√Radical] (Parenthesis) Add/Subtract: $\displaystyle d = \sqrt{(5)^2+(5)^2}$
[√Radical] Evaluate exponents: $\displaystyle d = \sqrt{25+25}$
[√Radical] Add: $\displaystyle d = \sqrt{50}$
[√Radical] Simplify: $\displaystyle d = 5\sqrt{2}$
[√Radical] Evaluate: $\displaystyle d = 7.07107$
Round: $\displaystyle d \approx 7.07$

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Given that the four points A(1,2), B(2, -5), C(7,0) and D(p,q) are the vertices of the rhombus ABCD

Serga [27]

Step-by-step explanation:

Three coordinates of a Rhombus is given to us. And the fourth coordinate is D(p,q) . Weneedtofind the value of p and q . We know that the diagonals of Rhombus bisect each other at right angles . So lets find the midpoint.

• Midpoint of AC :-

$\tt\to Midpoint_{(AC)}= \bigg(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\bigg)\\\\\tt\to Midpoint_{(AC)}= \bigg( \dfrac{1+7}{2},\dfrac{2+0}{2}\bigg)\\\\\tt\to Midpoint_{(AC)}= \bigg(\dfrac{8}{2},\dfrac{2}{2}\bigg)\\\\\tt\to \boxed{\orange{\tt Midpoint_{(AC)}= (4,1)}}$

_______________________________

• Midpoint of BD :-

$\tt\to Midpoint_{(BD)}= \bigg(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\bigg)\\\\\tt\to \boxed{\orange{\tt Midpoint_{(BD)}= \bigg( \dfrac{2+p}{2},\dfrac{q-5}{2}\bigg)}}$

_______________________________

Now since these two coordinates must be equal, therefore ,

$\tt\to \bigg( \dfrac{2+p}{2},\dfrac{q-5}{2}\bigg)= (4,1)\\\\\tt\to \dfrac{2+p}{2}=4 \qquad and \qquad \dfrac{q-5}{2}=1 \\\\\tt\to 2 + p = 8 \qquad and \qquad q-5 = 2 \\\\\tt\to p = 8 -2 \qquad and \qquad q = 2 +5 \\\\\tt\to\boxed{\orange{\tt (p,q) = (6,7) }}$

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3 years ago

The diameter of Mercury is approximately 4.9×103 kilometers. The diameter of Earth is approximately 1.3×104 kilometers. About ho

-Dominant- [34]

Hi there!

To get the answer to this question, let's divide the diameter of the Earth by the diameter of Mercury.

$\frac{1.3*10^4}{4.9*10^3}$

We can change this to

$\frac{1.3}{4.9}*\frac{10^4}{10^3}$

Let's solve for each individual fraction

$\frac{1.3}{4.9}$ ≈ 0.27

$\frac{10^4}{10^3}$ = 10

Now, multiply.

0.27 * 10 = 2.7

That's your answer.

Have an awesome day/night! :)

~collinjun0827, Junior Moderator

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3 years ago

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djverab [1.8K]

It’s 3/10 if it’s not then do 28

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3 years ago

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Casey wants to buy a gym membership. Gym A has a $150 joining fee and costs $35 per month. Gym B has no joining fee and costs $6

MariettaO [177]

Answer:

Step-by-step explanation:

Gym A has a $150 joining fee and costs $35 per month.

Assuming that Casey wants to attend for x months, the cost of using gym A will be

150 + 35 times x months. It becomes

150 + 35x

Gym B has no joining fee and costs $60 per month.

Again, assuming that Casey wants to attend for x months, the cost of using gym B will be

60 × x months = 60x

A) To determine the number of months that it will both gym memberships to be the same, we will equate them.

150 + 35x = 60x

60x - 35x = 150

25x = 150

x = 150/25 = 6

It will take 6 months for both gym memberships to be the same.

B) If Casey plans to only go to the gym for 5 months,

Plan A will cost 150 + 35×5 = $325

Plan B will cost 60 × 5 = $300

Plan B will be cheaper

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