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3 years ago

ASAP PLZ HELP I MARK BRIANLIST IF U GIVE ME RIGHT ANSWER

Mathematics

1 answer:

gayaneshka [121]3 years ago

7 0

Answer:

$d = \sqrt{13}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra II

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

Step-by-step explanation:

Step 1: Define

Point A (-2, 1)

Point B (1, -1)

Step 2: Find distance d

Substitute: $d = \sqrt{(1+2)^2+(-1-1)^2}$
Add/Subtract: $d = \sqrt{(3)^2+(-2)^2}$
Exponents: $d = \sqrt{9+4}$
Add: $d = \sqrt{13}$

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On a map, the distance from Happy Hill Park to Rainbow Valley Park is 4/1/2 inches. The scale is 1/2 inch: 3 miles. What is the

Studentka2010 [4]

we are given

The scale is

$\frac{1}{2} inch=3miles$

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we can change it into rational form

and we get

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so, we can multiply both sides by 9

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now, we can simplify it

$\frac{9}{2} inch=27 miles$

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3 years ago

A quantity with an initial value of 6200 decays continuously at a rate of 5.5% per month. What is the value of the quantity afte

ELEN [110]

Answer:

410.32

Step-by-step explanation:

Given that the initial quantity, Q= 6200

Decay rate, r = 5.5% per month

So, the value of quantity after 1 month, $q_1 = Q- r \times Q$

$q_1 = Q(1-r)\cdots(i)$

The value of quantity after 2 months, $q_2 = q_1- r \times q_1$

$q_2 = q_1(1-r)$

From equation (i)

$q_2=Q(1-r)(1-r) \\\\q_2=Q(1-r)^2\cdots(ii)$

The value of quantity after 3 months, $q_3 = q_2- r \times q_2$

$q_3 = q_2(1-r)$

From equation (ii)

$q_3=Q(1-r)^2(1-r)$

$q_3=Q(1-r)^3$

Similarly, the value of quantity after n months,

$q_n= Q(1- r)^n$

As 4 years = 48 months, so puttion n=48 to get the value of quantity after 4 years, we have,

$q_{48}=Q(1-r)^{48}$

Putting Q=6200 and r=5.5%=0.055, we have

$q_{48}=6200(1-0.055)^{48} \\\\q_{48}=410.32$

Hence, the value of quantity after 4 years is 410.32.

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3 years ago

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GAME SHOW GET READY... WHOEVER ANSWERS CORRECT GETS BRAINLIST

Mandarinka [93]

Answer:

1/2 (60)

Step-by-step explanation:

4 0

2 years ago

Please help. Thank you

Inessa [10]

Given:

$PQRS\sim TUVW$

In the given figure, PS=x, RS=35, UV=20, VW=25 and TW=15

To find:

The scale factor from PQRS to TUVW.

Solution:

We have,

$PQRS\sim TUVW$

We know that the corresponding sides of similar figures are proportional. The scale factor is the ratio of one side of image and corresponding side of preimage.

The scale factor is:

$k=\dfrac{VW}{RS}$

$k=\dfrac{25}{35}$

$k=\dfrac{5}{7}$

Therefore, the scale factor from PQRS to TUVW is $k=\dfrac{5}{7}$ .

6 0

3 years ago