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Ede4ka [16]
1 year ago
11

Please help its urgent

Mathematics
1 answer:
Harlamova29_29 [7]1 year ago
6 0

Answer:

above is the solution to the question hope it helps. dimensions are in cm, area is in cm^2

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Please help! Will give brainliest to correct answer
saveliy_v [14]

Answer:

Correct option is (C).

Step-by-step explanation:

Because in figure,

°.°point B is fixed at plane.

.°.Center is B.

& °.° Image is enlargement.

.°. scale factor will be greater than 1

Now,

scale factor = A'B'/AB

= 24/6

= 3

So correct option is (C).

#$# HOPE YOU UNDERSTAND #$#

#$¥ THANK YOU ¥$#

❤ ☺ ☺ ☺ ☺ ☺ ☺ ❤

5 0
3 years ago
Sam opened a money-market account that pays 2% simple interest. He started the account with $7,000 and made no further deposits.
I am Lyosha [343]
I = prt
420 = 7000 x .02 x t
420 = 140 x t
420/140 = 140 x t /140
t = 3 years
8 0
3 years ago
Choose the solution(s) of the following system of equations x2 + y2 = 6 x2 – y = 6
ElenaW [278]
<h2>Answer</h2>

x=\sqrt{6}, y=0\\ x=-\sqrt{6} , y=0\\x=\sqrt{5} , y=-1\\x=-\sqrt{5} , y=-1

Or as ordered pairs: (\sqrt{6} ,0),(-\sqrt{6} ,0),(\sqrt{5} ,-1),(-\sqrt{5} ,-1)

<h2>Explanation</h2>

Lets solve our system of equations  step by step

x^2+y^2=6 equation (1)

x^2-y=6 equation (2)

1. Solve for x^2 in equation (2)

x^2-y=6

x^2=6+y equation (3)

2. Replace equation (3) in equation (1) and solve for y

x^2+y^2=6

6+y+y^2=6

y^2+y=0

y(y+1)=0

y=0 or y=-1

3. Replace the values of y in equation (3) and solve for x

- For y=0

x^2=6+y

x^2=6

x=\sqrt{6} or x=-\sqrt{6}

- For y=-1

x^2=6+y

x^2=6-1

x^2=5

x=\sqrt{5} or x=-\sqrt{5}

So, the solutions of our system of equation are:

x=\sqrt{6}, y=0\\ x=-\sqrt{6} , y=0\\x=\sqrt{5} , y=-1\\x=-\sqrt{5} , y=-1

6 0
3 years ago
Read 2 more answers
Select all the expressions that are greater than 1.
natita [175]

Select the following expressions. I will just list the letter choice:

A, C, D

3 0
3 years ago
PLEASE HELP DUE NOW!!
Anarel [89]

Answer:

✓ C. f(x)= sqrt 1/x

Step-by-step explanation:

i hoep this answer your question

8 0
2 years ago
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