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Citrus2011 [14]

3 years ago

15

PLEASE ANSWER FAST IM FAILING MATH!!!

1 answer:

posledela3 years ago

4 0

Answer:

Step-by-step explanation:

From the figure attached,

Point B has been dilated to form point B'.

B(3, 1) → B'(6, 2)

→ B'[(2 × 3), (2 × 1)]

Since rule for the dilation of a point (x, y) by a factor of k is,

B(x, y) → B'(kx, ky)

By comparing the coordinates k = 2 is the scale factor by which the point B has been dilated about the origin.

Therefore, other vertices of the quadrilateral will be,

A(-2, 3) → A'(-4, 6)

C(1, -1) → C'(2, -2)

D(-3, -2) → D'(-6, -4)

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A) 101* (x+12) <br> B)63* (7x-14) <br> C)(2x+3) (x+41)* <br> Please help :)

IceJOKER [234]

Answer:

A. x+12 = 101

-12 -12

x=89

B. 7x-14 = 63

+14 +14

7x=77

/7 /7

x=11

C. 2x+3 = x+ 41

-x -3 -x -3

x=38

im a god *0*

4 0

2 years ago

Round each number to the nearest ten. 283

arsen [322]

283 rounded to the nearest ten is 280
When rounding to the nearest ten look at the number to the right of the place you are rounding. In this case it is the 3. If the number is 5 or higher you go up to the nearest ten. If the number is lower than 5 like it is in this number you round down to the nearest ten.

5 0

3 years ago

Help Please! And THANK YOU!

Semenov [28]

Answer:

$100$

Step-by-step explanation:

$\frac{3.75 \times 100}{1.50 \times 100} = \frac{375}{150}$

<h3>so the dividend and divisor were both multiplied by 100.</h3>

6 0

2 years ago

Part 1 Write your own real-world scenario where the Pythagorean Theorem can be applied to find a missing piece. You may choose t

kondor19780726 [428]

I think this answers all the parts

Hope this helped :)

7 0

3 years ago

in a AP the first term is 8,nth term is 33 and sum to first n terms is 123.Find n and common difference

allsm [11]

I believe there is no such AP...

Recursively, this sequence is supposed to be given by

$\begin{cases}a_1=8\\a_k=a_{k-1}+d&\text{for }k>1\end{cases}$

so that

$a_k=a_{k-1}+d=a_{k-2}+2d=\cdots=a_1+(k-1)d$

$a_n=a_1+(n-1)d$

$33=8+(n-1)d$

$21=(n-1)d$

$n$ has to be an integer, which means there are 4 possible cases.

Case 1: $n-1=1$ and $d=21$ . But

$\displaystyle\sum_{k=1}^2(8+21(k-1))=37\neq123$

Case 2: $n-1=21$ and $d=1$ . But

$\displaystyle\sum_{k=1}^{22}(8+1(k-1))=407\neq123$

Case 3: $n-1=3$ and $d=7$ . But

$\displaystyle\sum_{k=1}^4(8+7(k-1))=74\neq123$

Case 4: $n-1=7$ and $d=3$ . But

$\displaystyle\sum_{k=1}^8(8+3(k-1))=148\neq123$

8 0

3 years ago