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Gekata [30.6K]
2 years ago
14

Triangle DEF was tranformed to create triangle D'E'F'. Complete the statement about the transformation.

Mathematics
2 answers:
Tom [10]2 years ago
7 0

Answer:POINT F

Step-by-step explanation:

Got it right

wlad13 [49]2 years ago
5 0

Answer: Point F corresponds to point F'

Step-by-step explanation: I just took this test and made a 100%

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What is the solution to the equation <br><br> 7(3-x)=-7(x+3)-4
marissa [1.9K]

Answer:

No solution

Step-by-step explanation:

21−7x=−7x−21−4

21−7x=−7x−25

21=−25

No solution

4 0
2 years ago
£440 is divided between David, Mark &amp; Henry so that David gets twice as much as Mark, and Mark gets three times as much as H
Lesechka [4]

Mark received £ 132

<em><u>Solution:</u></em>

Given that £440 is divided between David, Mark & Henry

Let "d" be the share of david

Let "m" be the share of mark

Let "h" be the share of henry

Total amount is 440

Therefore,

share of david + share of mark + share of henry = 440

d + m + h = 440 ------- eqn 1

<em><u>David gets twice as much as Mark</u></em>

d = 2m ----- eqn 2

<em><u>Mark gets three times as much as Henry</u></em>

m = 3h

h = \frac{m}{3}  ------ eqn 3

<em><u>Substitute eqn 2 and eqn 3 in eqn 1</u></em>

2m + m + \frac{m}{3} = 440\\\\\frac{6m + 3m + m}{3} = 440\\\\10m = 1320\\\\m = 132

Thus Mark received £ 132

8 0
3 years ago
Suppose yoko borrows $3500 at an interest rate of 4% compounded each year Find the amount owed at the end of 1 year
olchik [2.2K]

Amount owed at the end of 1 year is 3640

<h3><u>Solution:</u></h3>

Given that yoko borrows $3500.

Rate of interest charged is 4% compounded each year

Need to determine amount owed at the end of 1 year.

In our case :

Borrowed Amount that is principal P = $3500

Rate of interest r = 4%

Duration = 1 year and as it is compounded yearly, number of times interest calculated in 1 year n = 1

<em><u>Formula for Amount of compounded yearly is as follows:</u></em>

A=p\left(1+\frac{r}{100}\right)^{n}

Where "p" is the principal

"r" is the rate of interest

"n" is the number of years

Substituting the values in above formula we get

\mathrm{A}=3500\left(1+\frac{4}{100}\right)^{1}

\begin{array}{l}{A=\frac{3500 \times 104}{100}} \\\\ {A=35 \times 104} \\\\ {A=3640}\end{array}

Hence amount owed at the end of 1 year is 3640

5 0
2 years ago
Which values of a and b make the following equation true? a = 11, b = 7 a = 11, b = 10 a = 28, b = 7 a = 28, b = 10
xz_007 [3.2K]

Answer:

The answer is

<h2>a = 11, b = 7</h2>

Step-by-step explanation:

( {5x}^{7}  {y}^{2} )( { - 4x}^{4}  {y}^{5} ) =  - 20 {x}^{a}  {y}^{b}

Multiply the terms on the left side of the equation

That's

- 20 {x}^{11}  {y}^{7}  =  - 20 {x}^{a}  {y}^{b}

Since the bases are the same we can equate the exponents

That's

{x}^{11}  =  {x}^{a}

a = 11

And

{y}^{7}  =  {y}^{b}

b = 11

Therefore

a = 11 and b = 7

Hope this helps you

7 0
3 years ago
Help me with the answer
rodikova [14]
Y=mx+b to solve its really easy good luck
4 0
3 years ago
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