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3 years ago

Hexagon DEFGHI is translated 8 units down and 3 units to the right. If the coordinates of the pre-image of point F are

Mathematics

2 answers:

Zarrin [17]3 years ago

6 0

The coordinates of F' is (-9 + 3, 2 - 8) = (-6, -6)

zhannawk [14.2K]3 years ago

5 0

Answer:

(-6,-6)

Step-by-step explanation:

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How much water should we add to 20 kg of seawater to change its concentration from 3% to 2%?

zhannawk [14.2K]

Answer:

10 kg

Step-by-step explanation:

In 3% concentration of 20 kg of seawater, there will be $\frac{20 \times 3}{100} = 0.6$ kg of salt.

So, if it is a 2% solution of seawater,

Then 2 kg salt will be there in 100 kg solution.

So, 0.6 kg salt will be there in $\frac{100 \times 0.6}{2}= 30$ kg of solution.

So, to make a solution from 3% to 2% we have to add (30 - 20) = 10 kg of water with a 20 kg solution. (Answer)

6 0

3 years ago

a boat takes 2 hours to travel 15 miles upriver against the current. if the rate of the boat in still water is 12 miles per hour

liq [111]

Answer:

7 or 8 miles per hour

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Mr. Neiland bought 12 tickets to a chili supper and spent a total of $60. He bought a combination of adult tickets for $10 each

Drupady [299]

Answer:

434.60

Step-by-step explanation:

7 0

3 years ago

What is the result of substituting for y in the bottom equation y=x+3 y=x^2+2x-4

8_murik_8 [283]

The solutions are (2.1925, 5.1925) and (-3.1925, -0.1925)

Solution:

Given that,

$y = x + 3 ------- eqn 1\\\\y = x^2 + 2x - 4 ----- eqn 2$

We have to substitute eqn 1 in eqn 2

$x + 3 = x^2 + 2x - 4$

$\mathrm{Switch\:sides}\\\\x^2+2x-4=x+3\\\\\mathrm{Subtract\:}3\mathrm{\:from\:both\:sides}\\\\x^2+2x-4-3=x+3-3\\\\\mathrm{Simplify}\\\\x^2+2x-7=x\\\\\mathrm{Subtract\:}x\mathrm{\:from\:both\:sides}\\\\x^2+2x-7-x=x-x\\\\\mathrm{Simplify}\\\\x^2+x-7=0$

$\mathrm{Solve\:with\:the\:quadratic\:formula}\\\\\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\mathrm{For\:}\quad a=1,\:b=1,\:c=-7\\\\x =\frac{-1\pm \sqrt{1^2-4\cdot \:1\left(-7\right)}}{2\cdot \:1}$

$x = \frac{-1 \pm \sqrt{ 1 + 28}}{2}\\\\x = \frac{ -1 \pm \sqrt{29}}{2}$

$x = \frac{ -1 \pm 5.385 }{2}\\\\We\ have\ two\ solutions\\\\x = \frac{ -1 + 5.385 }{2}\\\\x = 2.1925$

$Also\\\\x = \frac{ -1 - 5.385 }{2}\\\\x = -3.1925$

Substitute x = 2.1925 in eqn 1

y = 2.1925 + 3

y = 5.1925

Substitute x = -3.1925 in eqn 1

y = -3.1925 + 3

y = -0.1925

Thus the solutions are (2.1925, 5.1925) and (-3.1925, -0.1925)

6 0

4 years ago

Juan invest $3700 in a simple interest account at a rate of 4% for 15 years

OleMash [197]

Question:

Juan Invest $3700 In A Simple Interest Account At A Rate Of 4% For 15 Years. How Much Money Will Be In The Account After 15 Years?

Answer:

There will be $ 5920 in account after 15 years

Solution:

The simple interest is given by formula:

$S.I = \frac{p \times n \times r }{100}$

Where,

p is the principal

n is number of years

r is rate of interest

From given,

p = 3700

r = 4 %

t = 15 years

Therefore,

$S.I = \frac{3700 \times 4 \times 15 }{100}\\\\S.I = 37 \times 4 \times 15\\\\S.I = 2220$

How Much Money Will Be In The Account After 15 Years?

Total money = principal + simple interest

Total money = 3700 + 2220

Total money = 5920

Thus there will be $ 5920 in account after 15 years

8 0

4 years ago