It's math again! Please help!

Marie has a small copy of Rene Magritte's famous painting, The Schoolmaster. Her copy has dimensions 2 inches by 1.5 inches. The

Alex73 [517]

A. Find the dimensions of the original painting.

Since the scale is 1 in : 40 cm

2 in ( 40 cm / 1 in) = 80 cm

1.5 in ( 40 cm / 1 in) = 60 cm

80 cm by 60 cm

B. Find the area of the original painting.

A = 80 cm x 60 cm = 4800 sq cm

C. Since 1 inch is 2.54 cm, find the dimensions of the original painting in inches.

80 cm ( 1 in / 2.54 cm) = 31.5 in

60 cm ( 1 in / 2.54 cm) = 23.6 in

D, Find the area of the original painting in square inches.

A = 31.5 in x 23.6 in = 743.4 sq in

5 0

4 years ago

Reduced to lowest terms. 44 nickels to 11 dimes

Finger [1]

Answer:

4 nickels to 1 dime

Step-by-step explanation:

the steps are in the pic above I gotta get this to 20 characters bye!

3 0

3 years ago

Read 2 more answers

Colton has 32 blue marbles and 24 white marbles. If he wants to put them in identical groups without any left over, what is the

Alja [10]

The awnser is 50 because 32 divided by 47 is 50

8 0

3 years ago

Read 2 more answers

100 points! Mhanifa can you please help? Look at the picture attached. I will mark brainliest!

dimulka [17.4K]

Answer:

1 and 2.

Midpoints calculated, plotted and connected to make the triangle DEF, see the attached.

D= (-2, 2), E = (-1, -2), F = (-4, -1)

3.

As per definition, midsegment is parallel to a side.

Parallel lines have same slope.

Find slopes of FD and CB and compare.

m(FD) = (2 - (-1))/(-2 -(-4)) = 3/2
m(CB) = (1 - (-5))/(1 - (-3)) = 6/4 = 3/2
As we see the slopes are same

Find the slopes of FE and AB and compare.

m(FE) = (-2 - (- 1))/(-1 - (-4)) = -1/3
m(AB) = (1 - 3)/(1 - (-5)) = -2/6 = -1/3
Slopes are same

Find the slopes of DE and AC and compare.

m(DE) = (-2 - 2)/(-1 - (-2)) = -4/1 = -4
m(AC) = (-5 - 3)/(-3 - (-5)) = -8/2 = -4
Slopes are same

4.

As per definition, midsegment is half the parallel side.

We'll show that FD = 1/2CB

FD = $\sqrt{(2+1)^2+(-2+4)^2}$ = $\sqrt{3^2+2^2}$ = $\sqrt{13}$
CB = $\sqrt{(1 + 5)^2+(1+3)^2}$ = $\sqrt{6^2+4^2}$ = 2 $\sqrt{13}$
As we see FD = 1/2CB

FE = 1/2AB

FE = $\sqrt{(-4+1)^2+(-1+2)^2}$ = $\sqrt{3^2+1^2}$ = $\sqrt{10}$
AB = $\sqrt{(-5 -1)^2+(3-1)^2}$ = $\sqrt{6^2+2^2}$ = 2 $\sqrt{10}$
As we see FE = 1/2AB

DE = 1/2AC

DE = $\sqrt{(-2+1)^2+(2+2)^2}$ = $\sqrt{1^2+4^2}$ = $\sqrt{17}$
AC = $\sqrt{(-5 +3)^2+(3+5)^2}$ = $\sqrt{2^2+8^2}$ = 2 $\sqrt{17}$
As we see DE = 1/2AC

3 0

3 years ago

Read 2 more answers

A credit card company offers a cash back program on transactions. For every transaction over $100 the customer gets $3 back. F

Elena L [17]

The first equation is $3 times 9 = $27 the next equation iis $1 times 2 = $1
So you do 9 transactions over $100 and 2 transactions of $100 or lower.

5 0

3 years ago

Read 2 more answers