All categories

3 years ago

Helppppppppppppppppppppp

Mathematics

2 answers:

yawa3891 [41]3 years ago

8 0

Answer:

$7.20

Step-by-step explanation:

60% of 4.5 = 2.7

4.5+2.7 = $7.2

Hitman42 [59]3 years ago

8 0

Answer:

7.2

Step-by-step explanation:

So u do 60% of 4.5 and add ur answer to 4.5

You might be interested in

Help pls math ppl<br> Thank

Marianna [84]

Answer: you are correct

Step-by-step explanation:

7 0

2 years ago

Triangle ABC has the coordinates A(-3, -2), B(2, 4) and C(2,-2). What is the area of triangle ABC?

Snezhnost [94]

Answer: The area is 15 square units

Step-by-step explanation: It is helpful to sketch the coordinates on a graph to get an idea of what the triangle looks like so you can see which side will be the base, and where to get the height. You will see that AB is the hypotenuse, it is a right triangle, so AC can be the base and BC is the height.

Use the differences in x-values of points C & A to get the length of the base: 2-(-3) is 5

Use the differences in y-values of points B & C to get the height:

4-(-2) = 6

Area = bh/2

A = 5×6/2 = 30/2

A = 15 square unts

3 0

3 years ago

Rectangle A is a scale drawing of Rectangle B and has 25% of its area. If rectangle A has side lengths of 4 cm and 5 cm , what a

SpyIntel [72]

Answer:

8 cm and 10 cm

Step-by-step explanation:

Hello, <em> </em>I can help you with this.

Step 1

According to the question there are two rectangles A and B,

Rectangle A is a scale drawing of Rectangle B and has 25% of its area

in other words

$Area_{A}=0.25*Area_{B} (Equation\ 1)\\$

Step 2

Let

Rectangle A

length (1)= 4 cm

length (2)= 5 cm

$Area_{A}=4\ cm * 5\ cm\\Area_{A}=20\ cm^{2}$

put this value into equation 1

$Area_{A}=0.25*Area_{B} (Equation\ 1)\\\\20\ cm^{2} =0.25*Area_{B} \\divide\ each\ side\ by\ 0.25\\\frac{20\ cm^{2} }{0.25}=\frac{0.25}{0.25}*Area_{B}\\ Area_{B}=80\ cm^{2}$

Now, we know the area of rectangle B, to know its length we need to formule other equation

Step 3

$Area_{B}=80\ cm^{2}\\length (1B)*length (2B)=80\ cm^{2} (equation\ 2)\\$

the ratio between the lengths must be constant, so the ratio of A must be equal to ratio in B, then

$\frac{length(1A)}{length(2A)}=\frac{length(1B)}{length(2B)} \\\\\\frac{4}{5}= \frac{length(1B)}{length(2B)}\\0.8=\frac{length(1B)}{length(2B)}\\length(1B)=0.8*length(2B) (Equation 3)$

Step three

using Eq 1 and Eq 2 find the lengths

put the value of length(1B) into equation (2)

$length (1B)*length (2B)=80\ cm^{2} (equation\ 2)\\\(0.8*length(2B)) (*length (2B)=80\ cm^{2} \\\\0.8*(length (2B))^{2} =80\ cm^{2}\\(length (2B))^{2} =\frac{80\ cm^{2}}{0.8} \\(length (2B))^{2}=100\\\sqrt{(length (2B))^{2}}=\sqrt{100\ cm^{2}} \\ length (2B)=10\ cm$