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3 years ago

The ratio of 3 to 4 is written as 4/3. TrueFalse

Mathematics

2 answers:

irga5000 [103]3 years ago

8 0

False. It’s should be 3/4

Iteru [2.4K]3 years ago

5 0

Answer: False

Step-by-step explanation:

3:4

3/4

3to 4

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n △ABC, point P∈ AB is so that AP:BP=1:3 and point M is the midpoint of segment CP. Find the area of △ABC if the area of △BMP is

monitta

Answer: The area of ABC is 56 m².

Explanation:

It is given that in △ABC, point P∈ AB is so that AP:BP=1:3 and point M is the midpoint of segment CP.

Since point P divides the line AB in 1:3, therefore the area of triangle APC and BPC is also in ratio 1:3. To prove this draw a perpendicular h on AB from C.

$\frac{\text{Area of } \triangle BCP}{\text{Area of } \triangle ABC} =\frac{\frac{1}{2}\times BP\times CH}{\frac{1}{2}\times AB\times CH} =\frac{BP}{AB}= \frac{3}{4}$

Since the area of BPC is $\frac{3}{4}$ th part of total area, therefore area of APC is $\frac{1}{4}$ th part of total area.

The point M is the midpoint of CP, therefore the area of BMP and BMC is equal by midpoint theorem.

$\text{Area of } \triangle BMP=\text{Area of } \triangle BMC$

$21=\text{Area of } \triangle BMC$

Area of BPC is,

$\text{Area of } \triangle BPC=\text{Area of } \triangle BMP+\text{Area of } \triangle BMC$

$\text{Area of } \triangle BPC=21+21$

$\text{Area of } \triangle BPC=42$

Area of APC is,

$\text{Area of } \triangle APC=\frac{1}{3}\times \text{Area of } \triangle BPC$

$\text{Area of } \triangle APC=\frac{1}{3}\times 42$

$\text{Area of } \triangle APC=14$

Area of ABC is,

$\text{Area of } \triangle ABC=\text{Area of } \triangle APC+\text{Area of } \triangle BPC$

$\text{Area of } \triangle ABC=14+42=56$

Therefore, the area of ABC is 56 m².

5 0

3 years ago

Inez has saved $5. She doubles the amount she saves each week. Does this represent an exponential function?

lozanna [386]

Yes it is because the value is constantly raising.

4 0

2 years ago

720÷90=72 tens ÷ 9 tens = what?

Elden [556K]

720 divided by 72 is 80
7 tens divided by 9 is 8

6 0

3 years ago

Suppose both factors in a multiplication problem are multiples of 10. Why might the number of zeroes in the product be different

Lelu [443]

Answer:

Multiplying factors whose products are multiples of 10 add to the number of zeros when factors are multiplied as multiples of 10s.

Step-by-step explanation:

Let the first five multiples of ten be

10*1= 10

10*2= 20

10*3=30

10*4=40

10*5= 50

Suppose we chose 20 and 50.

Now multiplying 20 with 50 we get

20*50= 1000

IF we count the total number of zeros in the factors ( 20 and 50) they are 2.

But the number of zeros in the product (1000) are 3.

This is because when we multiply 2 with 5 we get 10 which adds to the existing number of zeros ( i.e 2) and we get a total of 3 zeros.

And multiplying 10 with 50 we get

10*50= 500

IF we count the total number of zeros in the factors ( 10 and 50) they are 2.

But the number of zeros in the product (500) are also 2.

This is because when we multiply 1 with 5 we get 5 which does not add to the existing number of zeros ( i.e 2) and the total number of zeros remain the same.

Similarly multiplying 20 with 30 we get

20*30= 600

IF we count the total number of zeros in the factors ( 20 and 30) they are 2.

But the number of zeros in the product (600) are also 2.

This is because when we multiply 2 with 3 we get 6 which does not have a zero and the total number of zeros remain the same as in the factors.

So we see that multiplying factors whose products are multiples of 10 add to the number of zeros when factors are multiplied as multiples of 10s.

5 0

3 years ago

Please help me with this!!

Aleonysh [2.5K]

The perimeter is 1/2 the perimeter of ABCD because the scale factor is 1/2. It would be 10 units.

7 0

3 years ago