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iVinArrow [24]
3 years ago
8

Which of the following verifies that A COWis similar to APIG?

Mathematics
1 answer:
pashok25 [27]3 years ago
6 0

Answer:

Step-by-step explanation:

In short, the answer is choice D. It cannot be determined. There isn't enough informatioIn short, the answer is choice D. It cannot be determined. There isn't enough information. n.

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Find the image of c under the translation described by the translation rule T<4,5> (C)
iVinArrow [24]

Answer:

A. Point E

Step-by-step explanation:


7 0
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13. The total cost of 3 equally-priced baseball caps is $48. How
S_A_V [24]
The answer to the question is A
8 0
3 years ago
3x + 2y = 4
MrRissso [65]

Answer:

B, "it has one solution"'

(2,-1)

8 0
4 years ago
Read 2 more answers
Kaia rewrote the sum 96+12 as 12 (8+1). She used the same method to rewrite the sum 95+15 as 5(19+3). Describe her method and te
djverab [1.8K]
When Kaira wrote 96 +12 as 12(8+1), she "factorized" 12, that is Karira found that 12 is a factor of both 96 and 12.


then she wrote 95+15=5(19+3), that is she factorized 5,

5 is a factor of both 95 and 15 because 95=5* 19 and 15=5*3



To use this method to calculate 38+11 we need 38 and 11 to have some common factor (different from 1),

but 11 is a prime number, that is, it has no factors except 1 and 11. We can only apply the method if 11 is a factor of 38, but it is not.


Answer: The method is called "factorization", and it cannot be applied to calculate 38+11
6 0
4 years ago
A townhouse in San Francisco was purchased for $80,000 in 1975. The appreciation of the building is modeled by the equation: A=8
jonny [76]

Answer:

In 1981 was the building worth double it’s value.

Step-by-step explanation:

Given : A townhouse in San Francisco was purchased for $80,000 in 1975. The appreciation of the building is modeled by the equation : A=80000(1.12)^t, where t represents time in years.

To find : In what year was the building worth double it’s value in 1975?

Solution :

The amount is $80,000.

The building worth double it’s value in 1975.

i.e. amount became A=2(80000).

Substitute in the model,

2(80000)=80000(1.12)^t

(1.12)^t=\frac{2(80000)}{80000}

(1.12)^t=2

Taking log both side,

t\log (1.12)=\log 2

t=\frac{\log 2}{\log (1.12)}

t=6.11

i.e. Approx in 6 years.

So, 1975+6=1981

Therefore, in 1981 was the building worth double it’s value.

4 0
4 years ago
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