All categories

3 years ago

I need help on this Kakooma puzzle. Please help me.

Mathematics

1 answer:

Katarina [22]3 years ago

5 0

Using my prior knowledge of the day and the time of the day and she said that I would do anything for you

You might be interested in

It is known that Ray AP bisects angle CAT. If m<CAT = 5x+17 and m<PAT = 3x-4, then determine the value of x.

Sonja [21]

Answer:

x = 25

Step-by-step explanation:

It is given that, ray Ray AP bisects angle CAT. It means,

$\angle CAT=\angle CAP+\angle PAT$

Also, $\angle CAP = \angle PAT$

It gives, $\angle CAT=2\angle PAT$

ATQ,

$\angle CAT = 5x+17 \ \text{and}\ \text{PAT}=3x-4$

So,

$5x+17=2(3x-4)\\\\5x+17=6x-8$

Taking like terms together,

$5x-6x=-8-17\\\\-x=-25\\\\x=25$

So, the value of x is 25.

8 0

3 years ago

Given the arithmetic sequence an = 2 − 5(n + 1), what is the domain for n?

Usimov [2.4K]

The correct answer among all the other choices is "All real numbers." Given the arithmetic sequence an = 2 − 5(n + 1), the domain for n could be all real numbers. <span>Thank you for posting your question. I hope this answer helped you. Let me know if you need more help. </span>

4 0

3 years ago

How does the area change?

kap26 [50]

Garden a is doubled to get garden b

6 0

2 years ago

If 14 g of a radioactive substance are present initially and 7 yr later only 7 g remain, how much of the substance will be pres

Delvig [45]

Answer:

2.60 gr

Step-by-step explanation:

We need to consider the function $P(t) = P_oe^{kt}$ where Po is the initial substance k is the rate of decay and t is the time

We know that $P_o = 14$ and at the 7 th year P(t) is 7

This means

$7 = 14e^{7k}$

We solve for k

$\frac{7}{14} = e^{7k}\\\\\ln (\frac{7}{14}) = 7k\\\\k = \frac{\ln (7/14)}{7} = -0.099\\\\Then \ P(t) = 14e^{-0.099t}\\\\Now \ we \ take \ t = 17\\\\P(17) = 14e^{(-0.099)(17)} = 2.60$

8 0

3 years ago

Please help with this

marta [7]

9514 1404 393

Answer:

see attached

Step-by-step explanation:

13. The difference of two matrices is the term-by-term difference. For example, the bottom right term is the difference of the bottom right terms ...

(-6) -(5) = -11

14. The scalar multiple of a matrix is the scalar multiplied by each term of the matrix. For example, the upper right term is ...

3(1) = 3

15. The matrix product is the dot product of the rows of the left matrix with the columns of the right matrix. For example, the product term in row two, column one is the dot product of row 2 of C and column 1 of B:

[9, -2, 4]·[-2, -7, 10] = 9(-2) +(-2)(-7) +4(10) = -18 +14 +40 = 36

_____

It is convenient to let an appropriate tool help with this. Many graphing calculators and spreadsheets are able to do these functions easily.

7 0

2 years ago