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kobusy [5.1K]
3 years ago
15

Which triangles are similar? which theorem do you use to prove the similarity?​

Mathematics
1 answer:
AURORKA [14]3 years ago
7 0
I think the similar triangle is d and b for this question
You might be interested in
Please i need help with this one ​
Vikentia [17]

Answer:

center: (1,4)

radius=3

8 0
3 years ago
Which way shows one way you can decompose 15 to find 42 x 15?
soldier1979 [14.2K]

Answer:

42 * 10 + 42 * 5

which agrees with the third option (C) shown on the list of possible answers

Step-by-step explanation:

Notice that using distributive property we can write:

42 * 15 = 42 (10 + 5) = 42 * 10 + 42 * 5

which agrees with the third option (C) shown on the list of possible answers.

5 0
3 years ago
Solve the equation by factoring: x^2-8x+15=0
blsea [12.9K]

Answer:

that is your solution

hope it helps

4 0
3 years ago
For what value(s) of k is the linear system consistent? (Enter your answers as a comma-separated list.)
AleksAgata [21]

Answer:

Step-by-step explanation:

Given

6x_1-9x_2=8

9x_1+kx_2=-1

The given system is AX=B can be represented by

\begin{bmatrix}6 &-9 \\ 9 & k\end{bmatrix}\begin{bmatrix}x_1\\ x_2\end{bmatrix}=\begin{bmatrix}8\\ -1\end{bmatrix}

The given system is consistent when determinant of A is not equal to zero

|A|

|A|=6k-(-81)=6k+81

k\neq \frac{-27}{2}

i.e. system is consistent for all value of k except k=\frac{-27}{2}

R-\frac{-27}{2}

                 

4 0
4 years ago
Read 2 more answers
In a standard normal distribution, what z value corresponds to 17% of the data between the mean and the z value?
hjlf
You're looking for a value z such that

\mathbb P(0

Because the distribution is symmetric, the value of z in either case will be the same.

Now, because the distribution is continuous, you have that

0.17=\mathbb P(0

The mean for the standard normal distribution is 0, and because the distribution is symmetric about its mean, it follows that \mathbb P(Z.

0.17=\mathbb P(Z

You can consult a z score table to find the corresponding score for this probability. It turns out to be z\approx0.4399.
4 0
3 years ago
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