Are angles 4 and 12 congruent supplementary or neither

7x - 5y = -24 -9x + 5y = 18

densk [106]

Answer:

Nothing further can be done with this topic. Please check the expression entered or try another topic.

7x−5y=−24−9x+5y=187x-5y=-24-9x+5y=18

6 0

3 years ago

Read 2 more answers

Consider the quadratic equation x squared = 4x - 5. how many solutions does the equation have?

snow_lady [41]

Answer:

The quadratic equation has two complex solutions

Step-by-step explanation:

we know that

The formula to solve a quadratic equation of the form $ax^{2} +bx+c=0$ is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$x^{2}=4x-5$

Equate to cero

$x^{2}-4x+5=0$

so

$a=1\\b=-4\\c=5$

substitute in the formula

$x=\frac{4(+/-)\sqrt{-4^{2}-4(1)(5)}} {2(1)}$

$x=\frac{4(+/-)\sqrt{16-20}} {2}$

$x=\frac{4(+/-)\sqrt{-4}} {2}$

Remember that

$i^{2}=-1$

so

$x=\frac{4(+/-)2i} {2}$

$x=2(+/-)i$

therefore

The quadratic equation has two complex solutions

5 0

4 years ago

Assume the probability is 1/2 that a child born at any time is a boy and births are independent. What is the probability of the

melisa1 [442]

There are only 23=823=8 possible ways to arrange the genders boy/girl, with repetition. Since the probability of boy and girl are equal, the probability of each of these arrangements are also equal. We can therefore count the number of possible "good" answers, and divide by 8.

There are 3 possible ways to have 1 boy, that being "first/second/third child is boy", so the probability of exactly 1 boy is indeed 3838.

Having at most 2 girls is the opposite of having three girls, so since there are only one way of having three girls, there must be 8−1=78−1=7 ways of having at most two girls, so the probability of at most two girls is indeed 78

8 0

3 years ago

You bike 1 mile the first day of your training, 1.3 miles the second day, 1.9 miles the third day, and 3.1 miles the fourth d

JulsSmile [24]

Answer:

19.9 miles

Step-by-step explanation:

In this problem we have:

$d_1=1 mi$ is the distance travelled during the 1st day

$d_2=1.3 mi$ is the distance travelled during the 2nd day

$d_3=1.9 mi$ is the distance travelled during the 3rd day

$d_4=3.1 mi$ is the distance travelled during the 4th day

We notice that the difference between the distance travelled on the (n+1)-th day and the distance travelled on the n-th day doubles every day. In fact:

$d_2-d_1=0.3\\d_3-d_2=2\cdot 0.3 = 0.6\\d_4-d_3=2\cdot 0.6 = 1.2$