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3 years ago

8

In Exercises 10–14, use the diagram to determine whether you can assume the statement.

1 answer:

lukranit [14]3 years ago

7 0

Answer:

it depends on what he means

Step-by-step explanation:

The friend needs to clarify the meaning of "if three lines intersect each other." If Line A intersects lines B and C, there will be two points of intersection, one at line B and one at line C.

If those lines are all in the same plane, and B and C are not parallel, so that line B intersects line C, then there will be a total of three points of intersection.

If the point of intersection of B and C is also the point where line A intersects them, then there will be only one point of intersection.

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So, if the meaning is "if there are three non-parallel lines in the same plane, and each intersects the other two", then the Line Intersection Postulate guarantees there will be 1 or 3 points of intersection.

If the meaning is "if there are three lines not necessarily in the same plane, and one intersects the other two (but those two don't intersect each other)", then there may be 1 or 2 points of intersection (allowing that all lines may intersect at the same point).

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Please help me i dont get it

Vladimir [108]

Answer:

a) (x+2)(x+3)

b) (x+2)(x+3)=0, x=-2 and x=-3

Step-by-step explanation:

3 0

3 years ago

Miki has 104 nickels and 88 dimes. She wants to divide her coins into groups where each group has the same number of nickels and

pochemuha

Given :

Miki has 104 nickels and 88 dimes.

She wants to divide her coins into groups where each group has the same number of nickels and the same number of dimes.

To Find :

Largest number of groups she can have .

Solution :

In the given question we need to find the largest number of groups she can have i.e we have to find the LCM of 104 and 88 .

Now , factorizing both of them , we get :

$104=1\times 2\times 2\times 2\times 13$

$88=1\times 2\times 2\times 2\times 11$

Form above , we can say that common factors are :

$HCF=2\times 2\times 2=8$

Therefore , the largest number of groups she can have is 8 .

Hence , this is the required solution .

6 0

4 years ago

Read 2 more answers

3(s-5)/4 -3 = 4s- (s-3)/2<br>pls answer fast

Jobisdone [24]

Answer:

$s = \frac{9± \sqrt{ - 39} }{4}$

Step-by-step explanation:

$\frac{3(s - 5)}{4 - 3} = \frac{4s \times (s - 3)}{2}$

➡️ $\frac{3(s - 5)}{1} = 2s \times (s - 3)$

➡️ $3s - 15 = 2 {s}^{2} - 6s$

➡️ $3s - 15 - 2 {s}^{2} + 6s = 0$

➡️ $9s - 15 - 2 {s}^{2} = 0$

➡️ $- 2 {s}^{2} + 9s - 15 = 0$

➡️ $2 {s}^{2} - 9s + 15 = 0$

➡️ $s = \frac{ - ( - 9)± \sqrt{( - 9 {)}^{2} - 4 \times 2 \times 15} }{2 \times 2}$

➡️ $s = \frac{9± \sqrt{81 - 120} }{4}$

➡️ $s = \frac{9± \sqrt{ - 39} }{4}$

➡️ $s∉ℝ$

5 0

3 years ago

A corporation must appoint a president, chief executive officer (CEO), chief operating officer (COO), and chief financial off

Travka [436]

Answer:

There are 524,560 ways that the committee can be appointed.

Step-by-step explanation:

There are 9 different positions, a president, 3 officers and five committee members.

So

P - O1 - O2 - O3 - C1 - C2 - C3 - C4 - C5

There are 17 candidates, any of them can be president. So P = 17.

All of the others can be an officers. So O1 = 16, O2 = 15(17 subtracted by the president and O1), O3 = 14.

The officers can be committee members, so, C1 = 16, C2 = 15, C3 = 14, C4 = 13, C5 = 12.

How many different ways can the committee be appointed?

This is the sequence C1...C5:

$N = 16*15*14*13*12 = 524,560$

There are 524,560 ways that the committee can be appointed.

6 0

3 years ago

Taxi charges $0.20 per mile and initial fee of $4. Taxi B charges $0.40 per mile and an initial fee of $2. Write an inequality t

Leni [432]

Answer :
0.20x + 4 < 0.40x + 2

3 0

3 years ago