All categories

3 years ago

Someone help me please

Mathematics

1 answer:

matrenka [14]3 years ago

8 0

Answer:

n=-2

Step-by-step explanation:

You might be interested in

The leg of a right triangle is 4 units and the hypotenuse is 9 units. What is the length, in units, of the other leg of the tria

Annette [7]

3) 65 square root is the answer

8 0

3 years ago

Read 2 more answers

Identifying Rays and Segments

hichkok12 [17]

The answer is a, b, and e.

5 0

3 years ago

Read 2 more answers

PLEASE EEEEE !!!!!!!!!!!! I NEED IT

tino4ka555 [31]

Answer:

1) ∠A=84°

2) ∠C=20°

Step-by-step explanation:

First, find ∠C:

(I'm assuming the exterior angle of 126° makes a straight line with ∠C)

The angles on a straight line always add up to 180. Therefore:

∠C+126=180

∠C=180-126

∠C=54

Then find ∠B:

We also know that all the angles in a triangle add up to 180. Therefore:

∠A+∠B+∠C=180

∠A+∠B+54=180

∠A+∠B=126

(we know ∠A=2(∠B))

2(∠B)+∠B=126

3(∠B)=126

∠B=42

Now, find ∠A:

∠A=2(∠B)

∠A=2(42)

∠A=84°

First, find ∠B:

(Again, I'm assuming the exterior angle of 100° makes a straight line with ∠B)

The angles on a straight line always add up to 180. Therefore:

∠B+100=180

∠B=180-100

∠B=80

Then find ∠A:

We also know that all the angles in a triangle add up to 180. Therefore:

∠A+∠B+∠C=180

∠A+80+∠C=180

∠A+∠C=100

(we know ∠A=4(∠C))

4(∠C)+∠C=100

5(∠C)=100

∠C=20°

6 0

3 years ago

How many 4 card hands that contain exactly 3 aces and 1 queen can be chosen from a 52 card deck?

sesenic [268]

There are 16 possible hands.

Choosing 3 aces from 4 possible is expressed by:
$_4C_3=\frac{4!}{3!1!} = 4$

Choosing 1 queen from 4 possible is expressed by:
$_4C_1=\frac{4!}{1!3!}=4$

This gives us 4*4 = 16 possible hands.

7 0

3 years ago

35 solve the following syste

blondinia [14]

Part A: The solution is $(-0.923,5.692)$

Part B: The point $(3,7)$ is not in the solution set.

Explanation:

Part A: The given inequalities are $3 x+4 y>20$ and $x$

The solution can be determined by solving the two inequalities by substitution method.

Changing inequalities to equality, we have,

$x=3 y-18$ and $3 x+4 y=20$

Let us substitute $x=3 y-18$ in the equation $3 x+4 y=20$ , we get,

$3 (3y-18)+4 y=20$

$9y-54+4y=20$

$13y=74$

$y=5.692$

Substituting $y=5.692$ in $x=3 y-18$ , we get,

$x=3 (5.692)-18$

$=17.076-18$

$x=-0.923$

Thus, the solution set is $(-0.923,5.692)$

Part B: Now, we shall determine whether the point $(3,7)$ is in the solution set.

Let us substitute the point $(3,7)$ in the inequalities $3 x+4 y>20$ and $x$ , we get,

$3 (3)+4 (7)>20$

$9+28>20$

$37>20$

Also, substituting $(3,7)$ in $x$ , we get,

$3$

Since, the point $(3,7)$ does not satisfy one of the inequality $x$ , the solution set does not contain the point $(3,7)$

Thus, the point $(3,7)$ is not in the solution set.

8 0

3 years ago