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2 years ago

Simplify the expression below

Mathematics

1 answer:

Degger [83]2 years ago

6 0

Answer:

jjjh;kn jhuj hnpjinp; inj in

Step-by-step explanation:

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Can someone explain how to do this cuz my computer ain’t doing such a good job

RideAnS [48]

Answer:

L=36

W=6

Step-by-step explanation:

If the length is six times the width than the width is equal to x. This would make the length 6x or 6 times x. The perimeter of the rectangle is all of the sides added together. So 6x+6x+x+x=14x

The perimeter equals 84 which gives you

14x=84

Divide by 14 on each side and x=6

The length is six times longer than the width which is 6x6 or 36 and the width would just remain 6.

5 0

3 years ago

Read 2 more answers

PQ⊥PS , m∠QPR=7x−9, m∠RPS=4x+22 Find : m∠QPR

emmainna [20.7K]

Given:

It is given that,

PQ ⊥ PS and

∠QPR = 7x-9

∠RPS = 4x+22

To find the value of ∠QPR.

Formula

As per the given problem PR lies between PQ and PS,

So,

∠QPR+∠RPS = 90°

Now,

Putting the values of ∠RPS and ∠QPR we get,

$7x-9+4x+22 = 90$

or, $11x = 90-22+9$

or, $11x = 77$

or, $x = \frac{77}{11}$

or, $x = 7$

Substituting the value of $x = 7$ in ∠QPR we get,

∠QPR = $7(7)-9$

or, ∠QPR = $40^\circ$

Hence,

The value of ∠QPR is 40°.

3 0

3 years ago

I’m 11 years old Jk I’m not I could rlly use a hand tho

horsena [70]

Answer:

880

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

One urn contains one blue ball (labeled B1) and three red balls (labeled R1, R2, and R3). A second urn contains two red balls (R

marusya05 [52]

Answer:

(a) See attachment for tree diagram

(b) 24 possible outcomes

Step-by-step explanation:

Given

$Urn\ 1 = \{B_1, R_1, R_2, R_3\}$

$Urn\ 2 = \{R_4, R_5, B_2, B_3\}$

Solving (a): A possibility tree

If urn 1 is selected, the following selection exists:

$B_1 \to [R_1, R_2, R_3]; R_1 \to [B_1, R_2, R_3]; R_2 \to [B_1, R_1, R_3]; R_3 \to [B_1, R_1, R_2]$

If urn 2 is selected, the following selection exists:

$B_2 \to [B_3, R_4, R_5]; B_3 \to [B_2, R_4, R_5]; R_4 \to [B_2, B_3, R_5]; R_5 \to [B_2, B_3, R_4]$

See attachment for possibility tree

Solving (b): The total number of outcome

For urn 1

There are 4 balls in urn 1

$n = \{B_1,R_1,R_2,R_3\}$

Each of the balls has 3 subsets. i.e.

$B_1 \to [R_1, R_2, R_3]; R_1 \to [B_1, R_2, R_3]; R_2 \to [B_1, R_1, R_3]; R_3 \to [B_1, R_1, R_2]$

So, the selection is:

$Urn\ 1 = 4 * 3$

$Urn\ 1 = 12$

For urn 2

There are 4 balls in urn 2

$n = \{B_2,B_3,R_4,R_5\}$

Each of the balls has 3 subsets. i.e.

$B_2 \to [B_3, R_4, R_5]; B_3 \to [B_2, R_4, R_5]; R_4 \to [B_2, B_3, R_5]; R_5 \to [B_2, B_3, R_4]$

So, the selection is:

$Urn\ 2 = 4 * 3$

$Urn\ 2 = 12$

Total number of outcomes is:

$Total = Urn\ 1 + Urn\ 2$

$Total = 12 + 12$

$Total = 24$

5 0

2 years ago

When angles are complementary the sum of their measures is 90 degrees. Two complementary angles have measured of 2x-10 degrees a

Vinvika [58]

56° and 34°

complementary angles sum to 90° thus

3x - 10 + 2x - 10 = 90

5x - 20 = 90

add 20 to both sides of the equation

5x = 90 + 20 = 110

divide both sides by 5

x = $\frac{110}{5}$ = 22

angle 2x - 10 = (2 × 22) - 10 = 44 - 10 = 34° and

angle 3x - 10 = (3 × 22) - 10 = 66 - 10 = 56°

3 0

3 years ago