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horrorfan [7]
2 years ago
5

Please help me with this math problem!! Will give brainliest!! NO LINKS PLEASE!! Thanks!! :)

Mathematics
1 answer:
iren2701 [21]2 years ago
3 0

Answer:

The sum of all the sides to this quadrilateral will equal a total of 20 inches, so you set up the equation:

2 + 2x + 7 + x + 2 = 20

Solve for x by combining like terms;

(2 + 7 + 2) + (2x + x) = 20

Which simplifies to;

11 + 3x = 20

Get rid of constants;

-11          -11

3x = 9

Isolate x by getting rid of it's coefficient;

/3  /3

<u>x = 3</u>

Now, to find the length of PQ, we plug in the value of x(3) into the equation that contains the line PQ.

x + 2

<u>3</u> + 2

= <u>5 inches.</u>

Now, to find the length of RS, we plug in the value of x(3) into the equation that contains the line RS.

2x

2(3)

= <u>6 inches.</u>

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Nastasia [14]

Answer:

3

Step-by-step explanation:

7 -4 = 3

5 0
3 years ago
2x+3y=-12<br> 2x+t=-16<br> X=. Y=<br> Help plz
8090 [49]

<u>To make this problem solvable, I have replaced the 't' in the second equation for a 'y'.</u>

Answer:

<em>x = -9</em>

<em>y = 2</em>

Step-by-step explanation:

<u>Solve the system:</u>

2x + 3y = -12       [1]

2x + y = -16       [2]

Subtracting [1] and [2]:

3y - y = -12 + 16

2y = 4

y = 4/2 = 2

From [1]:

2x + 3(2) = -12

2x + 6 = -12

2x = -18

x = -18/2 = -9

Solution:

x = -9

y = 2

8 0
3 years ago
Please help me to prove this!​
Sophie [7]

Answer:  see proof below

<u>Step-by-step explanation:</u>

Given: A + B = C                → A = C - B

                                          → B = C - A

Use the Double Angle Identity:     cos 2A = 2 cos² A - 1

                                             → (cos 2A + 1)/2 = cos² A

Use Sum to Product Identity: cos A + cos B = 2 cos [(A + B)/2] · 2 cos [(A - B)/2]

Use Even/Odd Identity: cos (-A) = cos (A)

<u>Proof LHS → RHS:</u>

LHS:                     cos² A + cos² B + cos² C

\text{Double Angle:}\qquad \dfrac{\cos 2A+1}{2}+\dfrac{\cos 2B+1}{2}+\cos^2 C\\\\\\.\qquad \qquad \qquad =\dfrac{1}{2}\bigg(2+\cos 2A+\cos 2B\bigg)+\cos^2 C\\\\\\.\qquad \qquad \qquad =1+\dfrac{1}{2}\bigg(\cos 2A+\cos 2B\bigg)+\cos^2 C

\text{Sum to Product:}\quad 1+\dfrac{1}{2}\bigg[2\cos \bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A-2B}{2}\bigg)\bigg]+\cos^2 C\\\\\\.\qquad \qquad \qquad =1+\cos (A+B)\cdot \cos (A-B)+\cos^2 C

\text{Given:}\qquad \qquad 1+\cos C\cdot \cos (A-B)+\cos^2C

\text{Factor:}\qquad \qquad 1+\cos C[\cos (A-B)+\cos C]

\text{Sum to Product:}\quad 1+\cos C\bigg[2\cos \bigg(\dfrac{A-B+C}{2}\bigg)\cdot \cos \bigg(\dfrac{A-B-C}{2}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =1+2\cos C\cdot \cos \bigg(\dfrac{A+(C-B)}{2}\bigg)\cdot \cos \bigg(\dfrac{-B-(C-A)}{2}\bigg)

\text{Given:}\qquad \qquad =1+2\cos C\cdot \cos \bigg(\dfrac{A+A}{2}\bigg)\cdot \cos \bigg(\dfrac{-B-B}{2}\bigg)\\\\\\.\qquad \qquad \qquad =1+2\cos C \cdot \cos A\cdot \cos (-B)

\text{Even/Odd:}\qquad \qquad 1+2\cos C \cdot \cos A\cdot \cos B\\\\\\.\qquad \qquad \qquad \quad =1+2\cos A \cdot \cos B\cdot \cos C

LHS = RHS: 1 + 2 cos A · cos B · cos C = 1 + 2 cos A · cos B · cos C   \checkmark

5 0
3 years ago
A baker makes apple tarts and apple pies each day. Each tart, t, requires 1 apple, and each pie, p, requires 8 apples. The baker
Lynna [10]
For this case, the first thing we must do is define variables.
 We have then:
 t: number of tarts
 p: number of pies
 We now write the system of equations:
 Each tart, t, requires 1 apple, and each pie, p, requires 8 apples. The baker receives a shipment of 184 apples every day:
 8p + t ≤ 184
 the baker makes no more than 40 tarts per day:
 t ≤ 40
 Answer:
 A system of inequalities that can be used to find the possible number of pies and tarts the baker can make is:
 D. t ≤ 40
 8p + t ≤ 184
5 0
3 years ago
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Answer:

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Step-by-step explanation:

5 0
3 years ago
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