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aalyn [17]
3 years ago
11

Need help/easy, please help

Mathematics
2 answers:
vagabundo [1.1K]3 years ago
6 0

what do you even need help with

hichkok12 [17]3 years ago
4 0

Answer:

where is the question

Step-by-step explanation:

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How to factor x^2+8x+12
Setler [38]
<span> x^2+8x+12 = </span>(x + 6)(x + 2)

5 0
3 years ago
Read 2 more answers
How to factor 4x^2+7x-15
GuDViN [60]
4x^2+7x-15

Break the expression into group:

(4x^2+5x)+(12x-15)

factor out x from 4x² - 5x : x (4x-5)
factor out 3 from 12x-15 : 3 (4x-5)

=x(4x-5)+3(4x-5)

factor out common term (4x-5) :

= (4x-5)(x+3)

hope this helps!

4 0
3 years ago
To run a mile, Jamal must run 4 laps around the track. His goal is to run 3 miles. Jamal has run 9 laps so far.
vodka [1.7K]
Don’t really know what your question is but 4 laps times 3 miles is 12 laps. He’s got 3 more laps to finish his goal.

Pro tip: Include the actual question
7 0
3 years ago
Read 2 more answers
What is the value of x in the equation one-fifthx – two-thirdsy = 30, when y = 15? 4 8 80 200
vova2212 [387]

Answer:

x = 200

Step-by-step explanation:

Given

\frac{1}{5} x - \frac{2}{3} y = 30 ← substitute y = 15 into the equation

\frac{1}{5} x - \frac{2}{3} × 15 = 30 , that is

\frac{1}{5} x - 10 = 30 ( add 10 to both sides )

\frac{1}{5} x = 40 ( multiply both sides by 5 to clear the fraction )

x = 200

5 0
3 years ago
Please prove this........​
Crazy boy [7]

Answer:  see proof below

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

Given: A + B + C = π    →     C = π - (A + B)

                                    → sin C = sin(π - (A + B))       cos C = sin(π - (A + B))

                                    → sin C = sin (A + B)              cos C = - cos(A + B)

Use the following Sum to Product Identity:

sin A + sin B = 2 cos[(A + B)/2] · sin [(A - B)/2]

cos A + cos B = 2 cos[(A + B)/2] · cos [(A - B)/2]

Use the following Double Angle Identity:

sin 2A = 2 sin A · cos A

<u>Proof LHS → RHS</u>

LHS:                        (sin 2A + sin 2B) + sin 2C

\text{Sum to Product:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-\sin 2C

\text{Double Angle:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-2\sin C\cdot \cos C

\text{Simplify:}\qquad \qquad 2\sin (A + B)\cdot \cos (A - B)-2\sin C\cdot \cos C

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

\text{Factor:}\qquad \qquad \qquad 2\sin C\cdot [\cos (A-B)+\cos (A+B)]

\text{Sum to Product:}\qquad 2\sin C\cdot 2\cos A\cdot \cos B

\text{Simplify:}\qquad \qquad 4\cos A\cdot \cos B \cdot \sin C

LHS = RHS: 4 cos A · cos B · sin C = 4 cos A · cos B · sin C    \checkmark

7 0
3 years ago
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