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Monica [59]
3 years ago
14

ALGEBRA 1:

Mathematics
1 answer:
gtnhenbr [62]3 years ago
3 0

Answer:

4x^2-14x+6

Step-by-step explanation:

(4x-2)(x-3)

*use box method

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A rectangles width is 5 feet less than its length. Write a quadratic function that expresses the rectangles area in terms of its
Ira Lisetskai [31]

Answer:

Area = L²-5L

Step-by-step explanation:

Area = length * width

Length = L

Width = L - 5

Area = L * (L-5) = L²-5L

4 0
3 years ago
what are the least common denominator of the four fractions listed below? 20 7\10 , 20 3\4 , 18 9\10 ,20 18\25. A:2. B: 40 C: 50
faust18 [17]
A, I think
none of them have anything higher than a 2, so yeah
8 0
3 years ago
Identify the part, whole and percent in the following statement: Find 15% of 750
erma4kov [3.2K]
You have to multiply 750*100 and then divide 75,000 with 15, which gives you 5,000
8 0
3 years ago
16c-24=4c-144<br><br> can you please explain how to do this equation
love history [14]

Answer:

c = -10

Step-by-step explanation:

16c -24=4c -144

Subtract 4c from both sides

16c - 4c - 24 = 4c - 4c - 144

12c - 24 = -144

Add 24 to both sides

12c -24 + 24 = -144 + 24

12c = -120

Note: -144 + 24 = -120

12c = -120

Divide both sides by 12

12c/12 = -120/12

c = -10

Note: 12/12 = 1 and -120/12 = -10

Checking :

16c -24=4c -144

Put c = -10

16(-10) - 24 = 4(-10) - 144

-160 - 24 = -40 - 144

-184 = -184

Therefore

c = -10

I hope this was helpful, please rate as brainliest

6 0
3 years ago
Read 2 more answers
Please help me thank you
Hoochie [10]

Answer:

\large\boxed{\sin2\theta=\dfrac{\sqrt3}{2},\ \cos2\theta=\dfrac{1}{2}}

Step-by-step explanation:

We know:

\sin2\theta=2\sin\theta\cos\theta\\\\\cos2\theta=\cos^2\theta-\sin^2\thet

We have

\sin\theta=\dfrac{1}{2}

Use \sin^2\theta+\cos^2\theta=1

\left(\dfrac{1}{2}\right)^2+\cos^2\theta=1\\\\\dfrac{1}{4}+\cos^2\theta=1\qquad\text{subtract}\ \dfrac{1}{4}\ \text{from both sides}\\\\\cos^2\theta=\dfrac{4}{4}-\dfrac{1}{4}\\\\\cos^2\theta=\dfrac{3}{4}\to\cos\theta=\pm\sqrt{\dfrac{3}{4}}\to\cos\theta=\pm\dfrac{\sqrt3}{\sqrt4}\to\cos\theta=\pm\dfrac{\sqrt3}{2}\\\\\theta\in[0^o,\ 90^o],\ \text{therefore all functions have positive values or equal 0.}\\\\\cos\theta=\dfrac{\sqrt3}{2}

\sin2\theta=2\left(\dfrac{1}{2}\right)\left(\dfrac{\sqrt3}{2}\right)=\dfrac{\sqrt3}{2}\\\\\cos2\theta=\left(\dfrac{\sqrt3}{2}\right)^2-\left(\dfrac{1}{2}\right)^2=\dfrac{3}{4}-\dfrac{1}{4}=\dfrac{3-1}{4}=\dfrac{2}{4}=\dfrac{1}{2}

3 0
3 years ago
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