3 years ago

anyone know the answer i worked it out but didn't know where i messed up so i am just gonna ask for someone to help me with it.

1 answer:

5 0

I also worked it out but I got -5/3

2(6p-5) ≥ 3(p-8) -1

12p-10 ≥ 3p-24-1

12p-10 ≥ 3p-25

12p-3p ≥ -25+10

9p ≥ -15

-15/9

P ≥ -5/3

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Solve for the smallest value of $x$ such that $24x^2 + 17x - 20 = 0.$ Express your answer as a simplified common fraction.

Gala2k [10]

X=5/8 or x=-4/3
here is the work if needed...
24x^2+17x-20=0
Step one: Factor left side of equation.
(8x-5)(3x-4)=0
Step two: Set factor equal to 0.
8x-5=0 or 3x-4=0
x=5/8 or x=-4/3

4 0

3 years ago

Find the values of x and y. Pls helpI’m helpless at math

Svetradugi [14.3K]

Answer:

x = 1

$y=\frac{\sqrt{3}}{2}$

Step-by-step explanation:

From the given right-angle triangle

The angle = ∠60°

The adjacent to the angle ∠60° is 1/2.

The opposite to the angle ∠60° is y.

The hypotenuse = x

<u>Determining the value of x:</u>

Using the trigonometric ratio

cos 60° = adjacent / hypotenuse

substituting adjacent = 1/2 and hypotenuse = x

$cos\:60^{\circ }=\:\:\frac{\frac{1}{2}}{x}$

$\cos \left(60^{\circ \:}\right)=\frac{1}{2x}$

$\frac{1}{2}=\frac{1}{2x}$ ∵ cos (60°) = 1/2

$2x=2$

Dividing both sides by 2

$\frac{2x}{2}=\frac{2}{2}$

Simplify

$x=1$

Thus, the value of hypotenuse x is:

x = 1

<u>Determining the value of y:</u>

Using the trigonometric ratio

sin 60° = opposite / hypotenuse

As we have already determined the value of hypotenuse x = 1

substituting opposite = y and hypotenuse = 1

sin 60° = y/1

y = 1 × sin 60°

$y=\frac{\sqrt{3}}{2}$ ∵ $\sin \left(60^{\circ \:}\right)=\frac{\sqrt{3}}{2}$

Therefore, the value of y is:

$y=\frac{\sqrt{3}}{2}$

Summary:

x = 1

$y=\frac{\sqrt{3}}{2}$

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3 years ago

Debra has ridden 57 miles of a bike course the course is 60 miles long what percentage of the course has Debra ridden so far

Reptile [31]

Hello!

To find our answer, we just divide.

57/60=0.95

We multiply by 100 to find our percent

100(0.95)-95

Therefore, our answer is 95%

I hope this helps!

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3 years ago

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