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Sloan [31]
3 years ago
8

~Have I found, everybody a fun assignment to do today?~

Mathematics
2 answers:
Serga [27]3 years ago
8 0

Answer:

um thanks i guess

Step-by-step explanation:

Furkat [3]3 years ago
8 0
Huh

so confused!!!!!!
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Kiana wants to cover a triangular area of her background with red, concrete patio stones. Each stone costs $0.42 and covers 29.2
Alex73 [517]
1. The problem says that the stone covers 29.26 square inches. So, the first thing you must do, is to convert 29.26 square inches (in²) to square feet (ft²):

 1 in²=0.00694 ft², then:

 29.26x0.00694= 0.20 ft²

 2. The triangular area has the following surface: 

 A=bxh/2
 A=(6 ft)(8 ft)/2
 A=24.0 ft²

 3. <span> Finally, the number of stones (N) that cover the triangular area is:
</span>
 N= 24.0 ft²/0.20 ft²
 N=120 stones

 Answer: <span>Kiana should buy 120 stones to cover the triangular area in her backyard.</span>


 


8 0
4 years ago
Write the fraction as a mixed number 33/8
Dominik [7]

Answer:

4 1/8

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
I do not understand this
iragen [17]

Answer:

D. 2 (1/3) hours

Step-by-step explanation:

'Of' means multiplication, so we are going to multiply

3 (1/2) x (2/3) = 2 (1/3)

or

3.5 x 0.66 = 2.333

I hope this helps!

4 0
2 years ago
The point P(1,1/2) lies on the curve y=x/(1+x). (a) If Q is the point (x,x/(1+x)), find the slope of the secant line PQ correct
lukranit [14]

Answer:

See explanation

Step-by-step explanation:

You are given the equation of the curve

y=\dfrac{x}{1+x}

Point P\left(1,\dfrac{1}{2}\right) lies on the curve.

Point Q\left(x,\dfrac{x}{1+x}\right) is an arbitrary point on the curve.

The slope of the secant line PQ is

\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{\frac{x}{1+x}-\frac{1}{2}}{x-1}=\dfrac{\frac{2x-(1+x)}{2(x+1)}}{x-1}=\dfrac{\frac{2x-1-x}{2(x+1)}}{x-1}=\\ \\=\dfrac{\frac{x-1}{2(x+1)}}{x-1}=\dfrac{x-1}{2(x+1)}\cdot \dfrac{1}{x-1}=\dfrac{1}{2(x+1)}\ [\text{When}\ x\neq 1]

1. If x=0.5, then the slope is

\dfrac{1}{2(0.5+1)}=\dfrac{1}{3}\approx 0.3333

2. If x=0.9, then the slope is

\dfrac{1}{2(0.9+1)}=\dfrac{1}{3.8}\approx 0.2632

3. If x=0.99, then the slope is

\dfrac{1}{2(0.99+1)}=\dfrac{1}{3.98}\approx 0.2513

4. If x=0.999, then the slope is

\dfrac{1}{2(0.999+1)}=\dfrac{1}{3.998}\approx 0.2501

5. If x=1.5, then the slope is

\dfrac{1}{2(1.5+1)}=\dfrac{1}{5}\approx 0.2

6. If x=1.1, then the slope is

\dfrac{1}{2(1.1+1)}=\dfrac{1}{4.2}\approx 0.2381

7. If x=1.01, then the slope is

\dfrac{1}{2(1.01+1)}=\dfrac{1}{4.02}\approx 0.2488

8. If x=1.001, then the slope is

\dfrac{1}{2(1.001+1)}=\dfrac{1}{4.002}\approx 0.2499

7 0
3 years ago
The Legs of a right triangle are lengths x and x square root of 3. The cosine of the smallest angle is
ki77a [65]

Answer:

Step-by-step explanation:

hypotenuse=√(x²+(x√3)²)=√(x²+3x²)=√(4x²)=2x

sides are x,x√3,2x

smallest side=x

smallest angle=opposite smallest side

let smallest angle=α

cos α=(x√3)/2x=√3/2

4 0
3 years ago
Read 2 more answers
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