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riadik2000 [5.3K]
3 years ago
12

A piece of paper is cut and then folded into rectangular prism as shown below.

Mathematics
1 answer:
CaHeK987 [17]3 years ago
5 0
A. C is on the bottom, Bis on the left
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Please help!<br> f(x)=x^2. What is g(x)
lorasvet [3.4K]

Answer:

D

Step-by-step explanation:

The standard form for a parabola is ax^2+bx+c. In this equation, b=0 and c=0. For g(x), you would plug in your y-value that was given (which is 4) and you would plug in your x-value that that given (which is 1). Use PEMDAS. The correct answer is the one that is true.

A. g(x) = (4x)^2

4 = (4(1))^2

4 = 16. (not true)

B. g(x) = 1/4 x^2

4 = 1/4 (1)^2

4 = 1/4 (not true)

C. g(x) = 16x^2

4 = 16(1)^2

4 = 16 (not true)

D. g(x) = 4x^2

4 = 4(1)^2

4 = 4. (true --> correct answer)

4 0
2 years ago
Find the value of sec(theta°)cos(theta°) for the following values of theta. <br> b. theta = 225
Nesterboy [21]

Answer:

sec(theta°)cos(theta°) = 1  

Step-by-step explanation:

given data

(theta°) = 225

to find out

sec(theta°)cos(theta°)

solution

as we know that given equation

(theta°) = 225

cos(theta°) will be

cos(225°) = -0.7071      .................................1

so we know

sec(theta°) = \frac{1}{cos(theta)}      ..............2

so put here value of cos(theta°)

sec(theta°) = \frac{1}{-0.7071}

sec(theta°) = - 1.4142

so

sec(theta°)cos(theta°) = -0.7071  × ( - 1.4142 )

sec(theta°)cos(theta°) = 1

so answer is sec(theta°)cos(theta°) = 1  

8 0
3 years ago
Value of P: 3/15(p+10)=-15
ArbitrLikvidat [17]
<span>Step  1  :</span> 1 Simplify — 5 <span>Equation at the end of step  1  :</span> 1 (— • (p + 10)) - -15 = 0 5 <span>Step  2  :</span><span>Equation at the end of step  2  :</span> (p + 10) ———————— - -15 = 0 5 <span>Step  3  :</span>Rewriting the whole as an Equivalent Fraction :

<span> 3.1 </span>  Subtracting a whole from a fraction 

Rewrite the whole as a fraction using <span> 5 </span> as the denominator :

-15 -15 • 5 -15 = ——— = ——————— 1 5

<span>Equivalent fraction : </span>The fraction thus generated looks different but has the same value as the whole 

<span>Common denominator : </span>The equivalent fraction and the other fraction involved in the calculation share the same denominator

Adding fractions that have a common denominator :

<span> 3.2 </span>      Adding up the two equivalent fractions 
Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

(p+10) - (-15 • 5) p + 85 —————————————————— = —————— 5 5 <span>Equation at the end of step  3  :</span> p + 85 —————— = 0 5 <span>Step  4  :</span>When a fraction equals zero :<span><span> 4.1 </span>   When a fraction equals zero ...</span>

Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero.

Now,to get rid of the <span>denominator, </span>Tiger multiplys both sides of the equation by the denominator.

Here's how:

p+85 ———— • 5 = 0 • 5 5

Now, on the left hand side, the <span> 5 </span> cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :
   p+85  = 0

Solving a Single Variable Equation :

<span> 4.2 </span>     Solve  :    p+85 = 0<span> 

 </span>Subtract  85  from both sides of the equation :<span> 
 </span>                     p = -85 

One solution was found :                  <span> p = -85</span>
7 0
3 years ago
I need this answer asap !!
igomit [66]

Answer:

i believe that it is angle J = 33.5°, angle K = 23.2°, angle L = 123.2°

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
A point P is moving along the curve whose equation is y = \sqrt x . Suppose that x is increasing at the rate of 4 units/s when x
Mice21 [21]

Answer:3 units/s

Step-by-step explanation:

Given

y=\sqrt{x}

Point P lie on this curve so any general point on curve can be written as (x,\sqrt{x})

and \frac{\mathrm{d} x}{\mathrm{d} t}=4 units/s

Distance between Point P and (2,0)

P=\sqrt{(x-2)^2+(\sqrt{x}-0)^2}

P at x=3 P=2

rate at which distance is changing is

\frac{\mathrm{d} P}{\mathrm{d} t}=\frac{\mathrm{d} \sqrt{(x-2)^2+(\sqrt{x}-0)^2}}{\mathrm{d} t}

\frac{\mathrm{d} P}{\mathrm{d} t}=\frac{2x-3}{\sqrt{(x-2)^2+(\sqrt{x}-0)^2}}\times \frac{\mathrm{d} x}{\mathrm{d} t}

\frac{\mathrm{d} P}{\mathrm{d} t}=\frac{2\times 3-3}{2\times 2}\times 4=3 units/s

8 0
3 years ago
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