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Salsk061 [2.6K]
3 years ago
11

What type of symmetry does each figure have? Check all of the boxes that apply.

Mathematics
1 answer:
larisa86 [58]3 years ago
7 0

Answer:

The oval is a line and rotational symmetry.

The line shapes rectangle has no symmetry.

The smiling face is a line symmetry.

Step-by-step explanation:

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I'M IN A TIMED TEST... HELP PLZ!!!The volume of this sphere is StartFraction 500 pi Over 3 EndFraction cubic inches. What is its
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It's B ( just did it )

Step-by-step explanation:

Looked up the answer real quick and the person had the same question with a picture.

It's on Ed2020 it's B

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Determine the range of the function.
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Solution: f(x)\ge-5\\Interval:[-5,\infty)

Step-by-step explanation:

Equation: \frac{1}{4} x^2-5 = y

1). \frac{1}{4} x^2-5 = y →\frac{x^2}{4}-5=y

2). \frac{x^2}{4}-5=y ∴ a=\frac{1}{4}, b=0, c=-5

3). x_v=-\frac{b}{2a}

         =-\frac{0}{2(\frac{1}{4})}

         =0

4). now plug x_v into y_v

y_v=\frac{0^2}{4}-5

    =-5

5). Minimum (0,-5) ∴ f(x)\ge-5

4 0
3 years ago
Evaluate the line integral, where c is the given curve. c y3 ds,
postnew [5]
Parameterizing \mathcal C by

\mathbf r(t)=(t^3,t)

with 0\le t\le2, we have

\mathrm ds=\|\mathbf r'(t)\|\,\mathrm dt=\sqrt{(3t^2)^2+1^2}\,\mathrm dt=\sqrt{9t^4+1}\,\mathrm dt

So

\displaystyle\int_{\mathcal C}y^3\,\mathrm ds=\int_{t=0}^{t=2}t^3\sqrt{9t^4+1}\,\mathrm dt
=\displaystyle\frac1{36}\int_0^236t^3\sqrt{9t^4+1}\,\mathrm dt
=\displaystyle\frac1{36}\int_{u=1}^{u=145}\sqrt u\,\mathrm du
=\dfrac{145^{3/2}-1}{54}
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3 years ago
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7 0
3 years ago
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I need help solving it
Oxana [17]
7. the last one

I = P R T

I = 15.75 P = 500 R = unknown T = 6

15.75 = 500 (r) (6)

divide the T and I

15.75 ÷ 6 = 500 (r) (6) ÷6

2.62 = 500 (r) *get rid of the six*

divide the P with new answer

2.62 ÷ 500 = 500 ÷ 500 *get rid of 500*

0.00524 = r

move decimal to make it in to a percentage

5.24% = R






6 0
2 years ago
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