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nalin [4]
3 years ago
11

Is -2×10 greater than 10?

Mathematics
1 answer:
Luba_88 [7]3 years ago
8 0
No, because it's -20 D:
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Which equation matched the graph ?
Nady [450]
D. y - 2 = -3(x - 2)
8 0
3 years ago
What is the simplified form of each expression? a. (6 + 3)2 − 4 b. 23 + (14 − 4) ÷ 2
liraira [26]

Answer:

a. 14

b. 28

Step-by-step explanation:

a. ex: (6+3)2 - 4 = 9(2) - 4

18 - 4 = <u>14</u>

b. ex: 23 + (14 - 4) / 2

Use PEMDAS

P 1st

10

D over A

5

A

<u>28</u>

<em><u>Hope this helps!</u></em>

5 0
3 years ago
A shirt normally sells for $65. It was on sale for 45% off the regular price for 3 days. On the forth day it increased by 10% mo
sergiy2304 [10]

Answer:

$39.33 (rounded to nearest cent)

Step-by-step explanation:

Cost of shirt normally = $65

Cost of shirt for 3 days = \frac{100-45}{100} × $65

                                      = \frac{55}{100} × $65

                                      = $35.75

Cost of shirt on 4th day = \frac{100+10}{100} × $35.75

                                       = \frac{110}{100} × $35.75

                                       = $39.33 (rounded to nearest cent)

3 0
3 years ago
Read 2 more answers
Suppose r(t) = (et cos t)i + (et sin t)j. Show that the angle between r and a never changes. What is the angle?
tatyana61 [14]

Answer:

Angle = Ф = cos^{-1}(0) = 0

Hence, it is proved that angle between position vector r and acceleration vector a = 0 and is it never changes.

Step-by-step explanation:

Given vector r(t) = e^{t}cost i + e^{t}sint j

As we know that,

velocity vector = v = \frac{dr}{dt}

Implies that

velocity vector = (e^{t} cost - e^{t} sint)i + (e^{t} sint - e^{t}cost )j

As acceleration is velocity over time so:

acceleration vector = a = \frac{dv}{dt}

Implies that

vector a =

(e^{t}cost - e^{t}sint - e^{t}sint - e^{t}cost )i + ( e^{t}sint + e^{t}cost + e^{t}cost - e^{t}sint )j

vector a = (-2e^{t}sint) i + ( 2e^{t}cost)j

Now scalar product of position vector r and acceleration vector a:

r. a = . \\

r.a = -2e^{2t}sintcost + 2e^{2t}sintcost

r.a = 0

Now, for angle between position vector r and acceleration vector a is given by:

cosФ = \frac{r.a}{|r|.|a|} = \frac{0}{|r|.|a|} = 0

Ф = cos^{-1}(0) = 0

Hence, it is proved that angle between position vector r and acceleration vector a = 0 and is it never changes.

4 0
3 years ago
What is the volume???
Shalnov [3]

Answer:

that would be 80

Step-by-step explanation:

Here it is

3 0
3 years ago
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