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11Alexandr11 [23.1K]

3 years ago

10

Linda scored 3 points more than twice Lisa's points on a driver's ed test. Lisa scored p points.

1 answer:

Flura [38]3 years ago

3 0

Ok so what I would do Is I would do 3 x p=m

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What is the probability of throwing a number greater than 4 when rolling a number cube that has sides labeled 1, 2, 3, 4, 5, 6?

Leona [35]

What is probability rolling a number greater than 4?

The only numbers there are is 5 and 6.
That means that there are 2 outcomes out of 6 total outcomes.
That would be 2/6.
Divide the the top and bottom by 2.
In simplest form it would be 1/3.
2/6=1/3

The answer is 1/3. The probability of throwing a number greater than 4 is 1/3.

4 0

3 years ago

For what value of x is line a parallel to line b? Enter your answer in the box. X =____

Nezavi [6.7K]

Step-by-step explanation:

115 = 5x + 15

115 - 15 = 5x

100 = 5x

20 = x

7 0

3 years ago

Use the disk method or the shell method to find the volumes of the solids generated by revolving the region bounded by the graph

diamong [38]

Answer:

a) 8π

b) 8/3 π

c) 32/5 π

d) 176/15 π

Step-by-step explanation:

Given lines : y = √x, y = 2, x = 0.

<u>a) The x-axis </u>

using the shell method

y = √x = , x = y^2

h = y^2 , p = y

vol = ( 2π ) $\int\limits^2_0 {ph} \, dy$

= $( 2\pi ) \int\limits^2_0 {y.y^2} \, dy$

∴ Vol = 8π

<u>b) The line y = 2 ( using the shell method )</u>

p = 2 - y

h = y^2

vol = ( 2π ) $\int\limits^2_0 {ph} \, dy$

= $( 2\pi ) \int\limits^2_0 {(2-y).y^2} \, dy$

= ( 2π ) * [ 2/3 * y^3 - y^4 / 4 ] ²₀

∴ Vol = 8/3 π

<u>c) The y-axis ( using shell method )</u>

h = 2-y = h = 2 - √x

p = x

vol = $(2\pi ) \int\limits^4_0 {ph} \, dx$

= $(2\pi ) \int\limits^4_0 {x(2-\sqrt{x} ) } \, dx$

= ( 2π ) [x^2 - 2/5*x^5/2 ]⁴₀

vol = ( 2π ) ( 16/5 ) = 32/5 π

<u>d) The line x = -1 (using shell method )</u>

p = 1 + x

h = 2√x

vol = $(2\pi ) \int\limits^4_0 {ph} \, dx$

Hence vol = 176/15 π

attached below is the graphical representation of P and h

3 0

3 years ago

Evaulate the function at the indicated values h(t)=t+(3/t)i find h(x-1)

alex41 [277]

<span>h<span>(t)</span>=<span>t<span>34</span></span>−3<span>t<span>14</span></span></span>

Note that the domain of h is <span>[0,∞]</span>.

By differentiating,

<span>h'<span>(t)</span>=<span>34</span><span>t<span>−<span>14</span></span></span>−<span>34</span><span>t<span>−<span>34</span></span></span></span>

by factoring out <span>34</span>,

<span>=<span>34</span><span>(<span>1<span>t<span>14</span></span></span>−<span>1<span>t<span>34</span></span></span>)</span></span>

by finding the common denominator,

<span>=<span>34</span><span><span><span>t<span>12</span></span>−1</span><span>t<span>34</span></span></span>=0</span>

<span>⇒<span>t<span>12</span></span>=1⇒t=1</span>

Since <span>h'<span>(0)</span></span> is undefined, <span>t=0</span> is also a critical number.

Hence, the critical numbers are <span>t=0,1</span>.

I hope that this was helpful.

6 0

3 years ago

melisa1 [442]

Answer:

$sin O=\dfrac{3\sqrt{13}}{13}\\cos O=\dfrac{2\sqrt{13}}{13}\\tan O=\dfrac{3}{2}$

Step-by-step explanation:

If the point (2,3) is on the terminal side of an angle in standard position.

Adjacent of O, x=2,

Opposite of O, y=3

Next, we determine the hypotenuse, r using Pythagoras Theorem.

$Hypotenuse =\sqrt{Opposite^2+Adjacent^2} \\r=\sqrt{3^2+2^2} \\r=\sqrt{13}$

Therefore:

$sin O=\dfrac{Opposite}{Hypotenuse} \\sin O=\dfrac{3}{\sqrt{13}} \\$Rationalizing\\sin O=\dfrac{3\sqrt{13}}{13}$

$cos O=\dfrac{Adjacent}{Hypotenuse} \\cos O=\dfrac{2}{\sqrt{13}} \\$Rationalizing\\cos O=\dfrac{2\sqrt{13}}{13}$

$Tan O=\dfrac{Opposite}{Adjacent} \\tan O=\dfrac{3}{2}$

4 0

3 years ago