hurry guys i need help! ! ! !

Rewrite the length of the movie as an improper fraction.

2 1/3 = 7/3

Now multiply the length of the movie by 3/4

7/3 x 3/4 = (7 x3) / (3 x 4) = 21/12 = 1 3/4 hours

Answer:

x =  1/2 or 0.5

Step-by-step explanation:

Brainliest Please!!

- Hermionia

Answer:

The answer is -1/2

Step-by-step explanation:

5/x = 15/x+20

5/x-15/x=20

-10/x=20

-10=20x

x= -10/20

x= -1/2

Answer:

B

Step-by-step explanation:

Answer:

Here's one :)

3 x 4=  12 =  1

8 x 3=  24 = 2

-by-step explanation:

The two methods of finding the quotient are flipping the divisor and then multiplying and the second one is converting the fractions into decimals and dividing them.

Fractions describe the part of something present out of the total amount

We know that the result of diving a fraction by a fraction is the same as multiplying that fraction by the reciprocal of the divisor.

∴ (3/8) ÷ (3/4) is,

= (3/8) × (4/3).

= 4/8.

= 1/2.

The second method is converting the fractions into decimals and then dividing them in a conventional way.

3/8 = 0.375 and 3/4 = 0.75.

∴ 0.375 ÷ 0.75 is,

= 0.5.

learn more about fractions here :

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Answer:

1/4096

Step-by-step explanation:

[tex](4*2)^{{(2-2*2)}^2}\\\\=8^{-2^2}\\\\=[\frac{1}{64}]^2\\\\=\frac{1}{4096}[/tex]

you work from bottom to top

Answer:

third option

Step-by-step explanation:

Differentiate using the product rule

Given f(x). g(x) then the derivative is

f(x). g'(x) + g(x). f'(x) ← product rule

Given

[tex]\frac{d}{dx}[/tex](x² secx )

with f(x) = x² and g(x) = secx , then

f'(x) = 2x and g'(x) = secxtanx, thus

[tex]\frac{d}{dx}[/tex](x² secx)

= x². secxtanx + secx. 2x ← factor out xsecx from each term

= xsecx(xtanx + 2) → third option

Answer:

1. 2x + 1 ≤ 3

2. 1 + 3x ≥ -5

3. 5x - 7 ≥ 17

4. 5 - 3x ≥ 11

Step-by-step explanation:

Ok

5x - 7 ≥ 17

subtract 7 and u get 5x ≥ 10

divide that and you get ≥ 2

so 3 goes with that one

1 + 3x ≥ -5

subtract 1 and you get 3x ≥ -6

divide that and youll get x ≥ -2

so 2 goes with this one

2x + 1 ≤ 3

subtract 1 and u get 2x ≤ 2

divide them and you get x ≤ 1

so 1 goes with this one

5 - 3x ≥ 11

subtract 5 and u get -3x  ≥ 6

divide it and u get x ≥ -2

so 4 goes with this one

Hoped this helps :)

also btw Brainly steals half the points you give out so instead of 15 ur giving me 25.

Answer:

alr i tried, i hope i got them right! <3 pls let me know if i got them wrong.

1. 108 | 6x12 = 72 9x12 = 108 |

2. 91 | 13x7 = 19

3. 19 | 108/6 90/6 = 19/15 |

4. 50. 400/8 = 50

5. 50. 300/6

hope this helped! <3

Step-by-step explanation:

Answer:

Step-by-step explanation:

Answer:

Step-by-step explanation:

Answers C, D are correct

TBh i really don't know about B because i also think its B but i'm not really that sure but it could be a possibility\

If a cheetah ran 196m is 8 seconds, then 196/8 = 24.5 meters per second.

Let’s verify all answers:

A) 24.5 x 6 = 147. 147m are ran in 6 seconds, not 125.

B) 24.5 x 10 = 245. This answer is correct!

C) We’ve establishes that a Cheetah ran 24.5 meters per second. Correct!

D) Since one minute equals 60 seconds, 24.5 x 60 = 1470. Correct!

E) 24.5 x 5 = 122.5. In this time frame, a cheetah wouldn’t run 125m.

Answer:

117.8

Step-by-step explanation:

i used heron's formula

√s(s-a)(s-b)(s-c)

s= (20+16+33)/2 = 34.5

√34.5(34.5-20)(34.5-16)(34.5-33)

= 117.8

Answer:

Part I)

[tex]a=-1\text{ and } c=2[/tex]

Giving the equation:

[tex]y=-x^3+x+2[/tex]

Part II)

(-1, 2)

Step-by-step explanation:

We have the equation:

[tex]y=ax^3+x+c[/tex]

Where a and c are constants.

We are given gradient/slope of the curve at the point (1, 2) is -2.

Part I)

We know that the gradient of the curve exactly at the point (1, 2) is -2.

So, we will differentiate our equation. We have:

[tex]\displaystyle y=ax^3+x+c[/tex]

Differentiate both sides with respect to x:

[tex]\displaystyle y^\prime=\frac{d}{dx}\big[ax^3+x+c\big][/tex]

Expand the right:

[tex]\displaystyle y^\prime=\frac{d}{dx}\big[ax^3]+\frac{d}{dx}\big[x\big]+\frac{d}{dx}\big[c\big][/tex]

Since a and c are simply constants, we can move them outside:

[tex]\displaystyle y^\prime=a\frac{d}{dx}\big[x^3]+\frac{d}{dx}\big[x\big]+c\frac{d}{dx}\big[1\big][/tex]

Differentiate. Note that the derivative of a constant is simply 0.

[tex]\displaystyle y^\prime=3ax^2+1[/tex]

We know that at the point (1, 2), the gradient/slope is -2.

Hence, when x is 1 and when y is 2, y’ is -2.

So (we don’t have a y so we can ignore it):

[tex]-2=3a(1)^2+1[/tex]

Solve for a. Simplify:

[tex]-2=3a+1[/tex]

Subtract 1 from both sides yield:

[tex]-3=3a[/tex]

So, it follows that:

[tex]a=-1[/tex]

Therefore, our derivative will be:

[tex]y=-3x^2+1[/tex]

Now, we can go back to our original equation:

[tex]y=ax^3+x+c[/tex]

Since we know that a is -1:

[tex]y=-x^3+x+c[/tex]

Recall that we know that a point on our curve is (1, 2).

So, when x is 1, y is 2. By substitution:

[tex]2=-(1)^3+1+c[/tex]

Solve for c. Simplify:

[tex]2=-1+1+c[/tex]

Then it follows that:

[tex]c=2[/tex]

Hence, our entire equation is:

[tex]y=-x^3+x+2[/tex]

Part II)

Our derivative is:

[tex]y^\prime=-3x^2+1[/tex]

If the gradient is -2, y’ is -2. Hence:

[tex]-2=-3x^2+1[/tex]

Solve for x. Subtracting 1 from both sides yields:

[tex]-3=-3x^2[/tex]

So:

[tex]x^2=1[/tex]

Then it follows that:

[tex]x=\pm1\\[/tex]

Since we already have the point (1, 2) where x is 1, our other point is where x is -1.

Using our equation:

[tex]y=-x^3+x+2[/tex]

We can see that our second point is:

[tex]y=-(-1)^3+(-1)+2[/tex]

Simplify:

[tex]y=1-1+2=2[/tex]

So, our second point where the gradient is -2 is at (-1, 2).

Answer:

x plus 3y minus 63

Step-by-step explanation:

The fifth term is 162 in the geometric sequence.

The geometric sequence defined as a series represents the sum of the terms in a finite or infinite geometric sequence. The successive terms in this series share a common ratio.

The nth term of a geometric progression is expressed as

Tₙ = arⁿ⁻¹

Where a is the first term, r is the common ratio

A arithmetic sequence is defined as an arrangement of numbers which is particular order.

The formula to find the general term of an arithmetic sequence is,

aₙ = a₁ + (n-1)d

Given data :

First term of sequence geometric (a₁) = 2

The common ratio of geometric (r) = 3

To determine the fifth term of geometric sequence

Let the fifth term of sequence = a₅    

The nth term of a geometric sequence is expressed as :

Tₙ = a₁rⁿ⁻¹

Substitute the values of n = 5, a₁ =2 and r = 3 in the formula,

T₅  = (2)(3)⁵⁻¹  

T₅  = (2)(3)⁴

T₅  = (2)(81)

T₅  = 162

Hence, the fifth term is 162 in the geometric sequence

Learn more about geometric sequence here :

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====================================================

Explanation:

Original list = {56,34,45,65,67,56,45,36,92,73,92,45,56}

Sorted list = {34,36,45,45,45,56,56,56,65,67,73,92,92}

We see that 45 and 56 both occur three times each, and this is the most occurring or most frequent. So the mode is the set {45, 56}. It is possible to have more than one mode. It is also possible to not have a mode at all if each value in a list only occurs one time.

Answer:

its between 56 and 45 and i havent rlly done this since 5th or 6th grade so

Step-by-step explanation:

Answer:

Step-by-step explanation:

Had the answer put don't remember

Answer:

The stack of blocks is 25.96 inches tall

Step-by-step explanation:

Volume of a Cube

Suppose we are given a cube of side L, the volume is calculated by:

[tex]V=L^3[/tex]

If the volume is given, the side can be found by solving for L:

[tex]L=\sqrt[3]{L}[/tex]

We know three cube-shaped blocks with volumes 375, 648, and 1029 cube inches. We'll find each of the side lengths and add them up to find the height that results when they stack up vertically.

Block 1, volume =375 cube inches:

[tex]L_1=\sqrt[3]{375}=7.21\ inch[/tex]

Block 2, volume= 648 cube inches:

[tex]L_2=\sqrt[3]{648}=8.65\ inch[/tex]

Block 3, volume =1029 cube inches:

[tex]L_3=\sqrt[3]{1029}=10.10\ inch[/tex]

Total height=[tex]7.21\ inch+8.65\ inch+10.10\ inch=25.96\ inch[/tex]

The stack of blocks is 25.96 inches tall

Answer:

To calculate the perimeter of a circle we have this formula:

[tex]P = 2\pi r[/tex]

And, if we're going to calculate for a semicircle, we should divide it by 2

We should also take into account the diameter, as, if it's a semicircle, we also have to add the base of it, which will have the diameter length

Knowing this, we can see the formula to calculate this specific perimeter

[tex]\frac{2\pi r}{2} + 2r[/tex]

Now, we can replace r by 22:

[tex]\frac{2\pi *22}{2} + 2*22 \approx 113.12[/tex]

So the perimeter is approximately 113.12 cm

Answer:

3 °F/h

Step-by-step explanation:

The rate per hour of temperature change = change in temperature per measurement/change in time per measurement.

Now, the change in temperature per measurement is constant = 3/4 °F and the change in time per measurement is constant = 15 minutes = 15/60 h = 1/4 h.

So, rate per hour of temperature change = change in temperature per measurement/change in time per measurement

= 3/4 °F ÷ 1/4 h

= 3/4 °F × 4/h

= 3 °F/h

So, the rate per hour of temperature change is 3 °F/h

Answer:

x = -84

y = -21

Step-by-step explanation:

4 - y = 17

4y = x

Use the first equation to solve for y

4 - y = 17

y = 4 -  17

y = -21

In the second equation, get y on its own.

4y = x

y = x/4

Substitute the new second equation into the new first equation

-21 = x/4

-21 * 4 = x

x = -84

Answer:

y = 4x/3 - 2

Step-by-step explanation:

Answer:

Phoebe surveyed teens and adults to find the average number of hours of sleep they get during the week. Her results are shown below. Which statement regarding her survey results is true?

Hours of Sleep for Teens

A box-and-whisker plot. The number line goes from 1 to 12. The whiskers range from 2.5 to 12, and the box ranges from 5 to 8.5. A line divides the box at 7.

adult have less variability in the sleep hours because well they work more then us and us teens we have school starting at 7:45 am and the adult have to wake up to go to work or having to stay up

hope this can help

Answer:

1- no

2- yes

3- no

4- yes

the ones that are no are because they either have x-values that repeat or they have y-values that repeat

Step-by-step explanation:

Answer:

this is really good but very sad so if thats what you're going for then you did a really good job!! :(

Step-by-step explanation:

Answer:

i love it!!! very cool

Step-by-step explanation:

Answer:

yes the solution is 8

Step-by-step explanation:

subtract 9 from both sides

1/2x+9-9=13-9

simplify

1/2x=4

multiply both sides by 2

2 x 1/2=4 x 2

simplify

x=8

Answer:

This be the ooogggaaaaa boooogaaaaa 696969

Step-by-step explanation:

uhhh

Answer:

Here is the summary:

Step-by-step explanation:

Given the equation

The points which satisfy the equation y=13x+5

y = 13x+5

Checking the point (0, 5)

5 = 13(0)+5

5 = 0 + 5

5 = 5

TRUE

Checking the point (-2, -21)

-21 = 13(-2)+5

-21 = -26 + 5

-21 = -21

TRUE

The points which satisfy the equation y=-3x

y = -3x

Checking the point (0, 0)

0 = -3(0)

0 = 0

TRUE

Checking the point (-5, -15)

y = -3x

-15 = -3(-5)

-15 = -15

TRUE

Checking the point (-3, 9)

y = -3x

9 = -3(-3)

9 = 9

TRUE

The points which satisfy the equation y=x-10

y = x-10

Checking (8, -2)

-2 = 8 - 10

-2 = -2

TRUE

Therefore, from the above calculations we conclude the summary:

Here is the summary:

Answer: x₁=7, x₂=-1

Step-by-step explanation:

(x-3)^2= x^2-6x+9

2(x^2-6x+9)=28+4

x^2-6x+9=16

x^2-6x-7=0

you use the quadratic equation will give you 2 roots,

x₁=7, x₂=-1

Answer:

[tex]39.5,44.5[/tex]

Step-by-step explanation:

[tex]Let\ the\ lower\ limit\ and\ the\ upper\ limit\ of\ the\ first\ interval\ be\ u_1,v_1.\\Let\ the\ lower\ limit\ and\ the\ upper\ limit\ of\ the\ second\ interval\ be\ u_2,v_2.\\Let\ the\ lower\ limit\ and\ the\ upper\ limit\ of\ the\ third\ interval\ be\ u_3,v_3.\\Hence,\\As\ the\ class\ marks\ are\ uniform\ with\ an\ a\ difference\ of\ 5, the\\ observation\ is\ continuous\ and\ the\ class\ size\ is\ 5\ too.\\Hence,\\v_1=u_2, v_2=u_3\\[/tex][tex]Now,\\Lets\ consider\ the\ first\ two\ classes.\\37=\frac{u_1+v_1}{2} \\42=\frac{u_2+v_2}{2}\\By\ adding\ the\ equations:\\79= \frac{u_1+v_1}{2}+\frac{u_2+v_2}{2}\\79=\frac{u_1+v_1+u_2+v_2}{2}\\79=\frac{(u_1+v_2)+(v_1+u_2)}{2}\\79=\frac{(u_1+v_2)}{2}+\frac{(v_1+u_2)}{2}\\Now,\\As\ the\ mid-point\ between\ two\ points\ can\ be\ calculated\ by\\ its\ average:\frac{u+v}{2} \\Hence,\\u_2\ lies\ in\ the\ mid-point\ of\ u_1\ and\ v_2.\\u_2=\frac{(u_1+v_2)}{2}\\[/tex]

[tex]Hence,\\By\ substituting\ u_2=\frac{(u_1+v_2)}{2},\\79=u_2+\frac{v_1+u_2}{2}\\As\ v_1=u_2[Proven],\\79=u_2+ \frac{u_2+u_2}{2}\\79=u_2+\frac{2u_2}{2}\\79=2u_2\\Hence,\\u_2=\frac{79}{2}\\u_2=39.5\\Now,\ as\ we\ already\ know\ that\ the\ class\ size=5,\\v_2-u_2=5\\Hence,\\v_2=5+u_2\\Here,\\v_2=5+39.5\\v_2=44.5\\Hence,\\The\ upper\ limit\ of\ the\ second\ interval=44.5\\The\ lower\ limit\ of\ the\ second\ interval=39.5\\[/tex]