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3 years ago

Callie read for 4 hours last week. Luis read 6 times as long. Write an equation to show how long Luis read.

Mathematics

2 answers:

zysi [14]3 years ago

7 0

Answer: 24

Step-by-step explanation: Multiply 4 by 6 to get 24

gtnhenbr [62]3 years ago

7 0

Answer:

4 times 6 = 24.

Step-by-step explanation:

Since Luis read 6 times as much as Callie, multiply the number of hours that Callie read, which is 4 hours, by 6. Therefore, Luis read for 24 hours last week.

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Tickets to Dundee crowns mr.dchs are $3 for students and $5 for adults. The number of adult tickets sold was 195 more than the n

Maurinko [17]

Answer: number of tickets sold to adult = 566

Step-by-step explanation:

Let x = number of tickets sold by DCHS to students.

Let y = number of tickets sold DCHS to adults.

Tickets to Dundee crowns mr.dchs are $3 for students and $5 for adults.

This means cost of total tickets sold to students and adults is $3x and $5y respectively.

If DCHS collected $3943 for the tickets, then we have

3x + 5y = 3943 - - - - - -1

The number of adult tickets sold was 195 more than the number of student tickets. This means

y = x +195.

Put y = x +195 in equation 1

3x + 5(x+195) = 3943

3x + 5x + 975= 3943

8x +975=3943

8x = 3943-975= 2968

x = 2968/8 = 371

y = 371 + 195= 566

4 0

3 years ago

Find the volume of this square based pyramid. 5ft 6 ft [? ]ft

Sav [38]

Answer:

48 ft³

Step-by-step explanation:

To find the height, you can use the slant height and the base.

Using the Pythagorean Theorem and 3 as the triangle's base,

a²+b²=c²

3²+b²=5²

9+b²=25

b²=16

b=4

The height of the pyramid is 4.

Volume formula for the pyramid is $\frac{1}{3} *b*h$

Inserting our measurements:

$\frac{1}{3} *6*6*4=\\2*6*4=\\12*4=\\48$

5 0

3 years ago

Help please Brainiest answer to whoever is first

sasho [114]

Answer:

84 inches squared

8 0

3 years ago

Read 2 more answers

A sample of 25 undergraduates reported the following dollar amounts of entertainment expenses last year:

Kaylis [27]

Answer:

mean = 724.2

Median = 715

Mode = 768

Range = 85

Standard deviation = 29.30

Interval = [665.6, 782.8]

Step-by-step explanation:

Number of samples n = 25

Summation X= 769 + 691 + 699 +730+711+ 765+ 702 718 +719 +712+ 768 +688 +757+695 768 +735 +709 +758 +708+ 693 +736 700+ 687 +772 +715 = 18105

1. Mean = 18105/25

= 724.2

2. Median is the middle value when arranged from the least value to the highest = 715

3. Mode is the number with the highest frequency = 768 (occured two times)

1. Range = highest value - lowest value

Highest value = 772

Lowest value = 687

772-687 = 85

2. Standard deviation = √(X-barX)²/n-1

= √20604/25-1

=√858.5

= 29.30

Please check attachment for the full calculation of the standard deviation

Interval

[Mean - 2(sd), mean + 2(sd)]

= [724.2-2x29.3, 724.2+2x29.3]

=[665.6, 782.8]

4 0

2 years ago

Can someone check whether its correct or no? this is supposed to be the steps in integration by parts

Gwar [14]

Answer:

$\displaystyle - \int \dfrac{\sin(2x)}{e^{2x}}\: \text{d}x=\dfrac{\sin(2x)}{4e^{2x}}+\dfrac{\cos(2x)}{4e^{2x}}+\text{C}$

Step-by-step explanation:

$\boxed{\begin{minipage}{5 cm}\underline{Integration by parts} \\\\$\displaystyle \int u \dfrac{\text{d}v}{\text{d}x}\:\text{d}x=uv-\int v\: \dfrac{\text{d}u}{\text{d}x}\:\text{d}x$ \\ \end{minipage}}$

Given integral:

$\displaystyle -\int \dfrac{\sin(2x)}{e^{2x}}\:\text{d}x$

$\textsf{Rewrite }\dfrac{1}{e^{2x}} \textsf{ as }e^{-2x} \textsf{ and bring the negative inside the integral}:$

$\implies \displaystyle \int -e^{-2x}\sin(2x)\:\text{d}x$

Using integration by parts:

$\textsf{Let }\:u=\sin (2x) \implies \dfrac{\text{d}u}{\text{d}x}=2 \cos (2x)$

$\textsf{Let }\:\dfrac{\text{d}v}{\text{d}x}=-e^{-2x} \implies v=\dfrac{1}{2}e^{-2x}$

Therefore:

$\begin{aligned}\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x & =\dfrac{1}{2}e^{-2x}\sin (2x)- \int \dfrac{1}{2}e^{-2x} \cdot 2 \cos (2x)\:\text{d}x\\\\& =\dfrac{1}{2}e^{-2x}\sin (2x)- \int e^{-2x} \cos (2x)\:\text{d}x\end{aligned}$

$\displaystyle \textsf{For }\:-\int e^{-2x} \cos (2x)\:\text{d}x \quad \textsf{integrate by parts}:$

$\textsf{Let }\:u=\cos(2x) \implies \dfrac{\text{d}u}{\text{d}x}=-2 \sin(2x)$

$\textsf{Let }\:\dfrac{\text{d}v}{\text{d}x}=-e^{-2x} \implies v=\dfrac{1}{2}e^{-2x}$

$\begin{aligned}\implies \displaystyle -\int e^{-2x}\cos(2x)\:\text{d}x & =\dfrac{1}{2}e^{-2x}\cos(2x)- \int \dfrac{1}{2}e^{-2x} \cdot -2 \sin(2x)\:\text{d}x\\\\& =\dfrac{1}{2}e^{-2x}\cos(2x)+ \int e^{-2x} \sin(2x)\:\text{d}x\end{aligned}$

Therefore:

$\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{2}e^{-2x}\sin (2x) +\dfrac{1}{2}e^{-2x}\cos(2x)+ \int e^{-2x} \sin(2x)\:\text{d}x$

$\textsf{Subtract }\: \displaystyle \int e^{-2x}\sin(2x)\:\text{d}x \quad \textsf{from both sides and add the constant C}:$

$\implies \displaystyle -2\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{2}e^{-2x}\sin (2x) +\dfrac{1}{2}e^{-2x}\cos(2x)+\text{C}$

Divide both sides by 2:

$\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{4}e^{-2x}\sin (2x) +\dfrac{1}{4}e^{-2x}\cos(2x)+\text{C}$

Rewrite in the same format as the given integral:

$\displaystyle \implies - \int \dfrac{\sin(2x)}{e^{2x}}\: \text{d}x=\dfrac{\sin(2x)}{4e^{2x}}+\dfrac{\cos(2x)}{4e^{2x}}+\text{C}$

5 0

2 years ago