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3 years ago

The product of 7/16, 4/3, and 1/2 is

1 answer:

4 0

First, we have two multiply two fractions (7/16 x 4/3) which we do top x top and bottom x bottom which gives us 28/48

Second, we do 28/48 x 1/2 which is 28/96.

Lastly, we simplify 28/96 which gives us 7/24

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Six times the difference of a number and three is equal to three less than the number.

icang [17]

Answer:

x=3

Step-by-step explanation:

lets take x as the number

6(x-3)=x-3

since it says "6 times" we know that we will have to multiply the difference by 6 and it says the difference of a number and 3 is equal to 3 less than the number so we know that we are subtracting 3 from x

6x-18=x-3

6x-x=18-3

5x=15

x=3

3 0

2 years ago

Use a diagram words, and angle relationships to explain your answer

tankabanditka [31]

Answer:

네, 죄송합니다. 알았어 알았어. 알았어. 그러면 너는 내 대답을 위해 이것을하고있다.

8 0

2 years ago

Okay, I'm having trouble with homework and it says on my worksheet: Well, the coupon is 10%. The digital camera costs $95. The q

mamaluj [8]

All you need to do is multiply 10% of 95 and you will get 9.5 and then subtract that same 9.5 off the 95 dollars and you get 85.5.

6 0

2 years ago

The Torrence family has a rectangular, in-ground pool in their yard. It holds 1700 ft of water. If it is 17 ft wide and 4 ft dee

xeze [42]

9514 1404 393

Answer:

25 ft

Step-by-step explanation:

The volume is given by the formula ...

V = LWD . . . . . the product of length, width, depth

Solving for L, we have ...

L = V/(WD) = (1700 ft³)/((17 ft)(4 ft)) = 100/4 ft = 25 ft

The length of the pool is 25 ft.

4 0

2 years ago

Yea I can do tomorrow it would take me to get the money back to you

Tems11 [23]

Given, the equation that represents the height of an object:

$y(t)=100t-16t^2$

First, we will find the velocity of the object which is the first derivative of the height using the method of the limits

$\frac{dy}{dt}=\lim_{h\to a}\frac{y(3+h)-y(3)}{(3+h)-(3)}$

We will find the value of the function y(t) when t = 3, and when t = 3+h

$\begin{gathered} y(3+h)=100(3+h)-16(3+h)^2 \\ y(3+h)=300+300h-16(9+6h+h^2) \\ y(3+h)=300+300h-144-96h-16h^2 \\ y(3+h)=156+4h-16h^2 \\ y(3)=100(3)-16(3)^2=156 \end{gathered}$

Substitute y(3+h) and y(3) into the expression of the limit

$\begin{gathered} \frac{dy}{dt}|_{t=3}=\lim_{h\to a}\frac{156+4h-16h^2-156}{3+h-3}=\lim_{h\to a}\frac{4h-16h^2}{h} \\ \\ \frac{dy}{dt}|_{t=3}=\lim_{h\to a}(4-16h) \end{gathered}$

Where a = 0

d) compute the instantaneous velocity at t = 3

$\frac{dy}{dt}|_{t=3}=100-16*2*3=4$

So, the answer will be:

$\begin{gathered} \frac{dy}{dt}|_{t=3}=\lim_{h\to a}(4-16h) \\ a=0 \\ \\ \frac{dy}{dt}|_{t=3}=4\text{ ft/sec} \end{gathered}$

5 0

1 year ago