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3 years ago

The area of the rectangular table is 18 square feet the width of the table is 3 ft what is its length

Mathematics

1 answer:

MrRa [10]3 years ago

6 0

Answer:

Step-by-step explanation:

Okay, In solving to find the square feet of a rectangle, you need to know the length and the width to be able to get the answer. Length x Width = area. In this case you already know the area, so just reverse the steps. 18 Ft is the area and the width is 3ft, you’re just missing length. 18/3 = 6. 6ft is the length

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Brent is designing a poster that has an area of 1 sq.ft. He is going to paste a photo collage on a section of the poster that is

ad-work [718]

Answer:54 th

Step-by-step explanation:

Given

Poster has an area of $A=1\ ft^2$

Size of photo is $\frac{1}{3}\ ft\times \frac{3}{5}\ ft$

and we know

$1\ ft\ \text{is equal to}\ 0.3048\ m$

so $\frac{1}{3}\ ft$ is $0.1016\ m$

$\frac{3}{5}\ ft$ is $0.1828\ m$

So area of photo is

$a=0.1016\times 0.1828$

$a=0.01858\ m^2$

ratio of area of photo and poster is

$\Rightarrow \frac{a}{A}=\frac{0.01858}{1}\approx \frac{1}{53.819}\approx \frac{1}{54}$

So photo will acquire $54^{th}$ Part of poster

7 0

4 years ago

Can anyone help me with this?

yawa3891 [41]

Answer:

Below.

Step-by-step explanation:

7) c^2 = a^2 + b^2

15^2 = 11^2 + b^2

b^2 = 225 - 121 = 104

b = √104

= √4√26

= 2√26.m

8). 4^2 = a^2 + (√6)^2

a^2 = 16 - 6

a^2 = 10

a = √10.

7 0

2 years ago

prove that if f is integrable on [a,b] and c is an element of [a,b], then changing the value of f at c does not change the fact

Neko [114]

Answer with Step-by-step explanation:

We are given that if f is integrable on [a,b].

c is an element which lie in the interval [a,b]

We have to prove that when we change the value of f at c then the value of f does not change on interval [a,b].

We know that limit property of an integral

$\int_{a}^{b}f dt=\int_{a}^{c}fdt+\int_{c}^{b} fdt$

$\int_{a}^{b} fdt=f(b)-f(a)$ ....(Equation I)

Using above property of integral then we get

$\int_{a}^{b}fdt=\int_{a}^{c}fdt+\int_{c}^{b} fdt$ ......(Equation II)

Substitute equation I and equation II are equal

Then we get

$\int_{a}^{b}fdt= f(c)-f(a)+{f(b)-f(c)}$

$\int_{a}^{b}fdt=f(c)-f(a)+f(b)-f(c)=f(b)-f(a)$

$\int_{a}^{c}fdt+\int_{c}^{b}fdt=f(b)-f(a)$

Therefore, $\int_{a}^{b}fdt=\int_{a}^{c}fdt+\int_{c}^{b}fdt$ .

Hence, the value of function does not change after changing the value of function at c.

6 0

3 years ago

Really need help real talk please respond

nadezda [96]

Answer:

1. B

2. A

3. C

4.A

Hope this helped!!!

5 0

3 years ago

Read 2 more answers

Use the digits 1-7 to create a seven-digit number that can be rounded to 6,300,000.

Alla [95]

There are many ways to figure this out. One way is 6,275,431.
You just need to start the number with a 6, followed by a 2 and 5 or 7, then you can mix up the rest of the numbers any way you want.

6 0

3 years ago