All categories

3 years ago

A rectangular note card is 4 inches tall and 7 inches wide. What is its area? square inches

Mathematics

1 answer:

Softa [21]3 years ago

7 0

Answer:

Step-by-step explanation:

just multiply width times length 4*7=28

You might be interested in

I need help I’m bad at math

Alina [70]

If you'd graph this function, you'd see that it's positive on [-1.5,0], and that it's possible to inscribe 3 rectangles on the intervals [-1.5,-1), (-1,-0.5), (-0.5, 1].

The width of each rect. is 1/2.

The heights of the 3 inscribed rect. are {-2.25+6, -1+6, -.25+6} = {3.75,5,5.75}.

The areas of these 3 inscribed rect. are (1/2)*{3.75,5,5.75}, which come out to:

{1.875, 2.5, 2.875}

Add these three areas together; you sum will represent the approx. area under the given curve on the given interval: 1.875+2.5+2.875 = ?

5 0

3 years ago

Three different units of measurement that could be used to identify area

vitfil [10]

Answer:

i hope it helps u........

8 0

3 years ago

Please help this is going to be due soon it’s a test

grandymaker [24]

Answer:

You are not allowed to use brainly for tests.. it is agianst their rules..

Step-by-step explanation:

8 0

3 years ago

What is the area of ABC?

Salsk061 [2.6K]

Answer:

Area of triangle ABC = 10

Option C is correct.

Step-by-step explanation:

We need to find area of triangle ABC

We are given A(2,1) B(4,7) and C(6,3)

The formula used to find $Area \ of \ triangle=\frac{1}{2}base\times height$ ABC is:

In the given triangle base= AC and height = BC

First we need to find the distance between A and C and B and C

The formula used to find distance between AC $Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

We have $x_1=2, y_1=1, x_2=6, y_2=3$

Putting values and finding distance

$Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\Distance=\sqrt{(6-2)^2+(3-1)^2}\\Distance=\sqrt{(4)^2+(2)^2}\\Distance=\sqrt{16+4}\\Distance=\sqrt{20}$

find distance between BC $Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

We have $x_1=4, y_1=7, x_2=6, y_2=3$

Putting values and finding distance

$Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\Distance=\sqrt{(6-4)^2+(3-7)^2}\\Distance=\sqrt{(2)^2+(4)^2}\\Distance=\sqrt{4+16}\\Distance=\sqrt{20}$

So, We have height = $\sqrt{20}$ and base = $\sqrt{20}$

Finding area of triangle

$Area \ of \ triangle=\frac{1}{2}base\times height\\Area \ of \ triangle=\frac{1}{2}\sqrt{20} \times \sqrt{20} \\Area \ of \ triangle=\frac{1}{2}\times 20\\Area \ of \ triangle=10$

So, Area of triangle ABC = 10

Option C is correct.

7 0

3 years ago

I will give brainliest to the right answer:)

Akimi4 [234]

Answer:

178.1

Step-by-step explanation:

In the first blank, write in 7 for x, and in the second blank, write in 8 for y. Then perform the indicated multiplication and addition.

12.3(7) + 11.5(8), or

86.1 + 92 = 178.1

5 0

3 years ago

A rectangular note card is 4 inches tall and 7 inches wide. What is its area? square inches​

A rectangular note card is 4 inches tall and 7 inches wide. What is its area? square inches