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3 years ago

Does anyone know how to set up 14.7 × 11.361

1 answer:

7 0

Just type it into a calculator. If I were to explain it in words, just line it up like
11.361
*14.7.
The answer is 167.0067

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Find the area of the circle to the nearest hundredths ( Use 3.14 for pi) (4cm)

Tpy6a [65]

Answer: 50.24 cm²

Step-by-step explanation:

<u>ASSUMING THE RADIUS OF THE CIRCLE IS THE 4 CM</u>

<u></u>

The formula for the area of a circle is pi(3.14) * r(4)^2

$3.14*4^2$

$3.14*16$

$50.24$

Hope it helps <3 :D

8 0

3 years ago

What is the average of 0,1,3,2

astraxan [27]

I'm gonna say the average is 1.5 because is the sum of the elements divided by the number of elements.

0+1+3+2=6
6/4= 1.5

3 0

3 years ago

Find the area of each parallelogram. What is the relationship between the areas?

castortr0y [4]

Answer:

Area of parallelogram is given by:

$A = bh$

where b is the base and h is the height of parallelogram.

In parallelogram TQRS.

Coordinate of TQRS are;

T(8, 16), Q(4, 4), R(16, 4) and S(20, 16)

Coordinate of T'Q'R'S' are;

T'(2, 4), Q'(1, 1), R'(4, 1) and S'(5, 4)

Find the length of QR and PT:

Using distance(D) formula:

$D = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

$QR = \sqrt{(4-16)^2+(4-4)^2} =\sqrt{(-12)^2+0}= \sqrt{144}= 12$ units

Similarly;

For PT:

From the graph:

P(8, 4) and T(8, 16), then

$PT = \sqrt{(8-8)^2+(4-16)^2} =\sqrt{(-0)^2+(-12)^2}= \sqrt{144}= 12$ units

In parallelogram TQRS

PT represents the height and QR represents the base of the parallelogram respectively.

then;

Area of parallelogram TQRS = $QR \cdot PT$

⇒Area of parallelogram TQRS = $12 \cdot 12 = 144$ unit square.

Now, in parallelogram T'Q'R'S'

Q'R' represents the base and P'T' represents the height of the parallelogram respectively.

here, P'(2, 1)

Find the length of Q'R' and P'T':

$Q'R' = \sqrt{(1-4)^2+(1-1)^2} =\sqrt{(-3)^2+0}= \sqrt{9}= 3$ units

$P'T' = \sqrt{(2-2)^2+(1-4)^2} =\sqrt{(-0)^2+(-3)^2}= \sqrt{9}= 3$ units

Then;

Area of parallelogram T'Q'R'S' = $Q'R' \cdot P'T'$

Area of parallelogram T'Q'R'S' = $9 \cdot 9= 81$ unit square.

Now, we have to find the relationship between the areas.

$\frac{\text{Area of parallelogram TQRS}}{\text{Area of parallelogram T'Q'R'S'}} = \frac{144}{81}$

then;

the relationship between the areas of TQRS and T'Q'R'S' is:

$\text{Area of parallelogram TQRS} = \frac{144}{81} \cdot {\text{Area of parallelogram T'Q'R'S'}$

7 0

3 years ago

Rewrite .86 in scientific notation

mr_godi [17]

This is the scientific notation >> 8.6 × 10-1

6 0

3 years ago

You can construct congruent triangles if you know which of the following? Select two answers.

lora16 [44]

Answer: E and D

Step-by-step explanation: E (SAS) allows you to find the other two angle measures and the last side length. D (ASA) lets you find the other angle and the other side lengths. The rest of the answers either don't account for both angles and sides, or don't give sufficient information.

7 0

3 years ago