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3 years ago

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A jewelry designer is making a pendant. The pendant will be a circular dis (center O) with a circular hold cut out of it, as sho

wn. The radius of the disc is 35 millimeters. Find the area of the pendant. Use 3.14 for π and round to the nearest tenth

1 answer:

antoniya [11.8K]3 years ago

6 0

Answer:

2884.8 millimeters squared

Step-by-step explanation:

Given:

The radius of the disc is 35 millimeters, so the area of it is:

π $r^{2}$ = 3.14* $35^{2}$ = 3846.5

Then, we find out the area of the circular hold cut out of the bigger one, its radius is a haft of the radius of the bigger circle = 35/2 = 17.5

π $r^{2}$ = 3.14* $17.5^{2}$ =961.6

=> the area of the pendant = 3846.5 - 961.6 =2884.8 millimeters squared

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Which of the following holds about 1 L of water

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Angelina_Jolie [31]

Answer:

0.294

Step-by-step explanation:

The number of possible outcomes for 2 consecutive days is 2 (Friday, Saturday and Saturday,Sunday).

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4 years ago

Brady made a scale drawing of a rectangular swimming pool on a coordinate grid. The points (-20, 25), (30, 25), (30, -10) and (-

djverab [1.8K]

Answer:

Length = 50 units

width = 35 units

Step-by-step explanation:

Let A, B, C and D be the corner of the pools.

Given:

The points of the corners are.

$A(x_{1}, y_{1}})=(-20, 25)$

$B(x_{2}, y_{2}})=(30, 25)$

$C(x_{3}, y_{3}})=(30, -10)$

$D(x_{4}, y_{4}})=(-20, -10)$

We need to find the dimension of the pools.

Solution:

Using distance formula of the two points.

$d(A,B)=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$ ----------(1)

For point AB

Substitute points A(30, 25) and B(30, 25) in above equation.

$AB=\sqrt{(30-(-20))^{2}+(25-25)^{2}}$

$AB=\sqrt{(30+20)^{2}}$

$AB=\sqrt{(50)^{2}$

AB = 50 units

Similarly for point BC

Substitute points B(-20, 25) and C(30, -10) in equation 1.

$d(B,C)=\sqrt{(x_{3}-x_{2})^{2}+(y_{3}-y_{2})^{2}}$

$BC=\sqrt{(30-30)^{2}+((-10)-25)^{2}}$

$BC=\sqrt{(-35)^{2}}$

BC = 35 units

Similarly for point DC

Substitute points D(-20, -10) and C(30, -10) in equation 1.

$d(D,C)=\sqrt{(x_{3}-x_{4})^{2}+(y_{3}-y_{4})^{2}}$

$DC=\sqrt{(30-(-20))^{2}+(-10-(-10))^{2}}$

$DC=\sqrt{(30+20)^{2}}$

$DC=\sqrt{(50)^{2}}$

DC = 50 units

Similarly for segment AD

Substitute points A(-20, 25) and D(-20, -10) in equation 1.

$d(A,D)=\sqrt{(x_{4}-x_{1})^{2}+(y_{4}-y_{1})^{2}}$

$AD=\sqrt{(-20-(-20))^{2}+(-10-25)^{2}}$

$AD=\sqrt{(-20+20)^{2}+(-35)^{2}}$

$AD=\sqrt{(-35)^{2}}$

AD = 35 units

Therefore, the dimension of the rectangular swimming pool are.

Length = 50 units

width = 35 units

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3 years ago

2) The triangles are similar. Find the similarity ratio of the first triangle to the 1 point

USPshnik [31]

Given:

Triangles FRI and DAY are similar.

To find:

Similarity ratio

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<em>If two triangles are similar then the corresponding angles are congruent and the corresponding sides are in proportion.</em>

Here, FR and DA are corresponding sides.

$$\Rightarrow \frac{FR}{DA} =\frac{4}{6}$

Cancel the common factors of 4 and 6, we get

$$\Rightarrow \frac{FR}{DA} =\frac{2}{3}$

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⇒ ΔFRI : ΔDAY = 2 : 3

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4 years ago