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4 years ago

Find the area of the shaded triangle

Mathematics

1 answer:

labwork [276]4 years ago

6 0

Area of ∆= ½b×h
=½54×33
=27 × 33
= 891cm²

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11. Health. Sherri is considering joining a health club. To be a member of Roman's Gym, she would

sergeinik [125]

Answer:

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Step-by-step explanation:

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3 years ago

What is the measure of arc CD?

CaHeK987 [17]

Option C:

The measure of arc CD is 40°.

Solution:

Given data:

m∠X = 11° and m(arc AB) = 18°

To find the measure of arc CD:

We know that,

<em>Angle formed by two intersecting secants outside the circle is equal to half of the difference between the intercepted arcs.</em>

$$m \angle X =\frac{1}{2}({arc \ CD}-{arc} \ AB)$

$$ 11^\circ =\frac{1}{2}({arc} \ CD-{18^\circ})$

Multiply by 2 on both sides.

22° = arc CD - 18°

Add 18° from both sides.

40° = arc CD

Switch the sides.

arc CD = 40°

Hence the measure of arc CD is 40°.

Option B is the correct answer.

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3 years ago

Find the comthon ratio of the geometric sequence 10,60, 360,...

Makovka662 [10]

Answer:

Step-by-step explanation:

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3 years ago

I need help please it’s for school

erma4kov [3.2K]

From Sweatcoin, London to the residence of Santa Claus is about 4,280 km away.

<h3>Where is the Santa Clause's residence? </h3>

The residence of Santa Clause is popularly known as the North Pole.

According to the above, we have to roughly calculate the distance from Sweatcoin in London to the North Pole. This distance is equivalent to about 4,280 km

Learn more about London in: brainly.com/question/7416097

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3 years ago

Which of the following functions f : {0, 1, 2, 3} ! {0, 1, . . . , 7} are one-to-one?

allsm [11]

Answer:

1. No.

$f(0)=0\\1^2=1; \text{then } f(1)=1\\2^2=4\equiv 4 \text{ mod 8}; \text{then } f(2)=4\\3^2=9\equiv 1 \text{mod 8 }; \text{then } f(3)=1$

Since f(1)=f(3) and $1\neq 3$ then f isn't one-to-one.

2. No

$f(0)=0\\1^3=1\equiv 1\text{ mod 8}; \text{then } f(1)=1\\2^3=8\equiv 0\text{ mod 8}; \text{then } f(2)=0\\3^3=27\equiv 3 \text{ mod 8}; \text{then } f(3)=3$

Since f(0)=f(2) and $0\neq 2$ then f isn't one-to-one.

3. No

$0^3-8=-8\equiv 0\text{ mod 8}; \text{then } f(0)=0\\1^3-8=-7\equiv 1\text{ mod 8}; \text{then } f(1)=1\\2^3-8=0\equiv 0 \text{ mod 8}; \text{then } f(2)=0\\3^3-8=27-8=19\equiv 3 \text{ mod 8}; \text{then } f(3)=3\\$

Since f(0)=f(2) and $0\neq 2$ then f isn't one-to-one.

4. Yes

$0^3+2*0=0; \text{then } f(0)=0\\1^3+2*1=3\equiv 3\text{ mod 8}; \text{then } f(1)=3\\2^3+2*2=8+4=12\equiv 4 \text{ mod 8}; \text{then } f(2)=4\\3^3+2*3=27+6=33\equiv 1\text{ mod 8}; \text{then } f(3)=1$

Since $f(0)\neq f(1)\neq f(2) \neq f(3)$ , then f is one-to-one

5. Since f(1)=f(3) and $1\neq 3$ then, f isn't one-to-one

3 0

4 years ago