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1 year ago

Please help me!!!!! Show your work 45-45-90 method <3

Mathematics

2 answers:

aivan3 [116]1 year ago

7 0

Answer:

hope it helps and u understand

check the steps in picture attached

saul85 [17]1 year ago

5 0

Pic attached shdjjdjdjrjrjrjrjrirititititt

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-3,-7 reflected over the x axis

Aleks [24]

Answer:

The reflected point would be (3,-7)

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2 years ago

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Please help with this question

Assoli18 [71]

Answer:

75°

Step-by-step explanation:

Please see attached picture for full solution.

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2 years ago

In a survey of 200 professors at Gotham University, professors were asked, “Regarding your weight, do you think you are: About r

DochEvi [55]

The probability that a randomly selected professor is female and feels "About Right" is 0.40

The distribution table :

____About right _ Oweight _ Uweight __ Total

Female _ 80 _____ 39 ______ 3 _______ 122

Male ___ 52 _____ 10 ______ 16 _______ 88

Total ___ 132 ____ 49 ______ 19 _______ 200

Oweight = Overweight

Uweight = Underweight

Recall :

Probability = required outcome / Total possible outcomes

Probability that a selected professor is female and feels 'About right' :

Required outcome = number of female professors who feels 'About right' = 80

Total number of professors = 200

Hence,

P(Female and 'About right') = 80/200 = 0.40

Learn more : brainly.com/question/11234923

3 0

2 years ago

PLEASE PLEASE ANSWER!

muminat

The area of the shaded region is $$8(\pi \ -\sqrt{3})\ \text{cm}^2$ .

Solution:

Given radius = 4 cm

Diameter = 2 × 4 = 8 cm

Let us first find the area of the semi-circle.

Area of the semi-circle = $\frac{1}{2}\times \pi r^2$

$$=\frac{1}{2}\times \pi\times 4^2$

$$=\frac{1}{2}\times \pi\times 16$

Area of the semi-circle = $$8\pi$ cm²

Angle in a semi-circle is always 90º.

∠C = 90°

So, ABC is a right angled triangle.

Using Pythagoras theorem, we can find base of the triangle.

$AC^2+BC^2=AB^2$

$AC^2+4^2=8^2$

$AC^2=64-16$

$AC^2=48$

$AC=4\sqrt{3}$ cm

Base of the triangle ABC = $4\sqrt{3}$ cm

Height of the triangle = 4 cm

Area of the triangle ABC = $\frac{1}{2}\times b \times h$

$$=\frac{1}{2}\times 4\sqrt{3} \times 4$

Area of the triangle ABC = $8\sqrt{3}$ cm²

Area of the shaded region

= Area of the semi-circle – Area of the triangle ABC

= $$8\pi \ \text{cm}^2-8\sqrt{3}\ \text{cm}^2$

= $$8(\pi \ -\sqrt{3})\ \text{cm}^2$

Hence the area of the shaded region is $$8(\pi \ -\sqrt{3})\ \text{cm}^2$ .

3 0

2 years ago

Please help ( 20 points )

ZanzabumX [31]

Answer:

1+pi/2, 5/2, square root of 12-1, 2.25. Let me know if this is correct!

Step-by-step explanation:

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2 years ago

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